Question

Find the value of $\lambda$ if the following lines are perpendicular to each other:
\[ l_1: \frac{1 - x}{-3} = \frac{3y - 2}{2\lambda} = \frac{z - 3}{3}, \quad l_2: \frac{x - 1}{3\lambda} = \frac{1 - y}{1} = \frac{2z - 5}{3} \]

Hint:

To check perpendicularity of two lines in 3D, take the dot product of their direction vectors. If it equals zero, the lines are perpendicular.

Answer 1

To determine if two lines are perpendicular, their direction vectors are compared. Perpendicularity is confirmed if the dot product of these vectors is zero. For line $l_1$, the direction ratios (DRs) are given by: \[ \vec{d_1} = \langle -3, 2\lambda, 3 \rangle \] For line $l_2$, the direction ratios are: \[ \vec{d_2} = \langle 3\lambda, -1, 3 \rangle \] The condition for perpendicularity is: \[ \vec{d_1} \cdot \vec{d_2} = 0 \] Substituting the DRs: \[ (-3)(3\lambda) + (2\lambda)(-1) + (3)(3) = 0 \] This simplifies to: \[ -9\lambda - 2\lambda + 9 = 0 \] \[ -11\lambda + 9 = 0 \] Solving for $\lambda$: \[ \lambda = \frac{9}{11} \] Re-evaluation of the direction vectors based on their denominators in the symmetric form of the line equations yields the same DRs: From $l_1$: DRs = $\langle -3, 2\lambda, 3 \rangle$
From $l_2$: DRs = $\langle 3\lambda, -1, 3 \rangle$
The dot product calculation remains: \[ (-3)(3\lambda) + (2\lambda)(-1) + (3)(3) = 0\\ -9\lambda - 2\lambda + 9 = 0\\ -11\lambda + 9 = 0\\ \Rightarrow \lambda = \frac{9}{11} \] Thus, the correct value is: \[ \boxed{\lambda = \frac{9}{11}} \]