Find the angle between two planes using their normal vectors

Question

The angle between two intersecting planes is defined to be the angle between their normal vectors. Find the angle between the planes $x – 2y + z = 0$ and $2x + 3y – 2z = 0$. Find the parametric equations of the line of intersection of the two planes above.

For the first plane I said $\overrightarrow n_0 =\langle 1, -2, 1 \rangle$ and for the second plane $\overrightarrow n_1 = \langle 2, 3, -2 \rangle$

Then using $\cos(\theta)= \frac{\mathbf A \bullet \mathbf B}{|\mathbf A||\mathbf B|}$ so $\theta = \cos^{-1}\left(\frac{\mathbf A \bullet \mathbf B}{|\mathbf A||\mathbf B|}\right)$

$\mathbf A \bullet \mathbf B = -6$

$|\mathbf A||\mathbf B|= \sqrt{102}$

$\theta = \cos^{-1}\left(\frac{-6}{\sqrt{102}}\right)$

Assuming this is correct so far, how do I find the parametric equations from here?

Answer 1

I don't think calculating the angle between two lines will help you to find the equation of the line of intersection of two lines.

It should be clear that the line of intersection is the line which is perpendicular to the normal of both the given planes.

Then the line will be along the cross product of the normal vector of both the planes.

$\therefore$ doing simple calculation gives us $$ \vec{n_{0}}\times\vec{n_{1}} = \langle1,4,7\rangle $$ So now we know the direction in which the required line is pointing.

Also it is not hard to guess that $(0,0,0)$ lies in both the plane. Thus the equation of line should pass through it.

Hence the parametric equation of the line will become:- $$ t = \dfrac{t}{4} = \dfrac{t}{7} $$ with $t$ as a parameter.

Answer 2

You can remember that $\vec a\times\vec b$ is perpendicular to both $\vec a$ and $\vec b$, so that $\vec n_0\times\vec n_1$ gives the direction of the intersection line. In addition, you know that both planes pass through $(0,0,0)$ and so does the intersection line.