Planes in 3D Math Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

hard

Find the angle between the planes

x + 2y - z = 5

and

2x - y + 3z = 1

Solution

1
Normal vectors: $\vec{n_1} = \langle 1, 2, -1 \rangle$ and $\vec{n_2} = \langle 2, -1, 3 \rangle$ . $\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|}$ .
2
$\vec{n_1} \cdot \vec{n_2} = 2 - 2 - 3 = -3$ . $|\vec{n_1}| = \sqrt{6}$ , $|\vec{n_2}| = \sqrt{14}$ . $\cos\theta = \frac{3}{\sqrt{84}} = \frac{3}{2\sqrt{21}}$ . $\theta = \cos^{-1}\left(\frac{3}{2\sqrt{21}}\right) \approx 70.9°$ .

Answer

\theta = \cos^{-1}\left(\frac{3}{2\sqrt{21}}\right) \approx 70.9°

The angle between two planes equals the angle between their normal vectors (or its supplement — we take the acute angle by using the absolute value of the dot product). Parallel planes have angle