Question

Find the equation of plane through the points $(2,2,1)$ and $(9,3,6)$ and perpendicular to the plane $2x+6y+6z=9$

Answer 1

As the required plane is perpendicular to $2x+6y+6z=9$, the normal vector $\overrightarrow{n_1}$ of plane $2x+6y+6z=9$ lies in the required plane.

The normal vector $\overrightarrow{n}$ of the required plane can be found by taking cross product of vector of given points with $\overrightarrow{n_1}$.

$\therefore \overrightarrow{n}=\overrightarrow{n_1}\times (2\hat{i}+2\hat{j}+\hat{k}-(9\hat{i}+3\hat{j}+6\hat{k}\left)\right)$

$\therefore \overrightarrow{n}=(2\hat{i}+6\hat{j}+6\hat{k})\times (-7\hat{i}-\hat{j}-5\hat{k})$

$\therefore \overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 6& 6\\ -7& -1& -5\end{array}\right|$

$\therefore \overrightarrow{n}=(-30+6)\hat{i}-(-10+42)\hat{j}+(-2+42)\hat{k}$

$\therefore \overrightarrow{n}=-24\hat{i}-32\hat{j}-40\hat{k}$

Now, equation of required plane is

$\overrightarrow{n}\cdot \overrightarrow{r}=\overrightarrow{n}\cdot (2\hat{i}+2\hat{j}+\hat{k})$

$\therefore (-24\hat{i}-32\hat{j}-40\hat{k})\cdot (x\hat{i}+y\hat{j}+z\hat{k})=(-24\hat{i}-32\hat{j}-40\hat{k})\cdot (2\hat{i}+2\hat{j}+\hat{k})$

$\therefore -24x-32y-40z=-48-64-40$

$\therefore 3x+4y+5z=19$

This is the required equation.