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

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Point $Q$ lies on the given line, so its coordinates can be expressed in terms of a scalar parameter $r$. Since the line joining $(6, 1, 3)$ and $Q$ is perpendicular to the given line, the dot product of their direction vectors must be zero.
Step 2: Expressing Q and the Direction Vector
1. Let the given line be $L: \frac{x-1}{2} = \frac{y}{1} = \frac{z-2}{3} = r$. 2. Coordinates of any point $Q$ on $L$: $(2r+1, r, 3r+2)$. 3. Let point $P$ be $(6, 1, 3)$. The direction vector of $PQ$ is: $\vec{d_{PQ}} = (2r+1-6, r-1, 3r+2-3) = (2r-5, r-1, 3r-1)$.
Step 3: Applying Perpendicularity Condition
The direction vector of the given line $L$ is $\vec{d_L} = (2, 1, 3)$. Since $PQ \perp L$: \[ (2r-5)(2) + (r-1)(1) + (3r-1)(3) = 0 \] \[ 4r - 10 + r - 1 + 9r - 3 = 0 \] \[ 14r - 14 = 0 \implies r = 1 \] Substituting $r=1$ into the coordinates of $Q$: $Q = (2(1)+1, 1, 3(1)+2) = (3, 1, 5)$.
Step 4: Final Answer
The coordinates of $Q$ are (3, 1, 5).