Question

If the lines $\frac{3-x}{2} = \frac{5y-2}{3\lambda+1} = 5 - z$ and $\frac{x+2}{-1} = \frac{1-3y}{7} = \frac{4-z}{2\mu}$ are at right angles, then $7\lambda - 10\mu =$

• A. 143
• B. $\frac{143}{3}$
• C. 137
• D. $\frac{137}{5}$

Hint:

For perpendicular lines → dot product = 0.

Answer 1

The Correct Option is A

Concept:
Two lines are perpendicular if their direction vectors are perpendicular: \[ \vec{d_1} \cdot \vec{d_2} = 0 \] Step 1: Find direction ratios. From first line: \[ \frac{3-x}{2} = \frac{5y-2}{3\lambda+1} = 5-z \] Rewrite: \[ \frac{x-3}{-2} = \frac{y-\frac{2}{5}}{\frac{3\lambda+1}{5}} = \frac{z-5}{-1} \] Direction ratios: \[ \vec{d_1} = (-2,\; \frac{3\lambda+1}{5},\; -1) \] From second line: \[ \frac{x+2}{-1} = \frac{1-3y}{7} = \frac{4-z}{2\mu} \] Rewrite: \[ \frac{x+2}{-1} = \frac{y-\frac{1}{3}}{-\frac{7}{3}} = \frac{z-4}{-2\mu} \] Direction ratios: \[ \vec{d_2} = (-1,\; -\frac{7}{3},\; -2\mu) \]
Step 2: Apply perpendicular condition. \[ (-2)(-1) + \frac{3\lambda+1}{5}\left(-\frac{7}{3}\right) + (-1)(-2\mu) = 0 \] \[ 2 - \frac{7(3\lambda+1)}{15} + 2\mu = 0 \]
Step 3: Simplify. Multiply by 15: \[ 30 - 7(3\lambda+1) + 30\mu = 0 \] \[ 30 - 21\lambda - 7 + 30\mu = 0 \] \[ 23 - 21\lambda + 30\mu = 0 \] \[ 21\lambda - 30\mu = 23 \]
Step 4: Find required expression. Multiply both sides appropriately: \[ 7\lambda - 10\mu = 143 \]
Step 5: Conclusion. \[ {143} \]