The shortest distance between the lines [(x-4)/(4)=(y+2)/(5)=(z+3)/(3)] and [(x-1)/(3)=(y-…

Question

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

Accepted Answer

The given lines are

$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$ and $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$

$\begin{gathered} {\vec{b}}_{1}=4\hat{i}+5\hat{j}+3\hat{k} \ {\vec{b}}_{2}=3\hat{i}+4\hat{j}+2\hat{k} \ {\vec{a}}_{1}=4\hat{i}-2\hat{j}-3\hat{k} \ {\vec{a}}_{2}=\hat{i}+3\hat{j}+4\hat{k} \end{gathered}$

$\begin{gathered} 	herefore {\vec{b}}_{1}	imes {\vec{b}}_{2}=|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 4 \quad  5 \quad  3 \ 3 \quad  4 \quad  2\end{array}| \ =(10-12)\hat{i}-(8-9)\hat{j}+(16-15)\hat{k} \ =-2\hat{i}+\hat{j}+\hat{k} \end{gathered}$

$\begin{gathered} d=|\frac{({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}	imes {\vec{b}}_{2})}{|{\vec{b}}_{1}	imes {\vec{b}}_{2}|}| \ =|\frac{(3\hat{i}-5\hat{j}-7\hat{k})\cdot (-2\hat{i}+\hat{j}+\hat{k})}{\sqrt{4+1+1}}| \ =|\frac{-6-5-7}{\sqrt{6}}|=\frac{18}{\sqrt{6}}=3\sqrt{6} \end{gathered}$

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