Question

Angle between the planes $x+y+2z=6$ and $2x-y+z=9$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $\frac{\pi }{3}$

Normal vector of plane $x+y+2z=6$ is $\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k}$

and normal vector to the plane $2x-y+z=9$ is $\overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k}$

Note: Angle between the plane is same as angle between their normals

Thus angle between above planes is $={\cos }^{-1}\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}={\cos }^{-1}\frac{2-1+2}{\sqrt{6}\sqrt{6}}={\cos }^{-1}\frac{1}{2}=\frac{\pi }{3}$