Thank you for reporting, we will resolve it shortly
Q.
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
Introduction to Three Dimensional Geometry
Solution:
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 $\bot$ 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),