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.

  • $\begingroup$ This is just a minor variation on the problems in the two questions you’re already asked. The only difference is how the vectors parallel to the plane you’re supposed to solve are described. It appears that you’re missing some fundamental understanding of how to solve these sorts of problems. I’d suggest reviewing the course material and, if you’re still having trouble after that, talking to your instructor or a TA. $\endgroup$ – amd Jan 14 '19 at 21:39

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:


  • $\begingroup$ Thanks a lot, it helped $\endgroup$ – Aliaksei Klimovich Jan 14 '19 at 22:35

Hint...find the cross product of the two normal vectors and this is the normal for the third plane.



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.


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 $\textbf{n}\times\textbf{n}_{1} = \textbf{n}\times\textbf{n}_{2} = \textbf{0}$.


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