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 gives: \[ Q(0, 0, 2) \]

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

Answer 2

Step 1: Given data.
The line \( L \) passes through the point \( P(2, -1, 3) \) and is perpendicular to the following two lines:
\[ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} \quad \text{and} \quad \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}. \]

Step 2: Find direction ratios (d.r.) of the given lines.
For the first line: \[ \text{Direction ratios} = (2, 1, -2). \] For the second line: \[ \text{Direction ratios} = (1, 3, 4). \] The line \( L \) is perpendicular to both these lines, so its direction ratios are given by the cross product of these two direction vectors.

Step 3: Find the cross product.
\[ \vec{d_L} = (2, 1, -2) \times (1, 3, 4) \] Using the determinant form:
\[ \vec{d_L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = \hat{i}(1 \times 4 - (-2) \times 3) - \hat{j}(2 \times 4 - (-2) \times 1) + \hat{k}(2 \times 3 - 1 \times 1) \] \[ \vec{d_L} = \hat{i}(4 + 6) - \hat{j}(8 + 2) + \hat{k}(6 - 1) \] \[ \vec{d_L} = 10\hat{i} - 10\hat{j} + 5\hat{k} \] So the direction ratios of line \( L \) are \( (10, -10, 5) \).

Step 4: Equation of line \( L \).
The line passes through \( P(2, -1, 3) \) and has direction ratios \( (10, -10, 5) \), so:
\[ \frac{x - 2}{10} = \frac{y + 1}{-10} = \frac{z - 3}{5} = t \] Hence, the parametric equations are:
\[ x = 2 + 10t, \quad y = -1 - 10t, \quad z = 3 + 5t. \]

Step 5: Find the point Q where the line intersects the yz-plane.
At the yz-plane, \( x = 0 \).
Substitute \( x = 0 \) in \( x = 2 + 10t \):
\[ 0 = 2 + 10t \Rightarrow t = -\frac{1}{5}. \] Substitute this value of \( t \) in \( y \) and \( z \):
\[ y = -1 - 10\left(-\frac{1}{5}\right) = -1 + 2 = 1, \] \[ z = 3 + 5\left(-\frac{1}{5}\right) = 3 - 1 = 2. \] Hence, \( Q(0, 1, 2) \).

Step 6: Distance between P and Q.
Using the distance formula:
\[ PQ = \sqrt{(2 - 0)^2 + (-1 - 1)^2 + (3 - 2)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3. \]

Final Answer:
\[ \boxed{3} \]