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.
On
FIRST APPROACH
The normal vectors to the planes $\pi_{1}$ and $\pi_{2}$ are respectively given by
\begin{cases} \textbf{n}_{1} = (1,1,1)\\ \textbf{n}_{2} = (1,-1,0)\\ \end{cases}
Therefore we can describe $\pi$ as the plane spanned by $\textbf{n} =\textbf{n}_{1}\times\textbf{n}_{2}$ passing through $P$. Precisely:
\begin{align*} X \in \pi \Longleftrightarrow \langle X - P,\textbf{n}\rangle = 0 \end{align*}
Notice that $\pi\perp\pi_{1}$ and $\pi\perp\pi_{2}$, because the normal vector to $\pi$ is given by $\textbf{n}_{1}\times\textbf{n}_{2}$, where $\times$ indicates the cross product.
SECOND APPROACH
In accordance to above-mentioned notation, another way to describe $\pi$ is given by \begin{align*} X = (x,y,z)\in \pi \Longleftrightarrow (x,y,z) = P + \alpha\textbf{n}_{1} + \beta\textbf{n}_{2} \end{align*}
where $(\alpha,\beta)\in\textbf{R}^{2}$. Once again, the normal vector to $\pi$ is given by $\textbf{n} = \textbf{n}_{1}\times\textbf{n}_{2}$, from whence we conclude that $\pi\perp\pi_{1}$ and $\pi\perp\pi_{2}$, given that $\textbf{n}\times\textbf{n}_{1} = \textbf{n}\times\textbf{n}_{2} = \textbf{0}$.
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:
$$(x-2)+(y-3)-2(z+6)=0$$