Question

If lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are mutually perpendicular, then $k$ is equal to:

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

Answer 1

The Correct Option is A

1. Identify direction vectors:

For the first line: \( \vec{d_1} = (-3, 2k, 2) \)

For the second line: \( \vec{d_2} = (3k, 1, -5) \)

2. Condition for perpendicularity:

Two lines are perpendicular if their direction vectors satisfy \( \vec{d_1} \cdot \vec{d_2} = 0 \).

Compute the dot product:

\[ (-3)(3k) + (2k)(1) + (2)(-5) = -9k + 2k - 10 = -7k - 10 = 0 \]

3. Solve for \( k \):

\[ -7k - 10 = 0 \implies k = -\frac{10}{7} \]

Correct Answer: (A) \( -\frac{10}{7} \)

Answer 2

The direction ratios of the lines are $(-3, 2k, 2)$ and $(3k, 1, -5)$. For the lines to be perpendicular, the dot product of their direction ratios must be zero: \[ -3(3k) + 2k(1) + 2(-5) = 0. \] Simplify: \[ -9k + 2k - 10 = 0 \implies -7k = 10 \implies k = -\frac{10}{7}. \]