Question

The value of $ k $ so that $ \frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2} = \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5} $ may be perpendicular is given by

• A. \( -10 \)
• B. \( \frac{10}{7} \)
• C. \( \frac{-10}{7} \)
• D. \( \frac{-7}{10} \)

Hint:

To solve for the value of \( k \) in problems involving direction ratios, use the condition for perpendicular vectors: their dot product must be zero.

Answer 1

The Correct Option is C

We are given a system of equations and asked to find the value of \( k \) such that the vectors are perpendicular.
Step 1: Interpret the equations as direction ratios
The general form of the direction ratios for a line is \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), where \( (a, b, c) \) is the direction vector.
From the given equations: The direction ratios of the line are \( \left( \frac{1}{-3}, \frac{1}{2k}, \frac{1}{2} \right) \).

The direction ratios for the second line are \( \left( \frac{1}{3k}, \frac{1}{1}, \frac{1}{-5} \right) \).


Step 2: Perpendicular condition
For two lines to be perpendicular, their direction ratios must satisfy the condition: \[ \vec{A} \cdot \vec{B} = 0 \] where \( \vec{A} \) and \( \vec{B} \) are the direction vectors of the two lines. Let the direction vectors be: \[ \vec{A} = \left( -\frac{1}{3}, \frac{1}{2k}, \frac{1}{2} \right), \quad \vec{B} = \left( \frac{1}{3k}, 1, -\frac{1}{5} \right) \] The dot product is: \[ \left( -\frac{1}{3} \times \frac{1}{3k} \right) + \left( \frac{1}{2k} \times 1 \right) + \left( \frac{1}{2} \times -\frac{1}{5} \right) = 0 \] Simplifying the equation: \[ -\frac{1}{9k} + \frac{1}{2k} - \frac{1}{10} = 0 \]
Step 3: Solve for \( k \)
Multiplying the equation by 90k to eliminate the denominators: \[ -10 + 45 - 9k = 0 \] \[ 35 = 9k \] Thus, \( k = \frac{-10}{7} \).
Step 4: Conclusion
The value of \( k \) is \( \frac{-10}{7} \), corresponding to option (c).