Question

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

• A. \( -\frac{1}{2} \)
• B. \( \frac{1}{2} \)
• 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: Convert lines to symmetric and parametric forms. For line \( L_1 \), the symmetric form is \( \frac{1 - x}{2} = \frac{y - 1}{3} = \frac{z}{1} \). The parametric form is \( x = 1 - 2t, \quad y = 1 + 3t, \quad z = t \), with direction ratios \( a_1 = -2, \, b_1 = 3, \, c_1 = 1 \). For line \( L_2 \), the symmetric form is \( \frac{2x - 3}{2p} = \frac{y}{-1} = \frac{z - 4}{7} \). The parametric form is \( x = \frac{3}{2} + pt, \quad y = -t, \quad z = 4 + 7t \), with direction ratios \( a_2 = p, \, b_2 = -1, \, c_2 = 7 \).

Step 2: Apply perpendicularity condition: \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \). Substitute direction ratios: \( (-2)(p) + (3)(-1) + (1)(7) = 0 \). Simplify: \( -2p - 3 + 7 = 0 \), which gives \( -2p + 4 = 0 \), so \( p = 2 \).

Step 3: Verify: With \( p = 2 \), \( L_2 \) direction ratios are \( a_2 = 2, \, b_2 = -1, \, c_2 = 7 \). The dot product is \( (-2)(2) + (3)(-1) + (1)(7) = -4 - 3 + 7 = 0 \). The lines are perpendicular.

Conclusion: The value of \( p \) is \( \mathbf{2} \).