Question

A straight line passing through \( (6,1,3) \) meets the line \( \frac{x-1}{2} = \frac{y}{1} = \frac{z-2}{3} \) at \( Q \). If the lines are perpendicular to each other, then the coordinates of \( Q \) are

• A. \( (2,1,3) \)
• B. \( (1,2,3) \)
• C. \( (3,1,5) \)
• D. \( (2,-1,3) \)
• E. \( (-1,2,3) \)

Hint:

Use parameter form and dot product = 0 for perpendicular conditions.

Answer 1

The Correct Option is C

Concept: For perpendicular lines: \[ \vec{PQ} \cdot \text{direction vector} = 0 \]

Step 1: Parametrize given line.
\[ x = 1 + 2t,\quad y = t,\quad z = 2 + 3t \]

Step 2: Coordinates of \( Q \).
\[ Q = (1+2t, t, 2+3t) \]

Step 3: Form vector.
\[ \vec{PQ} = (1+2t-6, t-1, 2+3t-3) = (-5+2t, t-1, -1+3t) \]

Step 4: Apply perpendicular condition.
\[ (-5+2t, t-1, -1+3t) \cdot (2,1,3) = 0 \] \[ -10 + 4t + t -1 -3 + 9t = 0 \] \[ 14t -14 = 0 \Rightarrow t = 1 \]

Step 5: Find coordinates.
\[ Q = (3,1,5) \]