Find the angle between each set of the following planes. 1) $5x - 3y + 2z = 11$ and $x + 3y + 2z = -5$

Question

Find the angle between each set of the following planes.

1) $5x - 3y + 2z = 11$ and $x + 3y + 2z = -5$

The $11$ and $-5$ I am confused on.

$n_1 = [5,-3,2]$ and $n_2 = [1,3,2]$

$\theta = cos^{-1}\frac{n_1 \cdot n_2}{|n_1||n_2|}$

$n_1 \cdot n_2 = 0$

so $\theta = cos^{-1}(0) = 90$

Answer 1

In general, the normal to the plane $ax+by+cz=d$ is $(a,b,c)$. Thus, since the dot product of the normals is zero, the planes are perpendicular.

Answer 2

As noticed by amd in the comments, note that the angle between the given planes is the same for the planes shifted at the origin that is

• $5x - 3y + 2z = 0$
• $x + 3y + 2z = 0$

and since $n_1\cdot n_2=0$ the angle between them is $90°$.