Question

Find the equation of the plane containing line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and point $(0,7,-7)$.

Answer 1

Let: $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}=k$

$L:x=-3k-1,y=2k+3,z=k-2$

$P_1=(0,7,-7)$

Put $k=0,1$ in equation of $L$:

$P_2=(-1,3,-2),P_3=(-4,5,-1)$

Vector from $P_1$ to $P_2$: $\overrightarrow{a}=(\hat{i}+4\hat{j}-5\hat{k})$

Vector from $P_1$ to $P_3$: $\overrightarrow{b}=(4\hat{i}+2\hat{j}-6\hat{k})$

Vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ lie in plane $P$

Equation of normal to plane $P$: $\overrightarrow{n}=\overrightarrow{a}\times \overrightarrow{b}$

$\overrightarrow{n}=\hat{i}+\hat{j}+\hat{k}$

Equation of plane is:

$\overrightarrow{n}.\left(\right(x-x_1)\hat{i}+(y-y_1)\hat{j}+(z-z_1\left)\hat{k}\right)$

or $P=\left(\right(\hat{i}+\hat{j}+\hat{k}).((x-0)\hat{i}+(y-7)\hat{j}+(z+7)\hat{k})$

$\therefore P=x+y+z=0$