Question

If the lines \(\frac{x-1}{-3} = \frac{y-2}{2 \lambda} = \frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda} = \frac{y-1}{2} = \frac{z-6}{-5}\) are perpendicular. Then the value of \(\lambda\) is:

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

Answer 1

The Correct Option is B

To determine the value of \(\lambda\) for which the given lines are perpendicular, we must analyze their direction ratios. The direction ratios for the first line \(\frac{x-1}{-3} = \frac{y-2}{2 \lambda} = \frac{z-3}{2}\) are \((-3, 2\lambda, 2)\). For the second line \(\frac{x-1}{3 \lambda} = \frac{y-1}{2} = \frac{z-6}{-5}\), the direction ratios are \((3\lambda, 2, -5)\).

For two lines to be perpendicular, the dot product of their direction vectors must be zero. Thus, we set up the dot product: \((-3) \times (3\lambda) + (2\lambda) \times 2 + 2 \times (-5) = 0\).

Simplifying, we have: \(-9\lambda + 4\lambda - 10 = 0\)

Combine like terms to get: \(-5\lambda - 10 = 0\)

Solve for \(\lambda\): \(-5\lambda = 10 \Rightarrow \lambda = -2\)

Therefore, the value of \(\lambda\) for which the lines are perpendicular is -2.