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

Question

If the shortest distance between the lines $$\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$$ and $$\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$$ is $$\frac{38}{3 \sqrt{5}} \mathrm{k}$$

Accepted Answer

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

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

${L}_{2}=\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$

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

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

$\begin{gathered} d=|\frac{(5\hat{i}+5\hat{j}-9\hat{k})\cdot (18\hat{i}-9\hat{k})}{\sqrt{324+81}}| \ |{\vec{b}}_{1}	imes \vec{b}||\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 2 \quad  3 \quad  4 \ 1 \quad  -3 \quad  2\end{array}| \ \Rightarrow \hat{i}(6+12)-\hat{j}(4-4)+\hat{k}(-6-3) \ \Rightarrow (18\hat{i}-9\hat{k}) \end{gathered}$

$\begin{gathered} d=|\frac{90+81}{9\sqrt{5}}| \ d=\frac{171}{9\sqrt{5}} \ \frac{38}{3\sqrt{5}}k=\frac{171}{9\sqrt{5}} \ \frac{38}{3\sqrt{5}}k=\frac{57}{3\sqrt{5}} \end{gathered}$

$\begin{gathered} k=\frac{57}{38}=\frac{3}{2} \ \int _{0}^{\frac{3}{2}}[{x}^{2}]dx \ \int _{0}^{1}0dx+\int _{0}^{\sqrt{}}1dx+\int _{\sqrt{2}}^{\frac{3}{2}}2dx \ 0+(\sqrt{2}-1)+2(\frac{3}{2}-\sqrt{2}) \ \sqrt{2}-1+3-2\sqrt{2} \ 2-\sqrt{2} \end{gathered}$

$\begin{gathered} \alpha =2 \ 6{\alpha }^{3}=6(2{)}^{3}=48 \end{gathered}$

Solutions · 1