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 vector is given by the denominators in the equation of the line.

• Line 1: $ \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} $. Direction vector: $ \mathbf{v_1} = \langle -3, 2k, 2 \rangle $.
• Line 2: $ \frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5} $. Direction vector: $ \mathbf{v_2} = \langle 3k, 1, -5 \rangle $.

Two lines are mutually perpendicular if their direction vectors are orthogonal (their dot product is zero):

$$ \mathbf{v_1} \cdot \mathbf{v_2} = 0 $$

Compute the dot product:

$$ (-3)(3k) + (2k)(1) + (2)(-5) = 0 $$ $$ -9k + 2k - 10 = 0 $$ $$ -7k = 10 $$ $$ k = -\frac{10}{7} $$

Therefore, if the lines are mutually perpendicular, $ k = -\frac{10}{7} $.