The shortest distance between the lines [(x-3)/(4)=(y+7)/(-11)=(z-1)/(5)] and [(x-5)/(3)=(…

Question

The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:

Accepted Answer

Given lines are

$\frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5}$ and

$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}$

Shortest distance between two lines,

$\begin{gathered} d=\frac{|({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}	imes {\vec{b}}_{2})|}{|({\vec{b}}_{1}	imes {\vec{b}}_{2})|} \ {\vec{a}}_{2}-{\vec{a}}_{1}=2\hat{i}+16\hat{j}-3\hat{k} \ {\vec{b}}_{1}	imes {\vec{b}}_{2}=|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 4 \quad  -11 \quad  5 \ 3 \quad  -6 \quad  1\end{array}|=19\hat{i}+11\hat{j}+9\hat{k} \ 	herefore \, d=\frac{187}{\sqrt{563}} \end{gathered}$

Solutions · 1