Vector Algebra

Find the angle between the planes $\;$ $2x + y -z = 9$ $\;$ and $\;$ $x + 2y + z = 7$

The given planes are

$2x + y -z = 9$ $\;\;\; \cdots \; (1a)$

$x + 2y + z = 7$ $\;\;\; \cdots \; (2a)$

The normals to the given planes are

$\overrightarrow{n_1} = 2 \hat{i} + \hat{j} - \hat{k}$ $\;\;\; \cdots \; (1b)$

$\overrightarrow{n_2} = \hat{i} + 2 \hat{j} + \hat{k}$ $\;\;\; \cdots \; (2b)$

If $\;$ $\theta$ $\;$ is the angle between the planes, then

$\begin{aligned} \cos \theta & = \dfrac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right| \left|\overrightarrow{n_2}\right|} \\\\ & = \dfrac{\left(2 \hat{i} + \hat{j} - \hat{k}\right) \cdot \left(\hat{i} + 2 \hat{j} + \hat{k}\right)}{\sqrt{\left(2\right)^2 + \left(1\right)^2 + \left(-1\right)^2} \; \sqrt{\left(1\right)^2 + \left(2\right)^2 + \left(1\right)^2}} \\\\ & = \dfrac{2 + 2 -1}{\sqrt{6} \; \sqrt{6}} \end{aligned}$

$\therefore$ $\;$ $\theta = \cos^{-1} \left(\dfrac{3}{6}\right) = \dfrac{\pi}{3}$