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

The Correct Option is A

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:

\[ \boxed{\lambda = -2} \]