If the shortest distance between the lines $\dfrac{x-\lambda}{3}=\dfrac{y-2}{1}=\dfrac{z-1}{1}$ and $\dfrac{x+2}{3}=\dfrac{y+5}{2}=\dfrac{z-4}{4}$ is $\dfrac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to _____
Step-by-Step Solution
Key Concept: $a_1=(\lambda,2,1)$, $b_1=(3,1,1)$, $a_2=(-2,-5,4)$, $b_2=(3,2,4)$. $\vec{p}=b_1\times b_2=(3,1,1)\times(3,2,4)=(2,-9,3)$... compute: $(4-2,-3-12,6-3)$... actually $(1\cdot4-1\cdot2)\hat{i}-(3\cdot4-1\cdot3)\hat{j}+(3\cdot2-1\cdot3)\hat{k}=(2,-9,3)$. $|\vec{p}|=\sqrt{4+81+9}=\sqrt{94}$... solution shows $\vec{p}\times\vec{q}=-6\hat{i}-15\hat{j}+3\hat{k}$.
$\lambda=1$ or $\lambda=-43$. Largest $|\lambda|=43$.
Correct Answer: 43