Question

If \( P(2, \beta, \alpha) \) lies on the plane \( x + 2y - z - 2 = 0 \) and \( Q (\alpha, -1, \beta) \) lies on the plane \( 2x - y + 3z + 6 = 0 \), then the direction cosines of the line \( PQ \) are:

• A. \( \left( \frac{-4}{\sqrt{17}}, 0, \frac{-1}{\sqrt{17}} \right) \)
• B. \( \left( \frac{4}{\sqrt{17}}, 0, \frac{1}{\sqrt{17}} \right) \)
• C. \( \left( \frac{1}{\sqrt{17}}, 0, \frac{-4}{\sqrt{17}} \right) \)
• D. \( \left( \frac{-1}{\sqrt{17}}, 0, \frac{4}{\sqrt{17}} \right) \)

Hint:

To find the direction cosines of a line joining two points, compute the direction ratios first and then divide each by the magnitude.

Answer 1

The Correct Option is A

Step 1: Finding the coordinates of \( P \) and \( Q \) 
Since \( P(2, \beta, \alpha) \) lies on the plane: \[ x + 2y - z - 2 = 0 \] Substituting \( P(2, \beta, \alpha) \): \[ 2 + 2\beta - \alpha - 2 = 0 \] \[ 2\beta - \alpha = 0 \quad \Rightarrow \quad \alpha = 2\beta. \] Similarly, for \( Q (\alpha, -1, \beta) \) on the plane: \[ 2x - y + 3z + 6 = 0 \] Substituting \( Q (\alpha, -1, \beta) \): \[ 2\alpha - (-1) + 3\beta + 6 = 0 \] \[ 2\alpha + 3\beta + 7 = 0. \] Step 2: Solving for \( \alpha \) and \( \beta \) 
Substituting \( \alpha = 2\beta \) in \( 2\alpha + 3\beta + 7 = 0 \): \[ 2(2\beta) + 3\beta + 7 = 0 \] \[ 4\beta + 3\beta + 7 = 0 \] \[ 7\beta = -7 \quad \Rightarrow \quad \beta = -1. \] \[ \alpha = 2(-1) = -2. \] Step 3: Finding the direction ratios of \( PQ \) 
\[ PQ = (\alpha - 2, -1 - \beta, \beta - \alpha) \] \[ = (-2 - 2, -1 - (-1), -1 - (-2)) \] \[ = (-4, 0, -1). \] Step 4: Finding the direction cosines 
\[ {Magnitude} = \sqrt{(-4)^2 + (0)^2 + (-1)^2} = \sqrt{16 + 0 + 1} = \sqrt{17}. \] \[ \text{Direction cosines} = \left( \frac{-4}{\sqrt{17}}, 0, \frac{-1}{\sqrt{17}} \right). \] Thus, the correct answer is: \[ \left( \frac{-4}{\sqrt{17}}, 0, \frac{-1}{\sqrt{17}} \right). \]