# How do I find the angle between the planes 5(x+1) + 3(y+2) + 2z = 0 and x + 3(y-1) + 2(z+4) = 0?

Sep 4, 2016

$\approx {38.69}^{\circ}$

#### Explanation:

Let the given planes be ${\pi}_{1} \mathmr{and} {\pi}_{2}$. Let ${\vec{n}}_{1} \mathmr{and} {\vec{n}}_{2}$

be their normals.

Recall that for a plane $: a x + b y + c z + d = 0$, its normal $\vec{n} = \left(a , b , c\right)$.

Clearly, ${\vec{n}}_{1} = \left(5 , 3 , 2\right) , \mathmr{and} , {n}_{2} = \left(1 , 3 , 2\right)$

the $\angle \alpha$ btwn. ${\pi}_{1} \mathmr{and} {\pi}_{2}$ is, by defn.,

$\alpha = a r c \cos \left\{| {\vec{n}}_{1} \cdot {\vec{n}}_{2} \frac{|}{| | {\vec{n}}_{1} | | | | {\vec{n}}_{2} | |}\right\}$.

$= a r c \cos \left\{| 5 \cdot 1 + 3 \cdot 3 + 2 \cdot 2 \frac{|}{\left(\sqrt{25 + 9 + 4} \sqrt{1 + 9 + 4}\right)}\right\}$

$= a r c \cos \left(\frac{18}{\sqrt{2 \cdot 19 \cdot 2 \cdot 7}}\right)$

$= a r c \cos \left(\frac{9}{\sqrt{133}}\right)$

$= a r c \cos \left(\frac{9}{11.53}\right)$

$\approx a r c \cos \left(0.7806\right)$

$\approx {38.69}^{\circ}$