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Vector Algebra — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVector Algebra5 Marks
Question
If $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}, 3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $\hat{\mathrm{i}}-6 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ are the position vectors of points A, B, C and D respectively, then find the angle between $\vec {AB}$ and $\vec {CD}$. Deduce that $\overrightarrow{A B}$ and $\overrightarrow{C D}$ are collinear.
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Answer

$A(\hat{i}+\hat{j}+\hat{k}), \mathrm{B}(2 \hat{i}+5 \hat{j}), \mathrm{C}(3 \hat{i}+2 \hat{j}-3 \hat{k})$ and $D(\hat{i}-6 \hat{j}-\hat{k})$
 $\begin{aligned} \overrightarrow{A B} &=(2-1) \hat{\imath}+(5-1) \hat{\jmath}+(0-1) \hat{k} \\ &=1 \hat{\imath}+4 \hat{\jmath}-\hat{k} \end{aligned}$
$\begin{aligned} \overrightarrow{C D} &=(1-3) \hat{\imath}+(-6-2) \hat{\jmath}+(-1-(-3)) \hat{k} \\ &=-2 \hat{\imath}-8 \hat{\jmath}+2 \hat{k} \end{aligned}$
Let $\theta$ be the angle between $\vec{\mathrm{AB}} $ and $\vec{\mathrm{CD}}$
So, $\cos \theta=\frac{\overrightarrow{A B} \cdot \overrightarrow{C D}}{|\overrightarrow{A B}||\overrightarrow{C D}|}$
$\cos \theta=\frac{(\hat{i}+4 \hat{j}-\hat{k}) \cdot(-2 \hat{i}-8 \hat{j}+2 \hat{k})}{\sqrt{1+16+1} \sqrt{4+64+4}}$
$\cos \theta=\frac{-2-32-2}{\sqrt{18} \sqrt{72}}$
$\cos \theta=\frac{-36}{\sqrt{2} \cdot 6 \sqrt{2}}$
cos $\theta$ = -1
cos $\theta$ = cos $\pi$
$\theta$ = $\pi$ 
Since $0 \leq \theta \leq \pi$ it follows that $\theta=\pi$. This shows that $\vec {AB}$ and $\vec {CD}$ are collinear.

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