رياضيات فصل ثاني

الدرس الثالث:الضرب القياسي

 أتحقق من فهمي 144 

أجد ناتج الضرب القياسي للمتجهين في كل مما يأتي :

 

$$\begin{gathered} \overline{)b} \stackrel{\rightharpoonup }{m}=-3\hat{i}+5\hat{j}-\hat{k} ,\stackrel{\rightharpoonup }{n}=-12\hat{i}+6\hat{j}-8\hat{k} \\ Solution: \\ since \stackrel{\rightharpoonup }{m}.\stackrel{\rightharpoonup }{n}=m_1{n}_1+m_2{n}_2+m_3{n}_3 \\ \stackrel{\rightharpoonup }{m}.\stackrel{\rightharpoonup }{n}=(-3\times -12)+(5\times 6)+(-1\times -8) \\ \stackrel{\rightharpoonup }{m}.\stackrel{\rightharpoonup }{n}=36+30+8=74 \end{gathered}$$ 

 

 أتحقق من فهمي 146

أجد قياس الزاوية $\theta$ بين المتجه $\stackrel{\rightharpoonup }{u}$ والمتجه $\stackrel{\rightharpoonup }{w}$ في كل مما يأتي. مُقربا الناتج إلى أقرب عُشر درجه :

$$\begin{gathered} \overline{)a} \stackrel{\rightharpoonup }{v}=-3\hat{i}+5\hat{j}-4\hat{k} ,\stackrel{\rightharpoonup }{w}=4\hat{i}+2\hat{j}-3\hat{k} \\ Solution: \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}{v}_{\mathit{1}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{v}_{\mathit{2}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{v}_{\mathit{3}}\mathit{\right)}}^{\mathit{2}}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}-3\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{5}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{4}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{50}} \\ \mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}{w}_{\mathit{1}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{w}_{\mathit{2}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{w}_{\mathit{3}}\mathit{\right)}}^{\mathit{2}}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{4}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{2}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{3}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{29}} \\ and \stackrel{\rightharpoonup }{v}.\stackrel{\rightharpoonup }{w}=v_1{w}_1+v_2{w}_2+v_3{w}_3 \\ \stackrel{\rightharpoonup }{v}.\stackrel{\rightharpoonup }{w}=(-3\times 4)+(5\times 2)+(-4\times -3) \\ =-12+10+12=10 \\ then \mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ 10=\sqrt{\mathit{50}} \times \sqrt{\mathit{29}} cos\theta \\ \Rightarrow cos\theta =\frac{10}{\sqrt{\mathit{1450}}} \mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{10}}{\sqrt{\mathit{1450}}}\mathit{\right)}\mathit{=}\mathit{74}\mathit{.}{\mathit{8}}^o \end{gathered}$$

 

$$\begin{gathered} \overline{)b}\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{=}\mathit{<}\mathit{2}\mathit{,}\mathit{-}\mathit{10}\mathit{,}\mathit{6}\mathit{>}\mathit{ }\mathit{,}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{<}\mathit{-}\mathit{3}\mathit{,}\mathit{15}\mathit{,}\mathit{-}\mathit{9}\mathit{>} \\ Solution\mathit{ } \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{2}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{10}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{6}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{140}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{3}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{15}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{9}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{315}} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}{v}_{\mathit{1}}{w}_{\mathit{1}}\mathit{+}{v}_{\mathit{2}}{w}_{\mathit{2}}\mathit{+}{v}_{\mathit{3}}{w}_{\mathit{3}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{(}\mathit{2}\mathit{\times }\mathit{-}\mathit{3}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{10}\mathit{\times }\mathit{15}\mathit{)}\mathit{+}\mathit{(}\mathit{6}\mathit{\times }\mathit{-}\mathit{9}\mathit{)} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{-}\mathit{6}\mathit{-}\mathit{150}\mathit{-}\mathit{54}\mathit{=}\mathit{-}\mathit{210} \\ then\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{-}\mathit{210}\mathit{=}\sqrt{\mathit{140}}\mathit{ }\mathit{\times }\sqrt{\mathit{315}}\mathit{ }cos\theta \mathit{ } \\ \mathit{\Rightarrow }cos\theta \mathit{ }\mathit{=}\frac{\mathit{-}\mathit{210}}{\sqrt{\mathit{4410}}}\mathit{=}\mathit{-}\mathit{1}\mathit{ }\mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{(}\mathit{-}\mathit{1}\mathit{)}\mathit{=}{\mathit{180}}^o \end{gathered}$$   

 

 أتحقق من فهمي 147

إذا كانت: $\stackrel{\rightharpoonup }{r}=\left(\begin{array}{c}3\\ -2\\ 1\end{array}\right) + t\left(\begin{array}{c}2\\ -5\\ -1\end{array}\right)$  معادلة متجهة للمستقيم $l_1$ وكانت $\stackrel{\rightharpoonup }{r}=\left(\begin{array}{c}5\\ 3\\ 0\end{array}\right) +u\left(\begin{array}{c}1\\ 0\\ -3\end{array}\right)$

معادلة متجهة للمستقيم $l_2$ فأجد قياس الزاوية الحادة بين المستقيمين إلى أقرب عشر درجة . 

$$\begin{gathered} \stackrel{\mathit{\rightharpoonup }}{v}\mathit{=}\mathit{<}\mathit{2}\mathit{,}\mathit{-}\mathit{5}\mathit{,}\mathit{-}\mathit{1}\mathit{>}\mathit{ }\mathit{,}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{<}\mathit{1}\mathit{,}\mathit{0}\mathit{,}\mathit{-}\mathit{3}\mathit{>} \\ Solution\mathit{ } \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{2}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{5}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{1}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{30}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{1}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{0}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{3}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{10}} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}{v}_{\mathit{1}}{w}_{\mathit{1}}\mathit{+}{v}_{\mathit{2}}{w}_{\mathit{2}}\mathit{+}{v}_{\mathit{3}}{w}_{\mathit{3}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{(}\mathit{2}\mathit{\times }\mathit{1}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{5}\mathit{\times }\mathit{0}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{1}\mathit{\times }\mathit{-}\mathit{3}\mathit{)} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{2}\mathit{+}\mathit{3}\mathit{=}\mathit{5} \\ then\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{5}\mathit{=}\sqrt{\mathit{30}}\mathit{ }\mathit{\times }\sqrt{\mathit{10}}\mathit{ }cos\theta \mathit{ } \\ \mathit{\Rightarrow }cos\theta \mathit{ }\mathit{=}\frac{5}{\sqrt{\mathit{300}}}\mathit{=}\frac{\mathit{1}}{\mathit{2}\sqrt{\mathit{3}}}\mathit{ }\mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{1}}{\mathit{2}\sqrt{\mathit{3}}}\mathit{\right)}\mathit{=}{\mathit{73}}^o \end{gathered}$$

 أتحقق من فهمي 149

أجد مساحة المثلث EFG الذي إحداثيات رؤوسه هي    G(6,-3,1) , F(5,1,7) ,    E(2,1,-1)

$$\begin{gathered} Solution: \\ \mathit{ }\mathit{ }{A}_{\mathit{△}EFG}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{EF}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{FG}\mathit{\right|}sin\theta \\ \stackrel{\mathit{\rightharpoonup }}{EF}\mathit{=}\mathit{<}\mathit{5}\mathit{-}\mathit{2}\mathit{,}\mathit{1}\mathit{-}\mathit{1}\mathit{,}\mathit{7}\mathit{-}\mathit{-}\mathit{1}\mathit{>}\mathit{=}\mathit{<}\mathit{3}\mathit{,}\mathit{0}\mathit{,}\mathit{8}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{FG}\mathit{=}\mathit{<}\mathit{6}\mathit{-}\mathit{5}\mathit{,}\mathit{-}\mathit{3}\mathit{-}\mathit{1}\mathit{,}\mathit{1}\mathit{-}\mathit{7}\mathit{>}\mathit{=}\mathit{<}\mathit{1}\mathit{,}\mathit{-}\mathit{4}\mathit{,}\mathit{-}\mathit{6}\mathit{>} \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{EF}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{0}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{8}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{73}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{FG}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{1}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{4}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{6}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{53}} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{EF}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{FG}\mathit{=}\mathit{(}\mathit{3}\mathit{\times }\mathit{1}\mathit{)}\mathit{+}\mathit{(}\mathit{0}\mathit{\times }\mathit{-}\mathit{4}\mathit{)}\mathit{+}\mathit{(}\mathit{8}\mathit{\times }\mathit{-}\mathit{6}\mathit{)} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{3}\mathit{-}\mathit{48}\mathit{=}\mathit{-}\mathit{45} \\ then\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{EF}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{FG}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{EF}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{FG}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{-}\mathit{45}\mathit{=}\sqrt{\mathit{73}}\mathit{ }\mathit{\times }\sqrt{\mathit{53}}\mathit{ }cos\theta \mathit{ } \\ \mathit{\Rightarrow }cos\theta \mathit{ }\mathit{=}\mathit{-}\frac{45}{\sqrt{\mathit{3869}}}\mathit{ }\mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{(}\mathit{-}\frac{\mathit{45}}{\sqrt{\mathit{3869}}}\mathit{)}\mathit{=}{\mathit{136}}^o \\ \mathit{ }But\mathit{ }\mathit{ }\mathit{ }{A}_{\mathit{△}EFG}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{EF}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{FG}\mathit{\right|}sin\theta \\ \mathit{ }then\mathit{ }{A}_{\mathit{△}EFG}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\sqrt{\mathit{73}}\mathit{\times }\sqrt{\mathit{53}}sin\mathit{(}{\mathit{136}}^o\mathit{)} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{21}\mathit{.}\mathit{6}\mathit{ }unit\mathit{ } \end{gathered}$$

 

 أتحقق من فهمي 151

إذا كانت: $\stackrel{\rightharpoonup }{r}=16\hat{i}+11\hat{j}-3\hat{k} +t(5\hat{i}+7\hat{j}-3\hat{k})$ معادلة متجهة للمستقيم l             

والنقطة  P(2,0,10/3) غير واقعة على المستقيم l  فأجيب عن السؤالين الآتيين:

$\overline{)a}$  أحدد مسقط العمود من النقطة P على المستقيم l .

          بفرض مسقط النقطة P على المستقيم l هو النقطة F .

     لذلك فإن F تحقق معادلة المستقيم على النحو التالي:

         $\stackrel{\rightharpoonup }{OF}=<16+5t,11+7t,-3-3t>$

    وكذلك فإن متجه الموقع OP هو :  $\stackrel{\rightharpoonup }{OP }=<2,0,\frac{10}{3}>$  

    وبتطبيق قاعدة المثلث لجمع المتجهات نجد أنَّ : 

   $$\begin{gathered} \stackrel{\rightharpoonup }{OP }+\stackrel{\rightharpoonup }{PF }=\stackrel{\rightharpoonup }{OF } \\ \stackrel{\rightharpoonup }{PF }=\stackrel{\rightharpoonup }{OF }-\stackrel{\rightharpoonup }{OP } \\ =<16+5t,11+7t,-3-3t>-<2,0,\frac{10}{3}> \\ \stackrel{\rightharpoonup }{PF }=<14+5t,11+7t,-\frac{19}{3}-3t> \\ but \stackrel{\rightharpoonup }{PF } \perp \stackrel{\rightharpoonup }{v} \Rightarrow \stackrel{\rightharpoonup }{PF }.\stackrel{\rightharpoonup }{v}=0 \\ 5\times (14+5t)+7\times (11+7t)-3\times (-\frac{19}{3}-3t)=0 \\ 70+25t+77+49t+19+9t=0 \\ 166+83t=0 \Rightarrow t=-2 \\ then \stackrel{\rightharpoonup }{OF }=<16+5t,11+7t,-3-3t> \\ \stackrel{\rightharpoonup }{OF }=<16+5(-2),11+7(-2),-3-3(-2)> \\ =<6,-3,3> \\ \Rightarrow F=(6,-3,3) \end{gathered}$$

  $\overline{)b}$ أجد البعد بين النقطة P على المستقيم l .

 $$\begin{gathered} Solution : \\ \stackrel{\rightharpoonup }{PF }=<14+5t,11+7t,-\frac{19}{3}-3t> when t=-2 \\ then \stackrel{\rightharpoonup }{PF }=<14+5(-2),11+7(-2),-\frac{19}{3}-3(-2)> \\ \stackrel{\rightharpoonup }{PF }=<4,-3,-\frac{1}{3}> \\ \left|\stackrel{\rightharpoonup }{PF }\right|=\sqrt{{\left(4\right)}^2+{(-3)}^2+{(-\frac{1}{3})}^2} \\ =\sqrt{25+\frac{1}{9}}=\frac{\sqrt{226}}{3} \end{gathered}$$

 أتحقق من فهمي 154

$\overline{)a}$ أجد قياس $\angle EDB$  في الهرم المبين في المثال السابق  .

$\overline{)b}$  أجد حجم الهرم

 

 

$$\begin{gathered} \overline{)a} Solution : \\ E(8,3,7) , D(1,-5,5) , B(9,-1,-3) \\ \stackrel{\rightharpoonup }{ED}=<-7,-8,-2>\Rightarrow \left|\stackrel{\rightharpoonup }{ED }\right|=\sqrt{{(-7)}^2+{(-8)}^2+{(-2)}^2} \\ \left|\stackrel{\rightharpoonup }{ED }\right|=\sqrt{49+64+4}=\sqrt{117} \\ \stackrel{\rightharpoonup }{DB }=<8,4,-8>\Rightarrow \left|\stackrel{\rightharpoonup }{BD }\right|=\sqrt{{\left(8\right)}^2+{\left(4\right)}^2+{(-8)}^2} \\ \left|\stackrel{\rightharpoonup }{BD }\right|=\sqrt{64+16+64}=12 \\ \stackrel{\rightharpoonup }{ED}.\stackrel{\rightharpoonup }{DB }=(-7\times 8)+(-8\times 4)+(-2\times -8) \\ =-56-32+16=-72 \\ Now \cos \theta =\frac{\stackrel{\rightharpoonup }{ED}.\stackrel{\rightharpoonup }{DB }}{\left|\stackrel{\rightharpoonup }{ED }\right|\left|\stackrel{\rightharpoonup }{BD }\right|}=\frac{-72}{12\sqrt{117}} \\ \cos \theta =\frac{\stackrel{\rightharpoonup }{ED}.\stackrel{\rightharpoonup }{DB }}{\left|\stackrel{\rightharpoonup }{ED }\right|\left|\stackrel{\rightharpoonup }{BD }\right|}=-\frac{6}{\sqrt{117}} \\ \theta ={\cos }^{-1}\left(-\frac{6}{\sqrt{117}}\right) \Rightarrow \theta ={124}^{°} \\ the complemt is ={56}^{°} \end{gathered}$$

 

$$\begin{gathered} \overline{)b} Solution : \\ A=\frac{1}{3}{\left(\left|\overline{AD}\right|\right)}^2\left|\overline{EM}\right| \\ Now : \left|\overline{AD}\right|=\sqrt{{(1-1)}^2+{(-5-1)}^2+{(5+1)}^2} \\ =6\sqrt{2} \\ to find M as a midpoint of D(1,-5,5) , B(9,-1,-3): \\ M=(\frac{1+9}{2},\frac{-5-1}{2},\frac{5-3}{2})=(5,-3,1) \\ Now : \left|\overline{EM}\right|=\sqrt{{(8-5)}^2+{(3+3)}^2+{(7-1)}^2} \\ =\sqrt{9+36+36}=\sqrt{81}=9 \\ A=\frac{1}{3}{\left(\left|\overline{AD}\right|\right)}^2\left|\overline{EM}\right| \\ A=\frac{1}{3}{\left(6\sqrt{2}\right)}^2\left(9\right)=216 unit \end{gathered}$$

تمارين ومسائل

أجد ناتج الضرب القياسي للمتجهين في كل مما يأتي : 

$$\begin{gathered} \overline{)1} \stackrel{\rightharpoonup }{u}=5\hat{i}-4\hat{j}+3\hat{k} ,\stackrel{\rightharpoonup }{v}=7\hat{i}+6\hat{j}-2\hat{k} \\ Solution: \\ since \stackrel{\rightharpoonup }{u}.\stackrel{\rightharpoonup }{v}=u_1{v}_1+u_2{v}_2+u_3{v}_3 \\ =(5\times 7)+(-4\times 6)+(3\times -2) \\ =35-24-6 =5 \\ \overline{)2} \stackrel{\rightharpoonup }{u}=4\hat{i}-8\hat{j}-3\hat{k} ,\stackrel{\rightharpoonup }{v}=12\hat{i}+9\hat{j}-8\hat{k} \\ Solution: \\ =(4\times 12)+(-8\times 9)+(-3\times -8) \\ =48-72+24=0 \\ \overline{)3} \stackrel{\rightharpoonup }{u}=<-5,9,17> ,\stackrel{\rightharpoonup }{v}=<4,6,-2> \\ Solution: \\ =(-5\times 4)+(9\times 6)+(17\times -2) \\ =-20+54-34=0 \\ \overline{)4} \stackrel{\rightharpoonup }{u}=<1,-4,12> ,\stackrel{\rightharpoonup }{v}=<3,10,-5> \\ Solution: \\ =(1\times 3)+(-4\times 10)+(12\times -5) \\ =3-40-60=-97 \end{gathered}$$ 

 

أجد قياس الزاوية $\theta$ بين المتجهين إلى أقرب عشر درجة في كلٌ مما يأتي :

$$\begin{gathered} \overline{)5} \stackrel{\rightharpoonup }{m}=4\hat{i}-2\hat{j}+5\hat{k} ,\stackrel{\rightharpoonup }{n}=3\hat{i}+4\hat{j}-2\hat{k} \\ Solution: \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{m}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{n}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{m}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{n}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{m}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}{m}_{\mathit{1}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{m}_{\mathit{2}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{m}_{\mathit{3}}\mathit{\right)}}^{\mathit{2}}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{m}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{4}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{2}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{5}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{45}} \\ \mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{n}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}{n}_{\mathit{1}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{n}_{\mathit{2}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{n}_{\mathit{3}}\mathit{\right)}}^{\mathit{2}}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{n}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{4}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{2}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{29}} \\ and \stackrel{\rightharpoonup }{m}.\stackrel{\rightharpoonup }{n}=m_1{n}_1+m_2{n}_2+m_3{n}_3 \\ \stackrel{\rightharpoonup }{v}.\stackrel{\rightharpoonup }{w}=(4\times 3)+(-2\times 4)+(5\times -2) \\ =12-8-10=-6 \\ then \mathit{ }cos\theta =\frac{\stackrel{\mathit{\rightharpoonup }}{m}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{n}}{\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{m}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{n}\mathit{\right|}} \\ \mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{-}\mathit{6}}{\sqrt{\mathit{1305}}}\mathit{\right)}\mathit{=}\mathit{99}\mathit{.}{\mathit{6}}^o \end{gathered}$$ 

 

$$\begin{gathered} \overline{)6} \stackrel{\rightharpoonup }{v}=<3,-2,9> ,\stackrel{\rightharpoonup }{w}=<5,3,-4> \\ Solution: \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}{v}_{\mathit{1}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{v}_{\mathit{2}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{v}_{\mathit{3}}\mathit{\right)}}^{\mathit{2}}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}3\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{2}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{9}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{94}} \\ \mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}{w}_{\mathit{1}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{w}_{\mathit{2}}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}{w}_{\mathit{3}}\mathit{\right)}}^{\mathit{2}}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{5}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{4}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{50}} \\ and \stackrel{\rightharpoonup }{v}.\stackrel{\rightharpoonup }{w}=v_1{w}_1+v_2{w}_2+v_3{w}_3 \\ \stackrel{\rightharpoonup }{v}.\stackrel{\rightharpoonup }{w}=(3\times 5)+(-2\times 3)+(9\times -4) \\ =15-6-36=-27 \\ then \mathit{ }\mathit{ }cos\theta \mathit{=}\frac{\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}}{\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}} \\ \Rightarrow cos\theta =\frac{-27}{\sqrt{\mathit{4700}}} \mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{-}\mathit{27}}{\sqrt{\mathit{4700}}}\mathit{\right)}\mathit{=}\mathit{113}\mathit{.}{\mathit{2}}^o \end{gathered}$$

 

$\overline{)7}$   إذا كانت  A(3,5,-4) و B(7,4,-3) و O نقطة الأصل فأجد $m\angle OAB$ إلى أقرب درجة . 

$\overline{)8}$   يمر المستقيم $l_1$بالنقطتين: (3,5,7-) و (1,4- ,2)

     ويمر المستقيم $l_2$ بالنقطتين: (1- ,1,2) و(6,-5,3)  

     أجد قياس الزاوية الحادة بين المستقيمين إلى أقرب عُشر درجة 

$$\begin{gathered} Solution: \\ l_1: (2,-1,4) ,(-3,5,7) \Rightarrow \stackrel{\mathit{\rightharpoonup }}{r_1}=<-5,6,3> \\ l_2: (6,-5,3) ,(1,2,-1) \Rightarrow \stackrel{\mathit{\rightharpoonup }}{r_2}=<-5,7,-4> \\ since\mathit{ }cos\theta \mathit{=}\frac{\mathit{ }\stackrel{\mathit{\rightharpoonup }}{r_{\mathit{1}}}\mathit{.}\mathit{ }\stackrel{\mathit{\rightharpoonup }}{r_{\mathit{2}}}}{\mathit{|} \stackrel{\mathit{\rightharpoonup }}{r_1}\mathit{|}\mathit{|} \stackrel{\mathit{\rightharpoonup }}{r_2}\mathit{|}} \\ \mathit{ }\mathit{|}\mathit{ }\stackrel{\mathit{\rightharpoonup }}{r_{\mathit{1}}}\mathit{|}\mathit{=}\sqrt{{\mathit{(}-5\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{6}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{70}} \\ \mathit{ }\mathit{|}\mathit{ }\stackrel{\mathit{\rightharpoonup }}{r_{\mathit{2}}}\mathit{|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{5}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{7}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{4}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{90}} \\ and \stackrel{\mathit{\rightharpoonup }}{r_1}. \stackrel{\mathit{\rightharpoonup }}{r_2}=(-5\times -5)+(6\times 7)+(3\times -4) \\ =25+42-12=55 \\ then \Rightarrow cos\theta =\frac{55}{\sqrt{\mathit{6300}}} \mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{55}}{\sqrt{\mathit{6300}}}\mathit{\right)}\mathit{=}\mathit{46}\mathit{.}{\mathit{1}}^o \end{gathered}$$

 

$\overline{)9}$  إذا كان المستقيم الذي له المعادلة المتجهة :$\stackrel{\rightharpoonup }{r}=<1,4,-5> +\lambda <-6,q+5,3>$

    والمستقيم الذي له المعادلة المتجهة : $\stackrel{\rightharpoonup }{r}=<1,4,-5> +\mu <5,q-6,-4>$ 

    متعامدين فما القيم الممكنة للثابت q ؟ 

$$\begin{gathered} Solution\mathit{:} \\ \stackrel{\mathit{\rightharpoonup }}{r}\mathit{=}\mathit{<}\mathit{1}\mathit{,}\mathit{4}\mathit{,}\mathit{-}\mathit{5}\mathit{>}\mathit{ }\mathit{+}\lambda \mathit{<}\mathit{-}\mathit{6}\mathit{,}q\mathit{+}\mathit{5}\mathit{,}\mathit{3}\mathit{>} \\ \mathit{\Rightarrow }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v_{\mathit{1}}}\mathit{=}\mathit{<}\mathit{-}\mathit{6}\mathit{,}q\mathit{+}\mathit{5}\mathit{,}\mathit{3}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{r}\mathit{=}\mathit{<}\mathit{1}\mathit{,}\mathit{4}\mathit{,}\mathit{-}\mathit{5}\mathit{>}\mathit{ }\mathit{+}\mu \mathit{<}\mathit{5}\mathit{,}q\mathit{-}\mathit{6}\mathit{,}\mathit{-}\mathit{4}\mathit{>} \\ \mathit{\Rightarrow }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v_{\mathit{2}}}\mathit{=}\mathit{<}\mathit{5}\mathit{,}q\mathit{-}\mathit{6}\mathit{,}\mathit{-}\mathit{4}\mathit{>} \\ since\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v_{\mathit{1}}}\mathit{\perp }\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v_{\mathit{2}}}\mathit{ }\mathit{\Rightarrow }\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v_{\mathit{1}}}\mathit{.}\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v_{\mathit{2}}}\mathit{=}\mathit{0} \\ \stackrel{\mathit{\rightharpoonup }}{v_{\mathit{1}}}\mathit{.}\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v_{\mathit{2}}}\mathit{=}\mathit{(}\mathit{-}\mathit{6}\mathit{\times }\mathit{5}\mathit{)}\mathit{+}\mathit{\left(}\mathit{\right(}q\mathit{+}\mathit{5}\mathit{)}\mathit{\times }\mathit{(}q\mathit{-}\mathit{6}\mathit{)}\mathit{+}\mathit{(}\mathit{3}\mathit{\times }\mathit{-}\mathit{4}\mathit{)}\mathit{=}\mathit{0} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }{q}^{\mathit{2}}\mathit{-}q\mathit{-}\mathit{72}\mathit{=}\mathit{0} \\ \mathit{(}q\mathit{-}\mathit{9}\mathit{)}\mathit{(}q\mathit{+}\mathit{8}\mathit{)}\mathit{=}\mathit{0} \\ \mathit{\Rightarrow }q\mathit{=}\mathit{9}\mathit{ }\mathit{ }amd\mathit{ }\mathit{ }q\mathit{=}\mathit{-}\mathit{8} \end{gathered}$$

إذا كانت: $\stackrel{\rightharpoonup }{r}=2\hat{j}-3\hat{k} +t(-\hat{i}+2\hat{j}+5\hat{k})$معادلة متجهة للمسستقيم l.                      

والنقطة (5 ,22 ,2-)P غير واقعة على المستقيم l فأجيب عن السؤالين الآتيين تباعا :

$\overline{)10}$  أحدد مسقط العمود من النقطة P على المستقيم l. 

      بفرض مسقط النقطة P على المستقيم l هو النقطة F .

     لذلك فإن F تحقق معادلة المستقيم على النحو التالي:

         $\stackrel{\rightharpoonup }{OF}=<-t,2+2t,-3+5t>$

    وكذلك فإن متجه الموقع OP هو :    $\stackrel{\rightharpoonup }{OP }=<-2,22,5>$

    وبتطبيق قاعدة المثلث لجمع المتجهات نجد أنَّ :  

 $$\begin{gathered} \stackrel{\rightharpoonup }{OP }+\stackrel{\rightharpoonup }{PF }=\stackrel{\rightharpoonup }{OF } \\ \stackrel{\rightharpoonup }{PF }=\stackrel{\rightharpoonup }{OF }-\stackrel{\rightharpoonup }{OP } \\ =<-t,2+2t,-3+5t>-<-2,22,5> \\ \stackrel{\rightharpoonup }{PF }=<-t+2,2t-20,5t-8> \\ but \stackrel{\rightharpoonup }{PF } \perp \stackrel{\rightharpoonup }{v} \Rightarrow \stackrel{\rightharpoonup }{PF }.\stackrel{\rightharpoonup }{v}=0 \\ -1\times (-t+2)+2\times (2t-20)+5\times (5t-8)=0 \\ t-2+4t-40+25t-40=0 \\ -82+30t=0 \Rightarrow t=\frac{41}{15} \\ then \stackrel{\rightharpoonup }{OF }=<-t,2+2t,-3+5t> \\ \stackrel{\rightharpoonup }{OF }=<-\frac{41}{15},2+2\left(\frac{41}{15}\right),-3+5\left(\frac{41}{15}\right)> \\ \stackrel{\rightharpoonup }{OF }=<-\frac{41}{15},\frac{112}{15},\frac{32}{3}> \\ \Rightarrow F=(-\frac{41}{15},\frac{112}{15},\frac{32}{3}) \end{gathered}$$

 

$\overline{)11}$ أجد البعد بين النقطة P والمستقيم l 

 $$\begin{gathered} Solution : \\ \stackrel{\rightharpoonup }{PF }=<-t+2,2t-20,5t-8> when t=\frac{41}{15} \\ then \stackrel{\rightharpoonup }{PF }=<-\left(\frac{41}{15}\right)+2,2\left(\frac{41}{15}\right)-20,5\left(\frac{41}{15}\right)-8> \\ \stackrel{\rightharpoonup }{PF }=<\frac{-11}{15},\frac{-218}{15},\frac{85}{15}> \\ \left|\stackrel{\rightharpoonup }{PF }\right|=\sqrt{{\left(\frac{-11}{15}\right)}^2+{\left(\frac{-218}{15}\right)}^2+{\left(\frac{85}{15}\right)}^2} \\ =\sqrt{\frac{121+47524+7225}{15}}=\sqrt{3658}=60.5 \end{gathered}$$

$\overline{)12}$ أجد مساحة المثلث ABC حيث: ، و $\stackrel{\rightharpoonup }{AC}=<9,1,4>$ .

$$\begin{gathered} Solution: \\ \stackrel{\mathit{\rightharpoonup }}{AB}\mathit{=}\mathit{<}\mathit{4}\mathit{,}\mathit{9}\mathit{,}\mathit{1}\mathit{>}\mathit{,}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{ AC}\mathit{=}\mathit{<}\mathit{9}\mathit{,}\mathit{1}\mathit{,}\mathit{4}\mathit{>} \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{4}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{9}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{1}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{98}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{9}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{1}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{4}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{98}} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{=}\mathit{(}\mathit{4}\mathit{\times }\mathit{9}\mathit{)}\mathit{+}\mathit{(}\mathit{9}\mathit{\times }\mathit{1}\mathit{)}\mathit{+}\mathit{(}\mathit{1}\mathit{\times }\mathit{4}\mathit{)}\mathit{ }\mathit{=}\mathit{49} \\ then\mathit{ }\mathit{ }\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }cos\theta =\frac{\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{AC}}{\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{\right|}} \\ \mathit{\Rightarrow }cos\theta \mathit{ }\mathit{=}\frac{\mathit{49}}{\sqrt{\mathit{98}}\sqrt{\mathit{98}}}\mathit{ }\mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{1}}{2}\mathit{\right)}\mathit{=}{\mathit{60}}^o \\ \mathit{ }But\mathit{ }\mathit{ }\mathit{ }{A}_{\mathit{△}ABC}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{\right|}sin\theta \\ \mathit{ }then\mathit{ }{A}_{\mathit{△}EFG}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\sqrt{98}\mathit{\times }\sqrt{\mathit{98}}sin\mathit{(}{\mathit{60}}^o\mathit{)} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\frac{\mathit{49}\sqrt{\mathit{3}}}{\mathit{2}}\mathit{ }unit\mathit{ } \end{gathered}$$

$\overline{)13}$  أجد مساحة المثلث ABC الذي إحداثيات رؤوسه هي:

       (2 ,5- ,4)C ,(3- ,7 ,2)B , A(1,3,1)‏.

$$\begin{gathered} Solution: \\ A(1,3,1) , B(2,7,-3) ,C(4,-5,2) \\ \stackrel{\mathit{\rightharpoonup }}{AB}\mathit{=}\mathit{<}\mathit{1}\mathit{,}\mathit{4}\mathit{,}\mathit{-}\mathit{4}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{BC}\mathit{=}\mathit{<}\mathit{-}\mathit{4}\mathit{,}\mathit{-}\mathit{12}\mathit{,}\mathit{5}\mathit{>} \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{1}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{4}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{4}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{33}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{4}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{12}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{5}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{185}} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{=}\mathit{(}\mathit{1}\mathit{\times }\mathit{-}\mathit{4}\mathit{)}\mathit{+}\mathit{(}\mathit{4}\mathit{\times }\mathit{-}\mathit{12}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{4}\mathit{\times }\mathit{5}\mathit{)}\mathit{=}\mathit{-}\mathit{72} \\ then\mathit{ }\mathit{ }cos\theta =\frac{\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{BC}}{\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{\right|}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }cos\theta \mathit{ }\mathit{=}\mathit{-}\frac{\mathit{72}}{\sqrt{\mathit{33}}\sqrt{\mathit{185}}} \\ \mathit{ }\mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{(}\mathit{-}\frac{\mathit{72}}{\sqrt{\mathit{6105}}}\mathit{)}\mathit{=}\mathit{22}\mathit{.}{\mathit{9}}^o \\ \mathit{ }But\mathit{ }\mathit{ }\mathit{ }{A}_{\mathit{△}ABC}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{\right|}sin\theta \\ \mathit{ }then\mathit{ }{A}_{\mathit{△}EFG}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\times }\sqrt{\mathit{33}}\mathit{\times }\sqrt{\mathit{185}}sin\mathit{(}\mathit{22}\mathit{.}{\mathit{9}}^o\mathit{)} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{15}\mathit{.}\mathit{2}\mathit{ }unit\mathit{ } \end{gathered}$$

$\overline{)14}$ حزام ناقل: يمثل المتجه: $\stackrel{\rightharpoonup }{F}=5\hat{i}-3\hat{j}+\hat{k}$ القوة التي يُولّدها حزام ناقل لتحريك حقيبة في مسار مستقيم؛

      من النقطة (1 ,1 ,1) إلى النقطة (7 ,4 ,9). أجد مقدار الشغل الذي تبذله القوّة  F  .

     علما بأن القوة بالنيوتن N والمسافة  بالمتر m ؛ ومقدار الشغل (w) المبذول بوحدة الجول (J)

     يساوي ناتج الضرب القياسي لمتجه القوة في متجه الإزاحة ؛ أي $\stackrel{\rightharpoonup }{w}=\stackrel{\rightharpoonup }{F}.\stackrel{\rightharpoonup }{d}$  

$\overline{)15}$  إذا كانت النقطة (1- ,17- ,27) R والنقطة (11 ,9- ,11) S  تقعان على المستقيم l .

     وكانت النقطة Q تقع على المستقيم l. حيث $\stackrel{\rightharpoonup }{OQ}$ عمودي على l فأجد متجه الموقع للنقطةَQ    . 

 

إذا كانت متجهات مواقع النقاط A و B و D هي $\left(\begin{array}{c}2\\ -29\\ 7\end{array}\right) ,\left(\begin{array}{c}4\\ 17\\ 14\end{array}\right) , \left(\begin{array}{c}-4\\ 13\\ 22\end{array}\right)$ على الترتيب .

 فأجيب عن الأسئلة الأربعة الآتية تباعًا:

$\overline{)16}$  أثبت أنً: $\stackrel{\_\_\_}{AB}\perp \stackrel{\_\_\_}{AD}$ :

$$\begin{gathered} Solution: \\ \mathit{ }\overline{AB}\mathit{\perp }\overline{AD}\mathit{ }when \stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{AD}\mathit{=}\mathit{0} \\ \mathit{ }\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{=}\mathit{<}\mathit{-}\mathit{4}\mathit{-}\mathit{4}\mathit{,}\mathit{13}\mathit{-}\mathit{17}\mathit{,}\mathit{22}\mathit{-}\mathit{14}\mathit{>}\mathit{=}\mathit{<}\mathit{-}\mathit{8}\mathit{,}\mathit{-}\mathit{4}\mathit{,}\mathit{8}\mathit{>} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{<}\mathit{-}\mathit{2}\mathit{,}\mathit{-}\mathit{1}\mathit{,}\mathit{2}\mathit{>} \\ \mathit{ }\stackrel{\mathit{\rightharpoonup }}{AD}\mathit{=}\mathit{<}\mathit{-}\mathit{4}\mathit{-}\mathit{2}\mathit{,}\mathit{13}\mathit{-}\mathit{-}\mathit{29}\mathit{,}\mathit{22}\mathit{-}\mathit{7}\mathit{>}\mathit{=}\mathit{<}\mathit{-}\mathit{6}\mathit{,}\mathit{42}\mathit{,}\mathit{15}\mathit{>} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{<}\mathit{-}\mathit{2}\mathit{,}\mathit{14}\mathit{,}\mathit{5}\mathit{>} \\ So\mathit{:}\mathit{ }\mathit{-}\mathit{2}\mathit{\times }\mathit{-}\mathit{2}\mathit{ }\mathit{+}\mathit{-}\mathit{1}\mathit{\times }\mathit{14}\mathit{ }\mathit{+}\mathit{2}\mathit{\times }\mathit{5}\stackrel{\mathit{?}}{\mathit{=}}\mathit{0} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{4}\mathit{-}\mathit{14}\mathit{+}\mathit{10}\mathit{=}\mathit{0}\mathit{ }\mathit{ }\Rightarrow \mathit{ }\overline{AB}\mathit{\perp }\overline{AD}\mathit{ } \end{gathered}$$

 

$\overline{)17}$  أجد متجه موقع النقطة C إذا كانABCD  مستطيلا 

      سنجد رأس المستطيل C أولاً : ولأنها تشكل شكلاً هندسياً مغلقاً

      فسيكون الجمع المتجهي يساوي صفر :  $\stackrel{\rightharpoonup }{AB}+\stackrel{\rightharpoonup }{BC}+\stackrel{\rightharpoonup }{CD}+\stackrel{\rightharpoonup }{DA}=0$

     

  

$\overline{)18}$  أجد مساحة المستطيلABCD   .‏

 $$\begin{gathered} Solution: \\ A(-4,13,22) , B(4,17,14) ,C(10,-25,-1) \\ \stackrel{\mathit{\rightharpoonup }}{AB}\mathit{=}\mathit{<}\mathit{8}\mathit{,}\mathit{4}\mathit{,}\mathit{-}\mathit{8}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{BC}\mathit{=}\mathit{<}\mathit{6}\mathit{,}\mathit{-}\mathit{42}\mathit{,}\mathit{-}\mathit{15}\mathit{>} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{8}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{4}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{8}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{144}}\mathit{=}\mathit{12} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{6}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{42}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{15}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{2025}}\mathit{=}\mathit{45} \\ \mathit{ }But\mathit{ }\mathit{ }\mathit{ }{A}_{\square ABCD}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{\right|} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }{A}_{\mathit{\square }ABCD}\mathit{=}\mathit{12}\mathit{\times }\mathit{45}\mathit{=}\mathit{540}\mathit{ }unit \end{gathered}$$

$\overline{)19}$  أجد متجه موقع مركز المستطيلABCD   .‏ 

     مركز المستطيل هي نقطة تقاطع قطريه وهي تنصف كل منهما :

$$\begin{gathered} Solution: \\ A(-4,13,22) ,C(10,-25,-1) \\ M=(\frac{-4+10}{2}, \frac{13-25}{2} , \frac{22-1}{2})=(3, -6 , \frac{21}{2}) \\ \stackrel{\mathit{\rightharpoonup }}{OM}\mathit{=}\mathit{<}\mathit{3}\mathit{,}\mathit{ }\mathit{-}\mathit{6}\mathit{ }\mathit{,}\mathit{ }\frac{\mathit{21}}{\mathit{2}}\mathit{>} \end{gathered}$$

تمثل: $\stackrel{\rightharpoonup }{r}=<-5,7,1> +t<3,1,4>$ معادلة متجهة للمستقيم l₁ .

وتمثل : $\stackrel{\rightharpoonup }{r}=<2,8,-1> +u<2,0,-3>$ معادلة متجهة للمستقيم l₂ .

وتمثل : $\stackrel{\rightharpoonup }{r}=<3,19,10> +v<-1,3,1>$معادلة متجهة للمستقيم l₃ . 

إذا تقاطع المستقيم l₂ والمستقيم l₁ في النقطة T وكانت النقطة F تقع على المستقيم l₃ حيث: $\overline{TF} \perp l_3$

فأجيب عن السؤالين الآتيين تباعا:

$\overline{)20}$  أجد إحداثيات النقطة F. 

$$\begin{gathered} Solution: \\ \stackrel{\rightharpoonup }{r}=<-5,7,1> +t<3,1,4> \\ \stackrel{\rightharpoonup }{r}=<2,8,-1> +u<2,0,-3> \\ now to find T: \\ -5+3t=2+2u \\ 3t-2u=7 ...\overline{) 1} \\ 7+t=8 \Rightarrow t=1 \\ then 3\left(1\right)-2u=7 \Rightarrow u=-2 \\ T=(-5+3,7+1,1+4)=(-2,8,5) \\ now to find \stackrel{\rightharpoonup }{OF} from \stackrel{\rightharpoonup }{r}=<3,19,10> +v<-1,3,1> \\ \stackrel{\rightharpoonup }{OF}=<3-v,19+3v,10+v> \\ So \stackrel{\rightharpoonup }{TF}=<3-v+2,19+3v-8,10+v-5> \\ =<5-v,11+3v,5+v> \\ now to find v: \\ but \stackrel{\rightharpoonup }{l_3} \perp \stackrel{\rightharpoonup }{TF } \Rightarrow \stackrel{\rightharpoonup }{l_3} .\stackrel{\rightharpoonup }{TF }=0 \\ -1\left(5-v\right)+3(11+3v)+1(5+v)=0 \\ -5+v+33+9v+5+v=0 \\ 33+11v=0 \Rightarrow v=-3 \\ then \stackrel{\rightharpoonup }{OF}=<3-v,19+3v,10+v> \\ =<3+3,19+3(-3),10-3> \\ =<6,10,7> \Rightarrow F\left(\right(6,10,7) \end{gathered}$$          

$\overline{)21}$ أجد البعد بين النقطة T والمستقيم l₃  

$$\begin{gathered} Solution: \\ \stackrel{\rightharpoonup }{TF}=<5-v,11+3v,5+v> and v=-3 \\ So: \stackrel{\rightharpoonup }{TF}=<5+3,11-9,5-3>=<8,2,2> \\ \left|\stackrel{\rightharpoonup }{TF}\right|=\sqrt{{\left(8\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2}=\sqrt{72}=6\sqrt{2} \end{gathered}$$

إذا كانت:  $\stackrel{\rightharpoonup }{r}=<5,3,0> +\lambda <-1,3,1>$ تمثل معادلة متجهة للمسستقيم l .

وكانت   A(3,-2,1) و B(5,3,0) . فأجيب عن السؤالين الآتيين تباعا:

$\overline{)22}$ أجد قياس الزاوية الحادة بين المستقيم $\stackrel{\rightharpoonup }{AB}$ والمستقيم  l .

$$\begin{gathered} Solution: \\ \stackrel{\rightharpoonup }{r}=<5,3,0> +\lambda <-1,3,1> \\ A(3,-2,1) , B(5,3,0) \\ \stackrel{\mathit{\rightharpoonup }}{AB}\mathit{=}\mathit{<}\mathit{2}\mathit{,}\mathit{5}\mathit{,}\mathit{-}\mathit{1}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{<}\mathit{-}\mathit{1}\mathit{,}\mathit{3}\mathit{,}\mathit{1}\mathit{>} \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{2}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{5}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{1}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{30}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{1}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{1}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{11}} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{(}\mathit{2}\mathit{\times }\mathit{-}\mathit{1}\mathit{)}\mathit{+}\mathit{(}\mathit{5}\mathit{\times }\mathit{3}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{1}\mathit{\times }\mathit{1}\mathit{)}\mathit{=}\mathit{12} \\ then\mathit{ }\mathit{ }cos\theta =\frac{\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{BC}}{\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{\right|}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }cos\theta \mathit{ }\mathit{=}\frac{\mathit{12}}{\sqrt{\mathit{30}}\sqrt{\mathit{11}}} \\ \mathit{ }\mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{12}}{\sqrt{\mathit{30}}\sqrt{\mathit{11}}}\mathit{\right)}\mathit{=}\mathit{48}\mathit{.}{\mathit{7}}^o \end{gathered}$$

$\overline{)23}$   تقع النقطة C على المستقيم $\overleftrightarrow{ AB}$ ، حيث: AC AB =  . أجد إحداثيات النقطة  C 

$$\begin{gathered} Solution: \\ A(3,-2,1) , B(5,3,0) \\ let c=(x , y , z) \\ \frac{ x+5}{2}=3 \Rightarrow x=1 \\ \frac{y\mathit{+}\mathit{3}}{\mathit{2}}\mathit{=}\mathit{-}\mathit{2}\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }y\mathit{=}\mathit{-}\mathit{7} \\ \frac{z\mathit{+}\mathit{0}}{\mathit{2}}\mathit{=}\mathit{1}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }z\mathit{=}\mathit{2} \\ \mathit{ }So\mathit{ }c\mathit{=}\mathit{(}\mathit{1}\mathit{ }\mathit{,}\mathit{ }\mathit{-}\mathit{7}\mathit{ }\mathit{,}\mathit{ }\mathit{2}\mathit{)} \end{gathered}$$

 تقع النقطة (9 ,4- ,7-)A والنقطة (3 ,5 ,8)B على المستقيم l_{1 .}

 وتقع النقطة (7 ,11 ,6)C على المستقيم l₂ الذي معادلته : $\stackrel{\rightharpoonup }{r}=<6,11,7> +t<-1,3,2>$

 $\overline{)24}$  أبين أن النقطة B تقع على المستقيم l₂ .  

$$\begin{gathered} Solution: \\ B(8,5,3) \\ \stackrel{\rightharpoonup }{r}=<6,11,7> +t<-1,3,2> \\ for x : \\ 8=6-t \Rightarrow t=-2 \\ for y : \\ 5=11+3t \Rightarrow t=-2 \\ for z : \\ 3=7+2t \Rightarrow t=-2 \end{gathered}$$ 

 

$\overline{)25}$ أبين أن المستقيم l₁  والمستقيم l₂ متعامدان

$$\begin{gathered} Solution: \\ {l}_1:\stackrel{\rightharpoonup }{r}=<-15,-9,6> \\ {l}_2:\stackrel{\rightharpoonup }{r}=<-1,3,2> \\ but \stackrel{\rightharpoonup }{l_1} \perp \stackrel{\rightharpoonup }{l_2} if \Rightarrow \stackrel{\rightharpoonup }{l_1} .\stackrel{\rightharpoonup }{l_2}=0 \\ \left( -15\times -1\right)+ \left( -9\times 3\right)+ \left( 6\times 2\right) \stackrel{?}{=0} \\ 15-27+12=0 \\ then \stackrel{\rightharpoonup }{l_1} \perp \stackrel{\rightharpoonup }{l_2} \end{gathered}$$

$\overline{)26}$  أجد $m\angle ABC$ . 

$$\begin{gathered} Solution: \\ A(-7,-4,9) , B(8,5,3) ,C(6,11,7) \\ \stackrel{\mathit{\rightharpoonup }}{AB}\mathit{=}\mathit{<}\mathit{15}\mathit{,}\mathit{9}\mathit{,}\mathit{-}\mathit{6}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{BC}\mathit{=}\mathit{<}\mathit{-}\mathit{2}\mathit{,}\mathit{6}\mathit{,}\mathit{4}\mathit{>} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{=}\mathit{(}\mathit{15}\mathit{\times }\mathit{-}\mathit{2}\mathit{)}\mathit{+}\mathit{(}\mathit{9}\mathit{\times }\mathit{6}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{6}\mathit{\times }\mathit{4}\mathit{)}\mathit{=}\mathit{-}\mathit{30}\mathit{+}\mathit{54}\mathit{-}\mathit{24}\mathit{=}\mathit{0} \\ then \stackrel{\mathit{\rightharpoonup }}{AB}\perp \stackrel{\mathit{\rightharpoonup }}{BC}\mathit{ }\mathit{ } \end{gathered}$$

$\overline{)27}$ أجد مساحة المثلث ABC . 

$$\begin{gathered} Solution: \\ A\mathit{(}\mathit{-}\mathit{7}\mathit{,}\mathit{-}\mathit{4}\mathit{,}\mathit{9}\mathit{)}\mathit{ }\mathit{,}\mathit{ }B\mathit{(}\mathit{8}\mathit{,}\mathit{5}\mathit{,}\mathit{3}\mathit{)}\mathit{ }\mathit{ }\mathit{,}C\mathit{(}\mathit{6}\mathit{,}\mathit{11}\mathit{,}\mathit{7}\mathit{)} \\ \stackrel{\mathit{\rightharpoonup }}{\mathit{A}\mathit{B}}\mathit{=}\mathit{<}\mathit{15}\mathit{,}\mathit{9}\mathit{,}\mathit{-}\mathit{6}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{\mathit{B}\mathit{C}}\mathit{=}\mathit{<}\mathit{-}\mathit{2}\mathit{,}\mathit{6}\mathit{,}\mathit{4}\mathit{>} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{15}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{9}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{6}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{342}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{2}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{6}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{4}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{56}} \\ But\mathit{ }\mathit{ }\mathit{ }{A}_{\mathit{△}ABC}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{\right|} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\times }\sqrt{\mathit{342}}\mathit{\times }\sqrt{\mathit{56}}\mathit{=}\mathit{69}\mathit{.}\mathit{2}\mathit{ }unit \end{gathered}$$

ABCD   هرم ثلاثي . إذا كانت إحداثيات رؤوسه هي:

 (19 ,11 ,10)D , (0 ,1- ,6)C , (2 ,5 ,4-)B ,(1- ,3 ,4)A

 فأجيب عن الأسئلة الثلاثة الآتية تباعا:

$\overline{)28}$  أجد مساحة المثلث ABC  في صورة $a\sqrt{6}$ 

$$\begin{gathered} Solution: \\ \stackrel{\mathit{\rightharpoonup }}{AB}\mathit{=}\mathit{<}\mathit{-}\mathit{8}\mathit{,}\mathit{2}\mathit{,}\mathit{3}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{AC}\mathit{=}\mathit{<}\mathit{2}\mathit{,}\mathit{-}\mathit{4}\mathit{,}\mathit{1}\mathit{>} \\ \mathit{ }But\mathit{ }\mathit{ }\mathit{ }{A}_{\mathit{△}ABC}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{\right|}sin\theta \\ \mathit{ }now\mathit{ }to\mathit{ }find\mathit{ }sin\mathit{ }\theta \mathit{ }from\mathit{ }cos\theta \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{8}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{2}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{77}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{2}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{4}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{1}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{21}} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{=}\mathit{(}\mathit{-}\mathit{8}\mathit{\times }\mathit{2}\mathit{)}\mathit{+}\mathit{(}\mathit{2}\mathit{\times }\mathit{-}\mathit{4}\mathit{)}\mathit{+}\mathit{(}\mathit{3}\mathit{\times }\mathit{1}\mathit{)}\mathit{=}\mathit{-}\mathit{21} \\ \mathit{\Rightarrow }cos\theta \mathit{ }\mathit{=}\mathit{-}\frac{21}{\sqrt{\mathit{77}}\mathit{\times }\sqrt{\mathit{21}}}\mathit{ }\mathit{=}\mathit{-}\sqrt{\frac{\mathit{3}}{\mathit{11}}} \\ \mathit{\Rightarrow }\mathit{sin}\theta \mathit{ }\mathit{=}\sqrt{\mathit{1}\mathit{-}\frac{\mathit{3}}{\mathit{11}}}\mathit{=}\sqrt{\frac{\mathit{8}}{\mathit{11}}} Hinit : si{n}^2\theta + co{s}^2\theta =1 \\ \mathit{ }So\mathit{ }\mathit{ }{A}_{\mathit{△}ABC}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{\right|}sin\theta \\ \mathit{ }then\mathit{ }{A}_{\mathit{△}ABC}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\sqrt{\mathit{77}}\mathit{\times }\sqrt{\mathit{21}}\mathit{\times }\sqrt{\frac{\mathit{8}}{\mathit{11}}} \\ =\frac{\mathit{1}}{\mathit{2}}\mathit{\times }\sqrt{\frac{\cancel{\mathit{11}}\mathit{\times }\mathit{7}\mathit{\times }\mathit{7}\mathit{\times }\mathit{3}\mathit{\times }\mathit{4}\mathit{\times }\mathit{2}}{\cancel{\mathit{11}}}}\mathit{ }\mathit{=}\mathit{7}\mathit{ }\sqrt{\mathit{6}}\mathit{ } \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ } \end{gathered}$$

$\overline{)29}$  أثبت أن: $m \angle AED={90}^{\circ }$ حيث (1 ,1,2) E

$$\begin{gathered} Solution: \\ A(4,3,-1) , E(1,2,1) ,D(10,11,19) \\ \stackrel{\mathit{\rightharpoonup }}{AE}\mathit{=}\mathit{<}\mathit{-}\mathit{3}\mathit{,}\mathit{-}\mathit{1}\mathit{,}\mathit{2}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{ED}\mathit{=}\mathit{<}\mathit{9}\mathit{,}\mathit{9}\mathit{,}\mathit{18}\mathit{>}\mathit{=}\mathit{<}\mathit{1}\mathit{,}\mathit{1}\mathit{,}\mathit{2}\mathit{>} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{AE}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{ED}\mathit{=}\mathit{(}\mathit{-}\mathit{3}\mathit{\times }\mathit{9}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{1}\mathit{\times }\mathit{9}\mathit{)}\mathit{+}\mathit{(}\mathit{2}\mathit{\times }\mathit{18}\mathit{)} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{-}\mathit{27}\mathit{-}\mathit{9}\mathit{+}\mathit{36}\mathit{=}\mathit{0} \\ then \stackrel{\mathit{\rightharpoonup }}{AE}\perp \stackrel{\mathit{\rightharpoonup }}{ED}\mathit{ }\mathit{ } \end{gathered}$$

$\overline{)30}$ إذا علمت أن النقطة E تقع في المستوى نفسه الذي يقع فيه المثلث ABC  فأجد حجم الهرم ABCD   

      $$\begin{gathered} Solution: \\ E(1,2,1) , D(10,11,19) \\ \stackrel{\mathit{\rightharpoonup }}{ED}\mathit{=}\mathit{<}\mathit{9}\mathit{,}\mathit{9}\mathit{,}\mathit{18}\mathit{>} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{\mathit{E}\mathit{D}}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{9}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{9}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{18}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\mathit{9}\sqrt{\mathit{6}} \\ \mathit{ }and\mathit{ }{A}_{\mathit{△}\mathit{A}\mathit{B}\mathit{C}}\mathit{=}\mathit{7}\sqrt{\mathit{6}}\mathit{ } \\ \mathit{ }But\mathit{ }\mathit{ }\mathit{ }{A}_{ABCD}\mathit{=}\frac{\mathit{1}}{\mathit{3}}{A}_{\mathit{△}ABC}\times \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{ED}\mathit{\right|} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\frac{\mathit{7}\sqrt{\mathit{6}}\mathit{\times }\mathit{9}\sqrt{\mathit{6}}}{\mathit{3}}\mathit{ }\mathit{=}\mathit{126}\mathit{ }unit \end{gathered}$$

 

إذا كانت (6- ,1 ,3)A و (0 ,2- ,5)B و (6- ,4- ,8)C. فأجيب عن الأسئلة الخمسة الآتية تباعا:‎

$\overline{)31}$  أبين أنّ $\stackrel{\rightharpoonup }{AC}=n\left(\begin{array}{c}5\\ -5\\ 0\end{array}\right)$   حيث n عدد صحيح  

$$\begin{gathered} Solution : \\ \stackrel{\rightharpoonup }{AC}=\left(\begin{array}{c}5\\ -5\\ 0\end{array}\right)=5\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right) , such that n=5 \end{gathered}$$

$\overline{)32}$  أبين أن قياس الزاوية ‎ ACBهو $co{s}^{-1}\left(\frac{5\sqrt{2}}{14}\right)$ 

 $$\begin{gathered} Solution: \\ A(3,1,-6) , B(5,-2,0) ,C(8,-4,-6) \\ \stackrel{\mathit{\rightharpoonup }}{AC}\mathit{=}\mathit{<}\mathit{5}\mathit{,}\mathit{-}\mathit{5}\mathit{,}\mathit{0}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{CB}\mathit{=}\mathit{<}\mathit{-}\mathit{3}\mathit{,}\mathit{2}\mathit{,}\mathit{6}\mathit{>} \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AC}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{5}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{5}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{0}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{50}}\mathit{=}\mathit{5}\sqrt{\mathit{2}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{CB}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{3}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{2}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{6}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{49}}\mathit{=}\mathit{7} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{=}\mathit{(}\mathit{5}\mathit{\times }\mathit{-}\mathit{3}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{5}\mathit{\times }\mathit{2}\mathit{)}\mathit{+}\mathit{(}\mathit{0}\mathit{\times }\mathit{6}\mathit{)}\mathit{=}\mathit{-}\mathit{25} \\ then\mathit{ }\mathit{ }cos\theta =\frac{\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{BC}}{\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BC}\mathit{\right|}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }cos\theta \mathit{ }\mathit{=}\mathit{-}\frac{\mathit{25}}{\mathit{5}\sqrt{\mathit{2}}\mathit{\times }\mathit{7}}\mathit{=}\mathit{-}\frac{\mathit{5}}{\mathit{7}\sqrt{\mathit{2}}}\mathit{\times }\frac{\sqrt{\mathit{2}}}{\sqrt{\mathit{2}}}\mathit{=}\frac{\mathit{5}\sqrt{\mathit{2}}}{\mathit{14}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }\theta \mathit{=}{\mathit{cos}}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{5}\sqrt{\mathit{2}}}{\mathit{14}}\mathit{\right)} \end{gathered}$$

$\overline{)33}$ أكتب معادلة متجهة للمستقيم AC 

$$\begin{gathered} Solution: \\ A(3,1,-6) ,C(8,-4,-6) \\ \stackrel{\mathit{\rightharpoonup }}{AC}\mathit{=}\mathit{<}\mathit{5}\mathit{,}\mathit{-}\mathit{5}\mathit{,}\mathit{0}\mathit{>}\mathit{=}\mathit{<}\mathit{1}\mathit{,}\mathit{-}\mathit{1}\mathit{,}\mathit{0}\mathit{>} \\ So :\stackrel{\rightharpoonup }{r}=<3,1,-6> +t<1,-1,0> \\ Or :\stackrel{\rightharpoonup }{r}=<8,-4,-6> +t<1,-1,0> \end{gathered}$$

$\overline{)34}$ إذا كانت ( p ,1- ,6)D وعُلم أن AC وBD  متقاطعان  فما قيمة p ؟ 

$$\begin{gathered} Solution: \\ A(3,1,-6) ,C(8,-4,-6) ,B(5,-2,0) ,D(6,-1,p) \\ \stackrel{\mathit{\rightharpoonup }}{AC}\mathit{=}\mathit{<}\mathit{5}\mathit{,}\mathit{-}\mathit{5}\mathit{,}\mathit{0}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{\mathit{A}\mathit{C}}\mathit{=}\mathit{<}\mathit{5}\mathit{,}\mathit{-}\mathit{5}\mathit{,}\mathit{0}\mathit{>}\mathit{=}\mathit{<}\mathit{1}\mathit{,}\mathit{-}\mathit{1}\mathit{,}\mathit{0}\mathit{>} \\ So\mathit{ }\mathit{:}\stackrel{\mathit{\rightharpoonup }}{\mathit{r}}\mathit{=}\mathit{<}\mathit{3}\mathit{,}\mathit{1}\mathit{,}\mathit{-}\mathit{6}\mathit{>}\mathit{ }\mathit{+}t\mathit{<}\mathit{1}\mathit{,}\mathit{-}\mathit{1}\mathit{,}\mathit{0}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{BD}\mathit{=}\mathit{<}\mathit{1}\mathit{,}\mathit{1}\mathit{,}p\mathit{>} \\ So\mathit{ }\mathit{:}\stackrel{\mathit{\rightharpoonup }}{\mathit{r}}\mathit{=}\mathit{<}\mathit{5}\mathit{,}\mathit{-}\mathit{2}\mathit{,}\mathit{0}\mathit{>}\mathit{ }\mathit{+}u\mathit{<}\mathit{1}\mathit{,}\mathit{1}\mathit{,}p\mathit{>} \\ for x : \\ 3+t=5+u \Rightarrow t-u=2 ...\overline{)1} \\ for y : \\ 1-t=-2+u \Rightarrow -t-u=-3 ...\overline{)2} \\ from 1 and 2 : \\ -2u=-1 \Rightarrow u=\frac{1}{2} \\ \Rightarrow t=\frac{5}{2} \\ for z : \\ -6+0t=0+up \\ -6=\frac{1}{2}p \Rightarrow p=-12 \end{gathered}$$

$\overline{)35}$ أبين أن الشكل ABCD معين. ثم أجد طول كل ضلع من أضلاعه

$$\begin{gathered} Solution: \\ A(3,1,-6) , B(5,-2,0) ,C(8,-4,-6) ,D(6,-1,-12) \\ \stackrel{\mathit{\rightharpoonup }}{AB}\mathit{=}\mathit{<}\mathit{2}\mathit{,}\mathit{-}\mathit{3}\mathit{,}\mathit{6}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{CD}\mathit{=}\mathit{<}\mathit{-}\mathit{2}\mathit{,}\mathit{-}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{>} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{2}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{3}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{6}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{49}}\mathit{=}\mathit{7} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{CD}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{2}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{3}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{6}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{49}}\mathit{=}\mathit{7} \\ \mathit{ }\mathit{متساويين}\mathit{ }\mathit{ }\mathit{متقابلين}\mathit{ }\mathit{ }\mathit{ }\mathit{ضلعين}\mathit{ } \\ now for \\ \stackrel{\mathit{\rightharpoonup }}{AD}\mathit{=}\mathit{<}\mathit{3}\mathit{,}\mathit{-}\mathit{2}\mathit{,}\mathit{-}\mathit{6}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{BC}\mathit{=}\mathit{<}\mathit{3}\mathit{,}\mathit{-}\mathit{2}\mathit{,}\mathit{6}\mathit{>} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{AB}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{2}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{6}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{49}}\mathit{=}\mathit{7} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{CD}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{2}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{6}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{49}}\mathit{=}\mathit{7} \\ \mathit{ }\mathit{متساويين}\mathit{ }\mathit{ }\mathit{متقابلين}\mathit{ }\mathit{ }\mathit{ }\mathit{ضلعين}\mathit{ } \end{gathered}$$

 $\overline{)36}$  أحل المسألة الواردة في بداية الدرس 

$$\begin{gathered} Solution: \\ \stackrel{\mathit{\rightharpoonup }}{v}\mathit{=}\mathit{<}\mathit{9}\mathit{-}\mathit{1}\mathit{,}\mathit{13}\mathit{-}\mathit{2}\mathit{,}\mathit{21}\mathit{-}\mathit{1}\mathit{>}\mathit{=}\mathit{<}\mathit{8}\mathit{,}\mathit{11}\mathit{,}\mathit{20}\mathit{>} \\ \stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{<}\mathit{14}\mathit{-}\mathit{4}\mathit{,}\mathit{1}\mathit{+}\mathit{3}\mathit{,}\mathit{18}\mathit{-}\mathit{2}\mathit{>}\mathit{=}\mathit{<}\mathit{10}\mathit{,}\mathit{4}\mathit{,}\mathit{16}\mathit{>} \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{8}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{11}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{20}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{585}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{10}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{4}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{16}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{372}} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{(}\mathit{8}\mathit{\times }\mathit{10}\mathit{)}\mathit{+}\mathit{(}\mathit{11}\mathit{\times }\mathit{4}\mathit{)}\mathit{+}\mathit{(}\mathit{20}\mathit{\times }\mathit{16}\mathit{)}\mathit{=}\mathit{444} \\ then\mathit{ }\mathit{ }cos\theta =\frac{\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}}{\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}}=\frac{444}{\sqrt{\mathit{585}}\times \sqrt{\mathit{372}}} \\ \mathit{ }\mathit{\Rightarrow }\theta \mathit{ }\mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{444}}{\sqrt{\mathit{585}}\mathit{\times }\sqrt{\mathit{372}}}\mathit{\right)}\mathit{=}\mathit{17}\mathit{.}{\mathit{9}}^o \end{gathered}$$

$\overline{)37}$  تبرير: إذا كانت (4 ,2- ,3)A و (9 ,5- ,1)B و (1- ,4,5-)C

      وكانت النقطة D تقع على المستقيم المار بالنقطة A والنقطة B

      وكانت الزاوية CDA قائمة؛ فما إحداثيات النقطة D ؟ أبرّر إجابتي . 

$$\begin{gathered} Solution: \\ A(3,-2,4) ,B(1,-5,9) , C(-4,5,-1) , D=(x,y,z) \\ \stackrel{\rightharpoonup }{AB}=<-2,-3,5> \\ \stackrel{\rightharpoonup }{r}=<3,-2,4> +t<-2,-3,5> \\ now let D=(x,y,z) which satisfied \stackrel{\rightharpoonup }{r}: \\ for x : \\ x=3-2t \\ for y : \\ y=-2-3t \\ for z : \\ z=4+5t \\ So \stackrel{\rightharpoonup }{CD}=<x+4,y-5,z+1> \\ =<3-2t+4,-2-3t-5,4+5t+1> \\ =<7-2t,-7-3t,5+4t> \\ So \stackrel{\rightharpoonup }{DA}=<x-3,y+2,z-4> \\ =<3-2t -3,-2-3t+2,4+5t-4> \\ =<-2t,-3t,5t> \\ and \stackrel{\rightharpoonup }{CD} \perp \stackrel{\rightharpoonup }{DA}\Rightarrow \stackrel{\rightharpoonup }{CD} . \stackrel{\rightharpoonup }{DA}=0 \\ \left(7-2t\right)\left(-2t\right)+\left(-7-3t\right)\left(-3t\right)+\left(5+5t\right)\left(5t\right)=0 divided by t\ne 0 \\ \left(7-2t\right)\left(-2\right)+\left(-7-3t\right)\left(-3\right)+\left(5+5t\right)\left(5\right)=0 \\ -14+4t+21+9t+25+25t=0 \\ 32 +38t=0 \Rightarrow t=-\frac{16}{19} \\ then D=(x,y,z)=(3-2t,-2-3t,4+5t ) \\ =(3-2\left(-\frac{89}{19}\right),-2-3\left(-\frac{16}{19}\right),4+5\left(-\frac{16}{19}\right) ) \\ =(\frac{89}{19},\frac{10}{19},\frac{-4}{19} ) \end{gathered}$$

تحد: إذا كانت: $\stackrel{\rightharpoonup }{r}=\left(\begin{array}{c}-8\\ 16\\ 1\end{array}\right) + t\left(\begin{array}{c}7\\ -3\\ -6\end{array}\right)$ معادلة متجهة للمستقيم l₁ . 

وكانت: $\stackrel{\rightharpoonup }{r}=\left(\begin{array}{c}-10\\ 31\\ -26\end{array}\right) + u\left(\begin{array}{c}3\\ -6\\ 7\end{array}\right)$معادلة متجهة  للمستقيم l₂  ،

وتقاطع هذان المستقيمان في النقطة P ، وكانت النقطة Q تقع على المستقيم l₁ ؛

حيث: t = 3 ، والنقطة R تقع على المستقيم l₂ حيث: u>3 ، PQ = PR  .

أجيب عن السؤالين الآتيين تباعا

$\overline{)38}$ إذا كان $m \angle RPQ=\theta$، فأبيّن أنّ : $cos\theta =-\frac{3}{94}$ 

$$\begin{gathered} Solution: \\ \stackrel{\mathit{\rightharpoonup }}{v}\mathit{=}\mathit{<}\mathit{7}\mathit{,}\mathit{-}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{>} \\ {l}_{\mathit{1}}\mathit{:}\stackrel{\mathit{\rightharpoonup }}{r}\mathit{=}\mathit{<}\mathit{-}\mathit{8}\mathit{,}\mathit{16}\mathit{,}\mathit{1}\mathit{>}\mathit{+}t\mathit{<}\mathit{7}\mathit{,}\mathit{-}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{>}\mathit{ } \\ \stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{<}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{,}\mathit{7}\mathit{>} \\ {l}_{\mathit{2}}\mathit{:}\stackrel{\mathit{\rightharpoonup }}{r}\mathit{=}\mathit{<}\mathit{-}\mathit{10}\mathit{,}\mathit{31}\mathit{,}\mathit{-}\mathit{26}\mathit{>}\mathit{+}u\mathit{<}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{,}\mathit{7}\mathit{>}\mathit{ } \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{7}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{3}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{6}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{94}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{6}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{7}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{94} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{(}\mathit{7}\mathit{\times }\mathit{3}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{3}\mathit{\times }\mathit{-}\mathit{6}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{6}\mathit{\times }\mathit{7}\mathit{)}\mathit{=}\mathit{-}\mathit{3} \\ then\mathit{ }\mathit{ }cos\theta =\frac{\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}}{\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }cos\theta \mathit{ }\mathit{=}\mathit{-}\frac{\mathit{3}}{\sqrt{\mathit{94}}\mathit{\times }\sqrt{\mathit{94}}}\mathit{=}\mathit{-}\frac{\mathit{3}}{\mathit{94}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }\theta \mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{(}\mathit{-}\frac{\mathit{3}}{\mathit{94}}\mathit{)} \end{gathered}$$

$\overline{)39}$ أبين أن مساحة المثلث PQR هي $2\sqrt{8827}$ وحدة مربعة .    

$$\begin{gathered} Solution: \\ \stackrel{\mathit{\rightharpoonup }}{v}\mathit{=}\mathit{<}\mathit{7}\mathit{,}\mathit{-}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{>} \\ {l}_{\mathit{1}}\mathit{:}\stackrel{\mathit{\rightharpoonup }}{r}\mathit{=}\mathit{<}\mathit{-}\mathit{8}\mathit{,}\mathit{16}\mathit{,}\mathit{1}\mathit{>}\mathit{+}t\mathit{<}\mathit{7}\mathit{,}\mathit{-}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{>}\mathit{ } \\ \stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{<}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{,}\mathit{7}\mathit{>} \\ {l}_{\mathit{2}}\mathit{:}\stackrel{\mathit{\rightharpoonup }}{r}\mathit{=}\mathit{<}\mathit{-}\mathit{10}\mathit{,}\mathit{31}\mathit{,}\mathit{-}\mathit{26}\mathit{>}\mathit{+}u\mathit{<}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{,}\mathit{7}\mathit{>}\mathit{ } \\ \mathit{ }at\mathit{ }P\mathit{:}\mathit{ }\mathit{ }{l}_{\mathit{1}}\mathit{=}{l}_{\mathit{2}}\mathit{ }so\mathit{:} \\ \mathit{-}\mathit{8}\mathit{+}\mathit{7}t\mathit{=}\mathit{-}\mathit{10}\mathit{+}\mathit{3}u\mathit{ }\mathit{ }\mathit{ } \\ \mathit{ }\mathit{7}t\mathit{-}\mathit{3}u\mathit{=}\mathit{-}\mathit{2}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\mathit{ }\overline{)\mathit{1}} \\ \mathit{16}\mathit{-}\mathit{3}t\mathit{=}\mathit{31}\mathit{-}\mathit{6}u\mathit{ } \\ \mathit{ }\mathit{-}\mathit{3}t\mathit{+}\mathit{6}u\mathit{=}\mathit{15}\mathit{ }\mathit{ }\mathit{ }\mathit{ } \\ \mathit{ }\mathit{-}t\mathit{+}\mathit{2}u\mathit{=}\mathit{5}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\mathit{ }\overline{)\mathit{2}} \\ from\mathit{ }\mathit{1}\mathit{ }and\mathit{ }\mathit{2}\mathit{ }\mathit{:}\mathit{ }\mathit{ }\mathit{ }u\mathit{=}\mathit{3}\mathit{ }\mathit{ }\mathit{,}\mathit{ }and\mathit{ }t\mathit{=}\mathit{1} \\ \mathit{ }So\mathit{ }P\mathit{=}\mathit{(}\mathit{-}\mathit{8}\mathit{+}\mathit{7}t\mathit{,}\mathit{16}\mathit{-}\mathit{3}t\mathit{,}\mathit{1}\mathit{-}\mathit{6}t\mathit{)}\mathit{\Rightarrow }P\mathit{=}\mathit{(}\mathit{-}\mathit{1}\mathit{,}\mathit{13}\mathit{,}\mathit{-}\mathit{5}\mathit{)} \\ now\mathit{ }to\mathit{ }find\mathit{ }Q\mathit{ }when\mathit{ }from\mathit{ }{l}_{\mathit{1}}\mathit{ }\mathit{ }when\mathit{ }t\mathit{=}\mathit{3}\mathit{ } \\ \stackrel{\mathit{\rightharpoonup }}{r}\mathit{=}\mathit{<}\mathit{-}\mathit{8}\mathit{,}\mathit{16}\mathit{,}\mathit{1}\mathit{>}\mathit{+}t\mathit{<}\mathit{7}\mathit{,}\mathit{-}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{>} \\ \mathit{ }Q\mathit{=}\mathit{(}\mathit{-}\mathit{8}\mathit{+}\mathit{7}\mathit{(}\mathit{3}\mathit{)}\mathit{,}\mathit{16}\mathit{-}\mathit{3}\mathit{(}\mathit{3}\mathit{)}\mathit{,}\mathit{1}\mathit{-}\mathit{6}\mathit{(}\mathit{3}\mathit{\left)}\mathit{\right)}\mathit{ }\mathit{\Rightarrow }Q\mathit{=}\mathit{(}\mathit{13}\mathit{,}\mathit{7}\mathit{,}\mathit{-}\mathit{17}\mathit{)} \\ \mathit{ }then\mathit{ }\stackrel{\mathit{\rightharpoonup }}{PQ}\mathit{=}\mathit{<}\mathit{14}\mathit{,}\mathit{-}\mathit{6}\mathit{,}\mathit{-}\mathit{12}\mathit{>} \\ now\mathit{ }to\mathit{ }find\mathit{ }R\mathit{ }when\mathit{ }from\mathit{ }{l}_{\mathit{2}}\mathit{ }\mathit{ }when\mathit{ }\mathit{ }PQ\mathit{=}PR \\ {l}_{\mathit{2}}\mathit{:}\stackrel{\mathit{\rightharpoonup }}{r}\mathit{=}\mathit{<}\mathit{-}\mathit{10}\mathit{,}\mathit{31}\mathit{,}\mathit{-}\mathit{26}\mathit{>}\mathit{+}u\mathit{<}\mathit{3}\mathit{,}\mathit{-}\mathit{6}\mathit{,}\mathit{7}\mathit{>}\mathit{ } \\ \mathit{ }R\mathit{=}\mathit{(}\mathit{-}\mathit{10}\mathit{+}\mathit{3}u\mathit{,}\mathit{31}\mathit{-}\mathit{6}u\mathit{,}\mathit{-}\mathit{26}\mathit{+}\mathit{7}u\mathit{)} \\ \mathit{ }then\mathit{ }\stackrel{\mathit{\rightharpoonup }}{PR}\mathit{=}\mathit{<}\mathit{-}\mathit{9}\mathit{+}\mathit{3}u\mathit{,}\mathit{18}\mathit{-}\mathit{6}u\mathit{,}\mathit{-}\mathit{21}\mathit{+}\mathit{7}u\mathit{>} \\ \mathit{ }but\mathit{ }\mathit{ }\mathit{ }PQ\mathit{=}PR\mathit{ }\mathit{:} \\ \mathit{ }for\mathit{ }x\mathit{ }\mathit{:}\mathit{ }\mathit{14}\mathit{=}\mathit{-}\mathit{11}\mathit{+}\mathit{3}u \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{PQ}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{14}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{6}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{12}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{376}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{PR}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{9}\mathit{+}\mathit{3}u\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{18}\mathit{-}\mathit{6}u\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{21}\mathit{+}\mathit{7}u\mathit{)}}^{\mathit{2}}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\sqrt{\mathit{81}\mathit{-}\mathit{54}u\mathit{+}\mathit{9}{u}^{\mathit{2}}\mathit{+}\mathit{324}\mathit{-}{\mathit{216}u\mathit{+}\mathit{36}u}^{\mathit{2}}\mathit{+}\mathit{441}\mathit{-}{\mathit{294}u\mathit{+}\mathit{49}u}^{\mathit{2}}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\sqrt{\mathit{846}\mathit{-}\mathit{564}u\mathit{+}\mathit{94}{u}^{\mathit{2}}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }PQ\mathit{=}PR\mathit{ }\mathit{ } \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\sqrt{\mathit{376}}\mathit{=}\mathit{ }\sqrt{\mathit{846}\mathit{-}\mathit{564}u\mathit{+}\mathit{94}{u}^{\mathit{2}}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{376}\mathit{ }\mathit{=}\mathit{846}\mathit{-}\mathit{564}u\mathit{+}\mathit{94}{u}^{\mathit{2}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{94}{u}^{\mathit{2}}\mathit{-}\mathit{564}u\mathit{+}\mathit{470}\mathit{=}\mathit{0}\mathit{ } \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }u\mathit{=}\mathit{5}\mathit{ }\mathit{,}\mathit{ }u\mathit{=}\mathit{1}\mathit{ }rejected\mathit{ } \\ \mathit{ }now\mathit{ }when\mathit{ }u\mathit{=}\mathit{ }\mathit{5}\mathit{ }then\mathit{ }\mathit{ }R\mathit{=}\mathit{(}\mathit{-}\mathit{10}\mathit{+}\mathit{3}u\mathit{,}\mathit{31}\mathit{-}\mathit{6}u\mathit{,}\mathit{-}\mathit{26}\mathit{+}\mathit{7}u\mathit{)} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }R\mathit{=}\mathit{(}\mathit{5}\mathit{,}\mathit{1}\mathit{,}\mathit{9}\mathit{)} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{PR}\mathit{\right|}\mathit{=}\sqrt{{\mathit{\left(}\mathit{6}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{12}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{14}\mathit{\right)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{376}} \\ {A}_{\mathit{△}PQR}\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{PQ}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{PR}\mathit{\right|}sin\theta \\ =\frac{1}{2}\sqrt{\mathit{376}}\times \sqrt{\mathit{376}}sin\theta \\ now to find sin\theta =\sqrt{1-{\left(\frac{3}{94}\right)}^2}=\frac{\sqrt{8827}}{94} \\ {A}_{\mathit{△}PQR}\mathit{=}\frac{1}{2}\sqrt{\mathit{376}}\times \sqrt{\mathit{376}}\times \frac{\sqrt{8827}}{94} \\ \mathit{=}\frac{\mathit{376}}{184}\times \sqrt{8827} =2\sqrt{8827} \end{gathered}$$

تحدٍّ: رسم متوازي المستطيلات الآتي باستعمال برمجية حاسوبية تعتمد    

في قياساتها على المتجهات. فكانت كالآتي: 

$\stackrel{\rightharpoonup }{AB}=2\hat{i}+4\hat{j}+4\hat{k} , \stackrel{\rightharpoonup }{AD }=-10\hat{i}+10\hat{j}-5\hat{k} , \stackrel{\rightharpoonup }{AE}=-6\hat{i}-3\hat{j}+6\hat{k} ,$

$\overline{)40}$  إذا كانت (2- ,3 ,8)B فأجد إحداثيات النقطة H .

$$\begin{gathered} Solution: \\ \stackrel{\mathit{\rightharpoonup }}{BH\mathit{ }}=\stackrel{\mathit{\rightharpoonup }}{BA\mathit{ }}+\stackrel{\mathit{\rightharpoonup }}{AD\mathit{ }}+\stackrel{\mathit{\rightharpoonup }}{DH\mathit{ }} but \stackrel{\mathit{\rightharpoonup }}{DH\mathit{ }}=\stackrel{\mathit{\rightharpoonup }}{AE} \\ \mathit{<}x\mathit{-}\mathit{8}\mathit{,}y\mathit{-}\mathit{3}\mathit{,}z\mathit{+}\mathit{2}\mathit{>} =\mathit{<}\mathit{-}\mathit{2}\mathit{,}\mathit{-}\mathit{4}\mathit{,}\mathit{-}\mathit{4}\mathit{>}\mathit{+}\mathit{<}\mathit{-}\mathit{10}\mathit{,}\mathit{10}\mathit{,}\mathit{-}\mathit{5}\mathit{>}\mathit{+}\mathit{<}\mathit{-}\mathit{6}\mathit{,}\mathit{-}\mathit{3}\mathit{,}\mathit{6}\mathit{>} \\ \mathit{ }\mathit{<}x\mathit{-}\mathit{8}\mathit{,}y\mathit{-}\mathit{3}\mathit{,}z\mathit{+}\mathit{2}\mathit{>}\mathit{=}\mathit{<}\mathit{-}\mathit{18}\mathit{,}\mathit{3}\mathit{,}\mathit{-}\mathit{3}\mathit{>} \\ x\mathit{-}\mathit{8}\mathit{=}\mathit{-}\mathit{18}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }x\mathit{=}\mathit{-}\mathit{10} \\ y\mathit{-}\mathit{3}\mathit{=}\mathit{3}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }y\mathit{=}\mathit{6} \\ z\mathit{+}\mathit{2}\mathit{=}\mathit{-}\mathit{3}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }z\mathit{=}\mathit{-}\mathit{5} \end{gathered}$$

$\overline{)41}$  أجد قياس الزاوية GAC مقربا إلى أقرب عشردرجة

$$\begin{gathered} Solution: \\ m\mathit{\angle }GAC\mathit{=}m\mathit{\angle }HBD \\ Solution: \\ H(-10,6,-5) , B(8,3,-2) , D(-4,9,-7) \\ since\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}cos\theta \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BD}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{12}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{6}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{5}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{\mathit{205}} \\ \mathit{ }\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{BH}\mathit{\right|}\mathit{=}\sqrt{{\mathit{(}\mathit{-}\mathit{18}\mathit{)}}^{\mathit{2}}\mathit{+}{\mathit{\left(}\mathit{3}\mathit{\right)}}^{\mathit{2}}\mathit{+}{\mathit{(}\mathit{-}\mathit{3}\mathit{)}}^{\mathit{2}}}\mathit{=}\sqrt{342} \\ and\mathit{ }\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{=}\mathit{(}\mathit{-}\mathit{12}\mathit{\times }\mathit{-}\mathit{18}\mathit{)}\mathit{+}\mathit{(}\mathit{6}\mathit{\times }\mathit{3}\mathit{)}\mathit{+}\mathit{(}\mathit{-}\mathit{5}\mathit{\times }\mathit{-}\mathit{3}\mathit{)}\mathit{=}\mathit{249} \\ then\mathit{ }\mathit{ }cos\theta =\frac{\stackrel{\mathit{\rightharpoonup }}{v}\mathit{.}\stackrel{\mathit{\rightharpoonup }}{w}}{\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{v}\mathit{\right|}\mathit{\left|}\stackrel{\mathit{\rightharpoonup }}{w}\mathit{\right|}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }cos\theta \mathit{ }\mathit{=}\frac{\mathit{249}}{\sqrt{\mathit{205}}\mathit{\times }\sqrt{\mathit{342}}}\mathit{ }\mathit{\Rightarrow }\theta \mathit{=}co{s}^{\mathit{-}\mathit{1}}\mathit{\left(}\frac{\mathit{249}}{\sqrt{\mathit{205}}\mathit{\times }\sqrt{\mathit{342}}}\mathit{\right)}=19.9^o \end{gathered}$$

 

$\overline{)42}$  إذا كان X نقطة منتصف الضلع EF فأجد جيب تمام الزاوية DXC .