Question

Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x+2}{4}=\frac{y−1}{2}=\frac{z+1}{3}$ be $P$, Then the distance of $P$ from the plane $3x+4y+12z+23=0$ is

• A. $5$
• B. $\frac{50}{13}$
• C. $4$
• D. $\frac{63}{13}$

Answer 1

Answer: (A)

$5$

Answer 2

$L:$ $\frac{x+2}{4}=\frac{y−1}{2}=\frac{z+1}{3}$

Let

$P=(4t−2,2t+1,3t−1)$

$∵ P$ is the foot of perpendicular of $(1, 2, 4)$

$∴ 4(4t – 3) + 2(2t – 1) + 3(3t – 5) = 0$

$⇒29t=29⇒t=1$

$∴ P = (2, 3, 2)$

Now, distance of $P$ from the plane

$3x \+ 4y \+ 12z \+ 23 = 0$, is

$\begin{vmatrix}\frac{6+12+24+23}{\sqrt{9+16+144}}\end{vmatrix}=\frac{65}{13}=5$