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 line equations to parametric form and extract direction ratios. For Line 1 (\( L_1 \)), the symmetric form is \[ \frac{1 - x}{2} = \frac{y - 1}{3} = \frac{z}{1} \]. In parametric form, this is \( x = 1 - 2t, \quad y = 1 + 3t, \quad z = t \). The direction ratios for \( L_1 \) are \( a_1 = -2, \, b_1 = 3, \, c_1 = 1 \). For Line 2 (\( L_2 \)), the symmetric form is \[ \frac{2x - 3}{2p} = \frac{y}{-1} = \frac{z - 4}{7} \]. In parametric form, this is \( x = \frac{3}{2} + pt, \quad y = -t, \quad z = 4 + 7t \). The direction ratios for \( L_2 \) are \( a_2 = p, \, b_2 = -1, \, c_2 = 7 \).

Step 2: Apply the perpendicularity condition. 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: \( (-2)(p) + (3)(-1) + (1)(7) = 0 \). Simplifying this equation yields \( -2p - 3 + 7 = 0 \), which simplifies further to \( -2p + 4 = 0 \). Solving for \( p \) gives \( p = 2 \).

Step 3: Verify the result. With \( p = 2 \), the direction ratios for \( L_2 \) are \( a_2 = 2, \, b_2 = -1, \, c_2 = 7 \). The dot product with \( L_1 \) is \( (-2)(2) + (3)(-1) + (1)(7) = -4 - 3 + 7 = 0 \). This confirms the lines are perpendicular.

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