Question

The position vectors of points A,B,C and D are

A=3i^+4j^+5k^, B=4i^+5j^+6k^, C=7i^+9j^+3k^ and D=4i^+6j^ then the displacement vectors AB and CD are

A) Perpendicular

B) Parallel

C) Antiparallel

D) Inclined at an angle of 60*

Answer 1

AB= i^+j^+k^, CD= -3i^-3j^-3k^
Now find the angle between them using
$\cos( \alpha ) = \frac{ab.cd}{ |ab| |cd| }$

Answer 2

Answer:

Vectors AB and CD are anti-parallel.

C is correct.

Explanation:

Given that,

Vector $OA=3\hat{i}+4\hat{j}+5\hat{k}$

Vector $OB=4\hat{i}+5\hat{j}+6\hat{k}$

Vector $OC=7\hat{i}+9\hat{j}+3\hat{k}$

Vector $OD=4\hat{i}+6\hat{j}$

The displacement vector AB is

$\vec{AB}=OB -OA$

$\vec{AB}=4\hat{i}+5\hat{j}+6\hat{k}-(3\hat{i}+4\hat{j}+5\hat{k})$

$\vec{AB}=4\hat{i}+5\hat{j}+6\hat{k}-3\hat{i}-4\hat{j}-5\hat{k}$

$\vec{AB}=1\hat{i}+1\hat{j}+1\hat{k}$

The displacement vector CD is

$\vec{CD}=OD -OC$

$\vec{CD}=4\hat{i}+6\hat{j}-(7\hat{i}+9\hat{j}+3\hat{k})$

$\vec{CD}=4\hat{i}+6\hat{j}-7\hat{i}-9\hat{j}-3\hat{k}$

$\vec{CD}=-3\hat{i}-3\hat{j}-3\hat{k}$

$\vec{CD}=-3(\hat{i}+\hat{j}+\hat{k})$

The angle between AB and CD is

$\cos\theta=\dfrac{1\hat{i}+1\hat{j}+1\hat{k}\cdot{-3(\hat{i}+\hat{j}+\hat{k})}}{\sqrt{1^2+1^2+1^2}\cdot(-3)\sqrt{(1)^2+(1)^2+(1)^2}}$

$\cos\theta=\dfrac{-9}{-9}$

$\cos\theta=1$

$\theta = 0^{\circ}$

CD= -3 AB

Hence, Vectors AB and CD are anti-parallel.