Question

The angle between the planes $2x-y+3z=6$ and $x+y+2z=7$ is:

• A. ${\cos }^{-1}\left(\sqrt{\frac{7}{3}}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{2}\sqrt{\frac{7}{3}}\right)$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{1}{2}\sqrt{\frac{7}{3}}\right)$

The $D.R^{\prime }s$ of the normals of the planes are $(2,-1,3)$ and $(1,1,2)$
Since angle between planes is same as the angle between normals. Hence
$\Rightarrow \cos \theta =\frac{(a_1{a}_2+b_1{b}_2+c_1{c}_2)}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\Rightarrow \cos \theta =\frac{2-1+6}{\sqrt{14}\sqrt{6}}=\sqrt{\frac{7}{12}}=\frac{1}{2}\sqrt{\frac{7}{3}}$
$\therefore \theta ={\cos }^{-1}\left(\frac{1}{2}\sqrt{\frac{7}{3}}\right)$