Question

Angle between the planes \(x+y+2 z=6\) and \(2 x-y+z=9\) is
(A) \(4^{\pi}\)
(B) \(6^{\pi}\) (C) \(3^{\pi}\)
(D) \(2 \pi\)

Answer 1

Ans: (C)
 \(\quad x+y+2 z=6 ; \quad 2 x-y+z=9\)
\(\therefore\) Angle between the planes \(=\) angle between the normals :
$$
\begin{aligned}
\theta=& \cos ^{-1}\left(\frac{1 \times 2+1(-1)+2 \times 1}{\sqrt{1^{2}+1^{2}+2^{2}} \cdot \sqrt{2^{2}+(-1)^{2}+1^{2}}}\right) \\
&=\cos ^{-1}\left(\frac{4-1}{6}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}
\end{aligned}
$$