Problem Statement
Find the coordinates of the foot of the perpendicular drawn from the point $P(0, 2, 3)$ to the line
$$\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$$
Also, find the perpendicular distance of the point $P$ from the given line.
Verified Solution & Marking Scheme
Express general point on the line in terms of parameter
Let $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = \lambda$.
A general point $Q$ on the line has coordinates:
$$Q(5\lambda - 3, \; 2\lambda + 1, \; 3\lambda - 4$$
Find direction ratios of vector PQ
Vector $\vec{PQ} = (5\lambda - 3 - 0)\hat{i} + (2\lambda + 1 - 2)\hat{j} + (3\lambda - 4 - 3)\hat{k} = (5\lambda - 3)\hat{i} + (2\lambda - 1)\hat{j} + (3\lambda - 7)\hat{k}$.
Apply perpendicularity condition with line direction vector
The direction ratios of the line are $\vec{b} = 5\hat{i} + 2\hat{j} + 3\hat{k}$. Since $PQ \perp \vec{b}$, $\vec{PQ} \cdot \vec{b} = 0$:
$$5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0$$
$$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0 \implies 38\lambda - 38 = 0 \implies \lambda = 1$$
Compute coordinates of foot and distance
Substituting $\lambda = 1$ into $Q$ gives the foot of the perpendicular:
$$F(5(1) - 3, \; 2(1) + 1, \; 3(1) - 4) = F(2, 3, -1)$$
Distance $PF = \sqrt{(2 - 0)^2 + (3 - 2)^2 + (-1 - 3)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}\text{ units}$.