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.

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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Assume that the probability that a bomb dropped from an aeroplane will strike a certain target is 0.2. If 6 bombs are dropped, find the probability that.
  1. exactly 2 will strike the target.
  2. at least 2 will strike the target.
Find the value of $\int_0^\pi \frac{\sec x}{\sec x+\tan x} d x$.
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$4\text{x}+3\text{y}=\log\big(4\text{x}-3\text{y}\big)$
If $\vec{\alpha}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\beta}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}},$ then express $\vec{\beta}$ in the form of $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2,$ where $\vec{\beta}_1$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$.
Find the points of local maxima or local minima and corresponding local maximum and local minimum values of the following functions. Also, find the points of inflection,
$\text{f}'(\text{x})=\text{x}^{4}-62\text{x}^{2}+120\text{x}+9$
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}|\text{x}-\text{a|}\sin(\frac{1}{\text{x}-\text{a}}), &\text{for} \text{ x} \neq\text{a}\\0,&\text{for} \text{ x} = \text{a}\end{cases}\text{ at x}=0$
Find the shortest distance between the following pairs of lines whose cartesian equation are:
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\text{z}$ and $\frac{\text{x}+2}{3}=\frac{\text{y}-2}{1};\text{z}=2$
Evaluate $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x$.
Solve the following differential equation:

$(\text{x}^{2} - 1 ) \frac{\text{dy}}{\text{dx}} + 2 \text{xy} = \frac{2}{\text{x}^{2} - 1 }.$

Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.