Let the shortest distance between the lines (x-3)/(3)=(y- )/(-1)=(z-3)/(1) and (x+3)/(-3)=…

Question

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

Accepted Answer

$\begin{gathered} {L}_{1}:\frac{x-3}{3}=\frac{y-\alpha }{-1}=\frac{z-3}{1} \ {L}_{2}:\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta }{4} \end{gathered}$

$\begin{gathered} {a}_{1}:(3,\alpha ,3)\, {a}^{2}=(-3,-7,\beta ) \ {\vec{b}}_{1}=3\hat{i}-\hat{j}+\hat{k}\, {\vec{b}}_{2}=-3\hat{i}+2\hat{j}+4\hat{k} \end{gathered}$

$d=\frac{|({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}	imes {\vec{b}}_{2})|}{|{\vec{b}}_{1}	imes {\vec{b}}_{2}|}=3\sqrt{30}$

$=\frac{\begin{vmatrix} 6 \quad  \alpha +7 \quad  3-\beta \ 3 \quad  -1 \quad  1 \ -3 \quad  2 \quad  4 \end{vmatrix}}{\sqrt{{6}^{2}+{15}^{2}+{3}^{2}}}=3\sqrt{30}$

$\begin{gathered} \Rightarrow |-15\alpha -3\beta -132|=270 \ |5\alpha +\beta +44|=90 \ \Rightarrow 5\alpha +\beta =90-44=46 \end{gathered}$

Solutions · 1