Question

Find the distance between the point $(2,3,-1)$ and foot or perpendicular drawn from $(3,1,-1)$ to the plane $x-y+3z=10$

Answer 1

Equation of plane-

$x-y+3z=10$

Direction ratio's of plane are- $1,-1,3$

Let the foot of perpendicular $L$ drawn from $P(3,1,-1)$ to the plane $x-y+z=10$ be $(x_1,y_1,z_1)$.

Therefore, the direction ratio's of $PL$ are-

$(x_1-3),(y_1-1)$ and $(z_1+1)$

Since perpendicular to plane is parallel to normal vector.

Direction ratio of plane and $\overrightarrow{PL}$ are proportional.

Therefore,

$\frac{x_1-3}{1}=\frac{y_1-1}{-1}=\frac{z_1+1}{3}=k\left(\text{Let}\right)$

$x_1=k+3$

$y_1=1-k$

$z_1=3k$

Since the point $L$ lie on the plane.

Thus,

$(k+3)-(1-k)+3(3k-1)=10$

$k+3-1+k+9k-3=10$

$11k=10+1$

$\Rightarrow k=\frac{11}{11}=1$

Thus,

$x_1=k+3=1+3=4$

$y_1=1-k=1-1=0$

$z_1=3k-1=3\left(1\right)-1=2$

Now, distance between the point $(2,3,-1)$ and foot of perpendicular $(4,0,2)$ is-

$d=\sqrt{{(4-2)}^2+{(0-3)}^2+{(2-(-1))}^2}$

$\Rightarrow d=\sqrt{4+9+9}=\sqrt{22}$

Hence the distance between the point $(2,3,-1)$ and foot of perpendicular $(4,0,2)$ is $\sqrt{22}$ units.