If the shortest distance between the lines $\dfrac{x-4}{1}=\dfrac{y+1}{2}=\dfrac{z}{-3}$ and $\dfrac{x-\lambda}{2}=\dfrac{y+1}{4}=\dfrac{z-2}{-5}$ is $\dfrac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is:
Step-by-Step Solution
Key Concept: Lines have direction vectors $(1,2,-3)$ and $(2,4,-5)$. $\vec{b_1}\times\vec{b_2}=(2,−5+6,4−4)=(2,1,0)$... compute carefully. Use SD formula and set equal to $6/\sqrt5$.
$\lambda=1$ or $7$. Sum $=8$.
Correct Answer: 2