How to Find an Equation of a Plane Containing Point Q and Perpendicular to Plane P

Question

Given point $Q(1, 0, -1)$ and plane $P: 2x + 3y - 2z = 6$, how do I find an equation of a plane that contains point $Q$ and is perpendicular to plane $P$?

Answer 1

$Q(1,0,-1);\;P:2x + 3y - 2z = 6$

The equation of a plane is $\pi:ax+by+cz=d$

$\pi\perp P\to (2,3,-2)\cdot (a,b,c)=0\to 2a+3b-2c=0$

Furthermore $Q\in\pi\to a\cdot 1+b\cdot 0+(-1)\cdot c=d\to a-c=d$

Thus we have the system $$ \begin{cases} 2a+3b-2c=0\\ a-c=d\\ \end{cases} $$ $$ \begin{cases} a-c=-\frac{3b}{2}\\ a-c=d\\ \end{cases} $$ the system has solutions only if $d=-\frac{3b}{2},\,c=a+\frac{3b}{2}$

$2 a x+2 b y+(2a+3 b) z+3 b=0$ for instance $$2 x+2 y+5 z+3=0$$

Answer 2

With $p = (x,y,z),\vec n_1 = (2,3,-2)$ we have

$$ P:2x+3y-2z=6\equiv p\cdot \vec n_1 = 6 $$

The orthogonal plane will accomplish

$$ \cases{ (p-Q)\cdot\vec n_2 = 0\\ \vec n_1\cdot \vec n_2 = 0\\ \|\vec n_2\| = 1 } $$

so there are infinite planes ...

Answer 3

Any plane satisfying the requirement can rotate around the line containing $Q(1,0,-1)$ and perpendicular to the plane $2x + 3y - 2z = 6$, while still satisfying the requirement.

So, let the plane cut the $x$-axis at $A(a,0,0)$ where $a$ serves as a parameter. Then, its normal vector is $$\vec n=\vec{AQ}\times (2,3,-2)=(1-a,0,-1)\times (2,3,-2)=(3,-2a,3-3a) $$

and the plane satisfying the requirement is $\vec n\cdot(x-1, y, z+1)=0$. Simply to obtain its equation parameterized with $a$ $$x-\frac{2a}3 y+(1-a)z=a$$