Question

Find the foot of the perpendicular from the point \( (0, 2, 3) \) on the line \( \frac{-x-3}{-5} = \frac{1-y}{-2} = \frac{3z+12}{9} \) and hence find the length of the perpendicular.

Hint:

Double-check your standard form conversion. The coefficients of \( x, y, z \) must always be \( +1 \). Here, signs and constants needed adjustment before extracting direction ratios.

Answer 1

Concept:
• Convert the line to standard form and write general coordinates for any point on the line in terms of a parameter \( \lambda \).
• The vector from the given point to the foot of the perpendicular must be orthogonal (perpendicular) to the line's direction vector.
• Solve for \( \lambda \) and calculate the distance between the points.

Step 1: Standardize the line equation and find a general point
Rewrite the line: \( \frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = \lambda \).
A general point \( P \) on the line is \( (5\lambda - 3, 2\lambda + 1, 3\lambda - 4) \).
The given point is \( Q(0, 2, 3) \).

Step 2: Set up the perpendicularity condition
Direction vector of the line: \( \vec{b} = 5\hat{i} + 2\hat{j} + 3\hat{k} \).
Vector \( \vec{QP} = (5\lambda - 3)\hat{i} + (2\lambda - 1)\hat{j} + (3\lambda - 7)\hat{k} \).
Since \( \vec{QP} \perp \vec{b} \), their dot product is zero:
\[ 5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0 \]
\[ 25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0 \implies 38\lambda = 38 \implies \lambda = 1 \]

Step 3: Find the foot and the length
Substituting \( \lambda = 1 \) into \( P \):
Foot of perpendicular = \( (5(1)-3, 2(1)+1, 3(1)-4) = (2, 3, -1) \).
Length of perpendicular = \( \sqrt{(2-0)^2 + (3-2)^2 + (-1-3)^2} \)
\[ = \sqrt{4 + 1 + 16} = \sqrt{21} \text{ units.} \]