Problem Statement
Find the coordinates of the foot of the perpendicular and the image of the point $P(1, 6, 3)$ in the line:
$\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$
Verified Solution & Marking Scheme
Write General Coordinates of a Point on the Line
Let $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda$.
General point $Q$ on the line is:
$Q = (\lambda, 2\lambda + 1, 3\lambda + 2)$
Determine Direction Ratios of Line Segment PQ
The coordinates of $P$ are $(1, 6, 3)$. Direction ratios of $PQ$ are:
$(\lambda - 1, \, (2\lambda + 1) - 6, \, (3\lambda + 2) - 3) = (\lambda - 1, \, 2\lambda - 5, \, 3\lambda - 1)$
Apply Perpendicularity Condition (Dot Product = 0)
The line has direction ratios $\vec{b} = (1, 2, 3)$. Since $PQ \perp$ line:
$1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) = 0$
$\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$
$14\lambda - 14 = 0 \implies 14\lambda = 14 \implies \lambda = 1$
Compute Foot of Perpendicular and Image Coordinates
Substituting $\lambda = 1$ into $Q$:
$Q = (1, \, 2(1) + 1, \, 3(1) + 2) = (1, 3, 5)$
So the foot of the perpendicular is $Q(1, 3, 5)$.
Let $P'(x', y', z')$ be the image of $P(1, 6, 3)$ in the line. Then $Q(1, 3, 5)$ is the midpoint of $PP'$:
$\frac{1 + x'}{2} = 1 \implies x' = 1$
$\frac{6 + y'}{2} = 3 \implies y' = 0$
$\frac{3 + z'}{2} = 5 \implies z' = 7$
Thus, the image is $P'(1, 0, 7)$.