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 check if two lines are perpendicular, we compare their direction vectors. If the dot product of direction vectors is zero, the lines are perpendicular. From $l_1$, the direction ratios (DRs) are: \[ \vec{d_1} = \langle -3, 2\lambda, 3 \rangle \] From $l_2$, the direction ratios are: \[ \vec{d_2} = \langle 3\lambda, -1, 3 \rangle \] Now, for the lines to be perpendicular: \[ \vec{d_1} \cdot \vec{d_2} = 0 \] \[ (-3)(3\lambda) + (2\lambda)(-1) + (3)(3) = 0 \] \[ -9\lambda - 2\lambda + 9 = 0 \Rightarrow -11\lambda + 9 = 0 \Rightarrow \lambda = \frac{9}{11} \] Wait — correction. On close inspection of the signs: \[ \vec{d_1} = \langle -3, 2\lambda, 3 \rangle,\quad \vec{d_2} = \langle \frac{1}{3\lambda}, -1, \frac{3}{2} \rangle \] But the correct approach is to just treat both in terms of direction vectors derived from denominators: From $l_1$: DRs = $\langle -3, 2\lambda, 3 \rangle$
From $l_2$: DRs = $\langle 3\lambda, -1, 3 \rangle$
Dot product: \[ (-3)(3\lambda) + (2\lambda)(-1) + (3)(3) = 0\\ -9\lambda - 2\lambda + 9 = 0\\ -11\lambda + 9 = 0\\ \Rightarrow \lambda = \frac{9}{11} \] Oops! There’s a misinterpretation in calculation — actually the correct DRs must come directly from the denominators in symmetric form (each line is in symmetric form). So the correct DRs are: - For $l_1$: $\langle -3, 2\lambda, 3 \rangle$
- For $l_2$: $\langle 3\lambda, -1, 3 \rangle$
Dot product: \[ (-3)(3\lambda) + (2\lambda)(-1) + (3)(3) = 0\\ -9\lambda - 2\lambda + 9 = 0\\ -11\lambda + 9 = 0\\ \Rightarrow \lambda = \frac{9}{11} \] Final boxed value: \[ \boxed{\lambda = \frac{9}{11}} \]