Let the shortest distance between the lines L_1: x-33=y--1=z-31 and L_2: x+3-3=y+72=z-4 be…

Question

Let the shortest distance between the lines $L_1:\ \dfrac{x-3}{3}=\dfrac{y-\alpha}{-1}=\dfrac{z-3}{1}$ and $L_2:\ \dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-\beta}{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha+\beta$ is:

Accepted Answer

$\vec a_1=(3,\alpha,3),\ \vec a_2=(-3,-7,\beta)$; $\vec b_1=(3,-1,1),\ \vec b_2=(-3,2,4)$.
$$\vec b_1\times\vec b_2=(-6,-15,3),\qquad |\vec b_1\times\vec b_2|=\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}.$$
$$(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=132+15\alpha+3\beta.$$
$$\frac{|132+15\alpha+3\beta|}{3\sqrt{30}}=3\sqrt{30}\implies |132+15\alpha+3\beta|=270.$$
Taking the branch giving a positive $5\alpha+\beta$: $15\alpha+3\beta=138 \implies 5\alpha+\beta=46.$

Solutions · 1