[Math] How to find an equation of a plane perpendicular to two another planes

analytic geometrylinear algebravectors

Find an equation of a plane $\pi$ that passes through point $P = (2, 3, −6)$ and is perpendicular to two planes $\pi_{1} : x + y + z − 5 = 0$ and $\pi_{2} : x − y + 2 = 0$.

Can someone help me with that with my example? I did not find any information for this on the internet.

Best Answer

The vector $\vec{n}_1$ perpendicular to $\pi_1$ is equal to:

$$\vec{n}_1=\vec i+\vec j+\vec k$$

The vector $\vec{n}_2$ perpendicular to $\pi_2$ is equal to:

$$\vec{n}_1=\vec i-\vec j$$

The vector perpendicular to both vectors can be obtained from the following expression:

$$\vec n = \vec n_1 \times \vec n_2=\vec i +\vec j - 2\vec k$$

So the equation of the plane perpendicular to $\pi_1,\pi_2$ and passing through $P$ is:

$$(x-2)+(y-3)-2(z+6)=0$$