Question

Find the equation of the plane passing through $(2,2,-2)$ and $(-2,-2,2)$ and perpendicular to the plane $2x-3y+z-7=0$.

Answer 1

Suppose required plane is $ax+by+ca+d=0$
$\therefore \overline{n}=(a,b,c)$ is normal to the plane
Required plane is perpendicular to $2x-3y+z-7=0$
$\therefore \overline{n}\cdot (2,3,1)=0$ $\left(1\right)$
Now $A(2,2,-2)$ and $B(-2,-2,2)$ are the points of plane.
$\therefore \overrightarrow{AB}=B-A=(-4,-4,4)$
$\therefore \overline{n}\cdot \overrightarrow{AB}=0$
$\therefore \overline{n}\cdot (-4,-4,4)=0$
$\therefore \overline{n}\cdot (-1,-1,1)=0$ .$\left(2\right)$

From result $\left(1\right)$ and $\left(2\right)$ we have
$\therefore \overline{n}=(2,-3,1)\times (-1,-1,1)$
$=\left|\begin{array}{ccc}i& j& k\\ 2& -3& 1\\ -1& -1& 1\end{array}\right|$
$=i(-3+1)-j(2+1)+k(-2,-2)$
$=-2\hat{i}-3\hat{j}-5\hat{k}$
$\therefore \overline{n}=(-2,-3,-5)$ OR
$\overline{n}=(2,3,5)$

If plane passes from point $A\left(\overline{a}\right)=(2,2,-1)$
Then we get its equation using $\overline{r}\cdot \overline{n}=\overline{a}\cdot \overline{n}$
$(x,y,z)\cdot (2,3,5)=(2,2,-2)\cdot (2,3,5)$
$\therefore 2x+3y+5z=4+6-10$
$\therefore 2x+3y+5z=0$
Which is required equation of plane.