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. $co{s}^{-1}\left(7\right)$
• B. $co{s}^{-1}(\frac{7}{\sqrt{155}})$
• C. $co{s}^{-1}(\frac{7}{\sqrt{465}})$
• D. $co{s}^{-1}(\frac{7}{465})$

Answer 1

The correct option is C $co{s}^{-1}(\frac{7}{\sqrt{465}})$

Let $A=(1,2,3),B=(-1,-2,-1),C=(2,3,2),D=(4,7,6)$
Direction ratios of diagonal $AC$ is $(1-2,2-3,3-2)=(-1,-1,1)$
Direction ratios of diagonal BD is $(-1-4,-2-7,-1-6)=(-5,-9,-7)$
Angle between two lines with direction ratios $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$\cos \theta =\frac{x_1{x}_2+y_1{y}_2+z_1{z}_2}{\sqrt{(x_1{)}^2+(y_1{)}^2+(z_1{)}^2}\sqrt{(x_2{)}^2+(y_2{)}^2+(z_2{)}^2}}$

$\cos \theta =\frac{(-1)(-5)+(-1)(-9)+\left(1\right)(-7)}{\sqrt{(-1{)}^2+(-1{)}^2+1^2}\sqrt{(-5{)}^2+(-9{)}^2+(-7{)}^2}}$

$=\frac{7}{465}$

$\therefore \theta =co{s}^{-1}\left(\frac{7}{465}\right)$
Hence, option C is correct.