Question

The angle between the diagonals of the parallelogram formed by the points $(1,2,3),(-1,-2,-1),(2,3,2),(4,7,6)$ is

• A. ${\cos }^{-1}\frac{1}{3}$
• B. ${\cos }^{-1}\frac{7}{\sqrt{155}}$
• C. ${\cos }^{-1}\frac{7}{\sqrt{465}}$
• D. ${\cos }^{-1}\frac{7}{465}$

Answer 1

The correct option is C ${\cos }^{-1}\frac{7}{\sqrt{465}}$

Let $A=(1,2,3),B=(-1,-2,-1),C=(2,3,2),D=(4,7,6)$
Direction ratios of diagonal $AC$ is $(1-2,2-3,3-2)=(-1,-1,1)=(a_1,b_1,c_1)$
Direction ratios of diagonal $BD$ is $(-1-4,-2-7,-1-6)=(-5,-9,-7)=(a_2,b_2,c_2)$
Angle between DR's of $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ is given by:
$\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\cos \theta =\frac{(-1)(-5)+(-1)(-9)+\left(1\right)(-7)}{\sqrt{1^2+1^2+1^2}\sqrt{5^2+9^2+7^2}}$
$\cos \theta =\frac{7}{\sqrt{465}}$
$\theta ={\cos }^{-1}\frac{7}{\sqrt{465}}$