Question

Find the angle between the planes
$2x+7y+11z-3=0\text{and}5x+3y+9z+1=0$

• A. ${\cos }^{-1}\left(\frac{130}{\sqrt{174}.\sqrt{115}}\right)$
• B. ${\cos }^{-1}\left(\frac{139}{\sqrt{174}.\sqrt{115}}\right)$
• C. ${\cos }^{-1}\left(\frac{123}{\sqrt{174}.\sqrt{115}}\right)$
• D. ${\cos }^{-1}\left(\frac{107}{\sqrt{174}.\sqrt{115}}\right)$

Answer 1

The correct option is A

${\cos }^{-1}\left(\frac{130}{\sqrt{174}.\sqrt{115}}\right)$

We know that the angle between two planes $a_{1}x+b_{1}y+c_{1}z+d_1=0\&a_{2}x+b_{2}y+c_{2}z+d_2=0$ is given by -

$\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}$

So, Here we’ll substitute the appropriate values.

$\cos \theta =\frac{2.5+7.3+11.9}{\sqrt{(2^2+7^2+{11}^2)(5^2+3^2+9^2)}}\cos \theta =\frac{130}{\sqrt{174}.\sqrt{115}}\theta ={\cos }^{-1}\left(\frac{130}{\sqrt{174}\sqrt{115}}\right)$