Question

The lines \( \frac{x + 3}{-2} = \frac{y}{1} = \frac{z - 4}{3} \) and \( \frac{x - 1}{\mu} = \frac{y - 1}{\mu + 1} = \frac{z}{\mu + 2} \) are perpendicular to each other. Then the value of \( \mu \) is:

• A. \( \frac{-5}{3} \)
• B. 3
• C. 4
• D. \( \frac{-1}{4} \)
• E. \( \frac{-7}{2} \)

Hint:

To check for perpendicular lines, set the dot product of the direction ratios equal to zero and solve for \( \mu \).

Answer 1

Answer: (E)

The direction ratios of the lines are given by the coefficients of \( x \), \( y \), and \( z \) in each equation. For the first line \( \frac{x + 3}{-2} = \frac{y}{1} = \frac{z - 4}{3} \), the direction ratios are \( \langle -2, 1, 3 \rangle \). For the second line \( \frac{x - 1}{\mu} = \frac{y - 1}{\mu + 1} = \frac{z}{\mu + 2} \), the direction ratios are \( \langle \mu, \mu + 1, \mu + 2 \rangle \). The condition for perpendicularity of two lines is that their direction ratios should satisfy: \[ \langle -2, 1, 3 \rangle \cdot \langle \mu, \mu + 1, \mu + 2 \rangle = 0 \] This gives: \[ -2\mu + 1(\mu + 1) + 3(\mu + 2) = 0 \] Simplifying this, we get: \[ -2\mu + \mu + 1 + 3\mu + 6 = 0 \quad \Rightarrow \quad 2\mu + 7 = 0 \quad \Rightarrow \quad \mu = \frac{-7}{2} \]