Question

The shortest distance between the lines $x=y=z$ and $x=1-y=\frac{z}{0}$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{6}}$

Answer 1

Answer: (D)

The lines are $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y-1}{-1}=\frac{z}{0}$,
Now, ${\vec{a}}_1=0$,
${\vec{b}}_1=\hat{i}+\hat{j}+\hat{k}$,
${\vec{a}}_2=\hat{j}$,
${\vec{b}}_2=\hat{i}-\hat{j}$
${\vec{b}}_1\times {\vec{b}}_2=\left(\hat{i}+\hat{j}+\hat{k}\right)\times \left(\hat{i}-\hat{j}\right)=\hat{i}+\hat{j}-2\hat{k}$
$S.D.=\frac{|\hat{j}\cdot \left(\hat{i}+\hat{j}-2\hat{k}\right)|}{|\hat{i}+\hat{j}-2\hat{k}|}$
$=\frac{1}{\sqrt{6}}$.