The foot of the perpendicular drawn from the point (1, 6, 3) to the l…

The correct answer is Option (3) → (1, 3, 5)

The line is

$\displaystyle \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=t$

So any point on the line is:

$x=t,\; y=1+2t,\; z=2+3t$

Direction vector: $\langle 1,\,2,\,3\rangle$

Point outside the line: $P(1,6,3)$

Foot of perpendicular = point $Q(t,\;1+2t,\;2+3t)$ such that $\overrightarrow{PQ}$ is perpendicular to the line direction.

Compute $\overrightarrow{PQ}$:

$\langle t-1,\;1+2t-6,\;2+3t-3\rangle = \langle t-1,\;2t-5,\;3t-1\rangle$

Dot product with direction $\langle 1,2,3\rangle$ must be $0$:

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

$t-1 + 4t -10 + 9t -3 = 0$

$14t -14 = 0$

$t = 1$

Substitute $t=1$ into line parametric equations:

$x = 1$

$y = 1 + 2\cdot 1 = 3$

$z = 2 + 3\cdot 1 = 5$

Foot of perpendicular = $(1,\;3,\;5)$