Question

The lines \[ \frac{1 - x}{2} = \frac{y - 1}{3} = \frac{z}{1} \] and \[ \frac{2x - 3}{2p} = \frac{y - 1}{1} = \frac{z - 4}{7} \] are perpendicular to each other for \( p \) equal to:

• A. \( -\frac12 \)
• B. \( \frac12 \)
• C. \( 2 \)
• D. \( 3 \)

Hint:

To check if two lines are perpendicular, calculate the dot product of their direction ratios. If the dot product is zero, the lines are perpendicular. Always express the lines in symmetric or parametric form to extract direction ratios easily.

Answer 1

The Correct Option is C

Step 1: Represent the lines in symmetric form:
For the first line (\( L_1 \)), the symmetric equation is given as: \[ \frac{1 - x}{2} = \frac{y - 1}{3} = \frac{z}{1} \] Rewriting this in parametric form: \[ x = 1 - 2t, \quad y = 1 + 3t, \quad z = t \] The direction ratios of \( L_1 \) are: \[ a_1 = -2, \quad b_1 = 3, \quad c_1 = 1 \]
For the second line (\( L_2 \)), the symmetric equation is given as: \[ \frac{2x - 3}{2p} = \frac{y - 1}{1} = \frac{z - 4}{7} \] Rewriting this in parametric form: \[ x = \frac{3}{2} + pt, \quad y = 1 - t, \quad z = 4 + 7t \] The direction ratios of \( L_2 \) are: \[ a_2 = p, \quad b_2 = -1, \quad c_2 = 7 \]
Step 2: Apply the condition for perpendicularity: Two lines are perpendicular if the dot product of their direction ratios is zero: \[ a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \] Substituting the direction ratios of \( L_1 \) and \( L_2 \): \[ (-2)(p) + (3)(-1) + (1)(7) = 0 \] Simplifying: \[ -2p - 3 + 7 = 0 \] \[ -2p + 4 = 0 \] \[ p = 2 \]
Step 3: Verify the result: For \( p = 2 \), the direction ratios of \( L_2 \) become: \[ a_2 = 2, \quad b_2 = -1, \quad c_2 = 7 \] The dot product with \( L_1 \) is: \[ (-2)(2) + (3)(-1) + (1)(7) = -4 - 3 + 7 = 0 \] Thus, the lines are perpendicular.
Conclusion: The value of \( p \) is \( \mathbf{2} \).