Question

The value of $\lambda$ for which the lines $\frac{2 - x}{3} = \frac{3 - 4y}{5} = \frac{z - 2}{3}$ and $\frac{x - 2}{-3} = \frac{2y - 4}{3} = \frac{2 - z}{\lambda}$ are perpendicular is:

• A. $-2$
• B. $2$
• C. $\frac{8}{19}$
• D. $\frac{19}{8}$

Answer 1

Answer: (A)

$-2$

Answer 2

To determine when the lines are perpendicular, we need to find the dot product of their direction vectors and set it equal to zero.

For the first line: Direction vector $\vec{d_1} = \langle -3, -4, 3 \rangle$.

For the second line: Direction vector $\vec{d_2} = \langle -3, 3, -\lambda \rangle$.

The dot product of $\vec{d_1}$ and $\vec{d_2}$ is given by:

$\vec{d_1} \cdot \vec{d_2} = (-3)(-3) + (-4)(3) + (3)(-\lambda)$

Simplifying:

$\vec{d_1} \cdot \vec{d_2} = 6 - 12 - 3\lambda$

For the lines to be perpendicular, we require:

$6 - 12 - 3\lambda = 0$

Simplifying further:

$-6 - 3\lambda = 0 \implies -3\lambda = 6 \implies \lambda = -2$

Therefore, the value of $\lambda$ that makes the lines perpendicular is:

$\boxed{\lambda = -2}$