Question

Find the foot of the perpendicular from the point \( (0, 2, 3) \) on the line \( \frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} \) and hence find the length of the perpendicular.

Hint:

Drawing a small sketch of the point and line helps visualize the vector \(\vec{PQ}\) correctly.
This method is more direct and less error-prone than using projection formulas for general lines.

Answer 1

Concept:
• Any point on the line can be represented as a function of parameter \( \lambda \).
• The vector from the given point to the foot of the perpendicular is orthogonal to the line's direction vector.
• Orthogonality condition: Dot product of vectors is zero.

Step 1: Find a general point on the line
Let the foot of the perpendicular be \( Q \). Setting the line equation to \( \lambda \): \[ x = 5\lambda - 3, y = 2\lambda + 1, z = 3\lambda - 4 \] Coordinates of \( Q = (5\lambda-3, 2\lambda+1, 3\lambda-4) \).

Step 2: Find the direction vector \( \vec{PQ} \) and use orthogonality
Given point \( P(0, 2, 3) \). \( \vec{PQ} = (5\lambda-3 - 0)\hat{i} + (2\lambda+1 - 2)\hat{j} + (3\lambda-4 - 3)\hat{k} \) \( \vec{PQ} = (5\lambda-3)\hat{i} + (2\lambda-1)\hat{j} + (3\lambda-7)\hat{k} \). The direction vector of the line is \( \vec{b} = 5\hat{i} + 2\hat{j} + 3\hat{k} \). Since \( \vec{PQ} \perp \vec{b} \), their dot product is zero: \[ 5(5\lambda-3) + 2(2\lambda-1) + 3(3\lambda-7) = 0 \] \[ 25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0 \Rightarrow 38\lambda - 38 = 0 \Rightarrow \lambda = 1 \]

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