Question

Let a straight line \( L \) pass through the point \(P(2, -1, 3)\) and be perpendicular to the lines \[ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} \quad \text{and} \quad \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}. \] If the line \(L\) intersects the yz-plane at the point Q, then the distance between the points P and Q is:

• A. 2
• B. \(\sqrt{10}\)
• C. 3
• D. \(2\sqrt{3}\)

Hint:

For intersection points on coordinate planes, set the appropriate coordinate (e.g., \(x = 0\) for yz-plane) to simplify calculations.

Answer 1

The Correct Option is C

Step 1: Finding the direction vector for the required line. Using the cross product of direction vectors: \[ \mathbf{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 1 & -2
1 & 3 & 4 \end{vmatrix} = 10\hat{i} - 10\hat{j} + 5\hat{k} = 5(2\hat{i} - 2\hat{j} + \hat{k}) \]

Step 2: Equation of the line. The parametric form is: \[ \frac{x - 2}{2} = \frac{y + 1}{-2} = \frac{z - 3}{1} = \lambda \]

Step 3: Finding intersection with yz-plane. In the yz-plane, \(x = 0\). Setting \(x = 0\) in the line equation: \[ \frac{0 - 2}{2} = \lambda \implies \lambda = -1 \] Substituting \(\lambda = -1\) into the parametric equations: \[ Q(0, 0, 2) \]

Step 4: Calculating the distance. \[ d(P, Q) = \sqrt{4 + 3 + 2} = \sqrt{9} = 3 \]