Problem Statement
Find the foot of perpendicular and distance from $P(0, 2, 3)$ to line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$. Also find image of $P$.
Verified Solution & Marking Scheme
General Point Q
$Q(5\lambda - 3, 2\lambda + 1, 3\lambda - 4)$.
Dot Product with Direction Vector
$5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0 \implies 38\lambda - 38 = 0 \implies \lambda = 1$
Foot, Distance, Image
Foot: $Q(2, 3, -1)$. Distance: $\sqrt{4 + 1 + 16} = \sqrt{21}$.
Image $P'(x_1, y_1, z_1)$: $\frac{x_1 + 0}{2} = 2 \implies x_1 = 4$; $\frac{y_1 + 2}{2} = 3 \implies y_1 = 4$; $\frac{z_1 + 3}{2} = -1 \implies z_1 = -5$. Image is $(4, 4, -5)$.