Let the foot of the perpendicular from the point (1,2,4) on the line x+24=y-12=z+13 be P.

Question

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

Accepted Answer

$$L:{{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3} = t$$

Let P = (4t - 2, 2t + 1, 3t - 1)

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

$$\therefore$$ $4(4t - 3) + 2(2t - 1) + 3(3t - 5) = 0$

$$ \Rightarrow 29t = 29 \Rightarrow t = 1$$

$$\therefore$$ P = (2, 3, 2)

Now, distance of P from the plane

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

$$\left| {{{6 + 12 + 24 + 23} \over {\sqrt {9 + 16 + 144} }}} \right| = {{65} \over {13}} = 5$$

Solutions · 1