 # How to Find the Angle Between Two Planes

In the figure below, you can see two planes that intersect. They are drawn as lines because we’re looking at them from the side. To find the angle between two planes, you do similar things to when you find the angle between a line and a plane. The difference is that in this case, you find the angle between the normal vectors to the planes ${\stackrel{\to }{n}}_{\alpha }$ and ${\stackrel{\to }{n}}_{\beta }$. Take note of this:

• If $\gamma <90\text{°}$, you have the correct angle, and $\alpha =\gamma$.

• if $\gamma >90\text{°}$, the real angle between the two planes is $\alpha =180\text{°}-\gamma$.

The angle between two planes is always $\le 90\text{°}$.

Example 1

You have two planes $\alpha$ and $\beta$ with normal vectors ${\stackrel{\to }{n}}_{\alpha }=\phantom{\rule{-0.17em}{0ex}}\left(1,1,1\right)$ and ${\stackrel{\to }{n}}_{\beta }=\phantom{\rule{-0.17em}{0ex}}\left(-2,3,1\right)$. The angle between the two normal vectors is $\begin{array}{llll}\hfill \mathrm{cos}\gamma & =\frac{\phantom{\rule{-0.17em}{0ex}}\left(1,1,1\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-2,3,1\right)}{\phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(1,1,1\right)|\cdot \phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(-2,3,1\right)|}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1\cdot \phantom{\rule{-0.17em}{0ex}}\left(-2\right)+1\cdot 3+1\cdot 1}{\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}\cdot \sqrt{\phantom{\rule{-0.17em}{0ex}}{\left(-2\right)}^{2}+{3}^{2}+{1}^{2}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-2+3+1}{\sqrt{3}\cdot \sqrt{14}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2}{\sqrt{42}},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \gamma & ={\mathrm{cos}}^{-1}\phantom{\rule{-0.17em}{0ex}}\left(\frac{2}{\sqrt{42}}\right)\approx 72.02\text{°}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

As $72.02\text{°}<90\text{°}$, you have the correct angle, and $\alpha =\gamma =72.02\text{°}$.