Question

Angle between planes $2x-y+z$ $=$ $6$ and $x+y+2z$ $=$ $7,$ is -

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

Answer 1

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

Plane $1$: $2x-y+z=6$

normal vector is $\overline{n_1}=2\hat{i}-\hat{j}+\hat{k}$

Plane $2$: $x+y+2z=7$

normal vector is $\overline{n_2}=\hat{i}+\hat{j}+2\hat{k}$

Angle between planes is same as the angle between their normal.

$\Rightarrow \cos \theta =\frac{\overline{n_1}\cdot \overline{n_2}}{|\overline{n_1}||\overline{n_2}|}$

$=\frac{(2\hat{i}-\hat{j}+\hat{k})\cdot (\hat{i}+\hat{j}+2\hat{k})}{\left(\sqrt{4+1+1}\right)\sqrt{1+1+4}}$

$=\left|\frac{2-1+2}{\sqrt{6}\cdot \sqrt{6}}\right|$

$=\frac{3}{6}$

$=\frac{1}{2}$

$\Rightarrow \cos \theta =\frac{1}{2}$

$\Rightarrow \theta =\frac{2}{3}$.