If d_1 is the shortest distance between the lines x+1=2y=-12z,x=y+2=6z-6 and d_2 is the sh…

Question

If ${d}_{1}$ is the shortest distance between the lines $x+1=2y=-12z,x=y+2=6z-6$ and ${d}_{2}$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5},\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$, then the value of $\frac{32\sqrt{3}{d}_{1}}{{d}_{2}}$ is :

Answer: 16

Accepted Answer

${L}_{1}:\frac{x+1}{1}=\frac{y}{1/2}=\frac{z}{-1/12},{L}_{2}:\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$

${d}_{1}=$ shortest distance between ${L}_{1}\&{L}_{2}$

$\begin{gathered} =|\frac{({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}	imes {\vec{b}}_{2})}{|({\vec{b}}_{1}	imes {\vec{b}}_{2})|}| \ {d}_{1}=2 \ {L}_{3}:\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5},{L}_{4}:\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3} \ {d}_{2}={L}_{3}{L}_{4} \ {d}_{2}=\frac{12}{\sqrt{3}} \ =\frac{32\sqrt{3}{d}_{1}}{{d}_{2}}=\frac{32\sqrt{3}	imes 2}{\frac{12}{\sqrt{3}}}=16 \end{gathered}$

Solutions · 1