Question

The foot of the perpendicular from $(0,2,3)$ to the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ is

• A. (-2, 3, 4)
• B. (2, -1, 3)
• C. (2, 3, -1)
• D. (3, 2, -1)

Answer 1

Answer: (C)

Let N be the foot of the perpendicular from the point (0, 2, 3) on the given line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ ...(1)
Any pt. on (1) is $(5r-3,2r+1,3r-4)$. If this point is N, then direction numbers of NP are
$<5r-3-0,2r+1-2,3r-4-3>$
$i.e.,<5r-3,2r-1,3r-7>$. Since PN is $\perp$ to the given line.
$\therefore$ $5(5r-3)+2(2r-1)+3(3r-7)$ = 0
$\Rightarrow$ $38r-38=0\Rightarrow r=1$
$\therefore$ N is (5 - 3, 2 + 1, 3 - 4) $i.e..$ (2, 3, -1),