Question

Find the equation to the line through $(-1,3,2)$ and perpendicular to the plane $x+2y+2z=3$, then the length of perpendicular and the coordinates of its foot.

Answer 1

Acc. to given equation of plane ,dr's of normal is (1,2,2) .

Let P be (-1,3,2) and Q be required point .

Equation of plane from point P and having dr's (1,2,2) is

$\frac{x_1+1}{1}=\frac{y_1-3}{2}=\frac{z_1-2}{2}=\lambda$

$\Rightarrow x_1=\lambda -1,y_1=2\lambda +3,z_1=2\lambda +2$

for some value of $\lambda$ it lies on given plane s.t.

$(\lambda -1)+2(2\lambda +3)+2(2\lambda +2)=3$

$\Rightarrow \lambda -1+4\lambda +6+4\lambda +4=3$

$\Rightarrow \lambda =\frac{-2}{3}$

$\Rightarrow d{r}^{\prime }sofPQ\left(\right(\frac{-5}{3}+1),(\frac{5}{3}-3),(\frac{2}{3}-2\left)\right)\Rightarrow (\frac{2}{3},\frac{-4}{3},\frac{-4}{3})$

$\Rightarrow length=\sqrt{(\frac{2}{3}{)}^2+(\frac{-4}{3}{)}^2+(\frac{-4}{3}{)}^2}$

$=\frac{6}{3}=2units$