If the lines $\frac{1-x}{3}=\frac{y-2}{2λ}=\frac{z-3}{2}$ and $\frac{x-1}{3λ}=\frac{y-1}{1}=\frac{6-z}{5}$ are perpendicular, then $λ$ is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{-10}{7}$
The given lines are:
$\frac{1-x}{3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}$ and $\frac{x-1}{3\lambda}=\frac{y-1}{1}=\frac{6-z}{5}$
For the first line, the direction ratios are obtained from the denominators:
$\Rightarrow$ direction ratios of line 1: $(-3,\;2\lambda,\;2)$
For the second line, rewriting $\frac{6-z}{5}$ as $\frac{z-6}{-5}$ gives direction ratios:
$\Rightarrow$ direction ratios of line 2: $(3\lambda,\;1,\;-5)$
Since the lines are perpendicular, their dot product is zero:
$(-3)(3\lambda)+(2\lambda)(1)+(2)(-5)=0$
$-9\lambda+2\lambda-10=0$
$-7\lambda-10=0$
Hence,
$\lambda=-\frac{10}{7}$