Question

The shortest distance between the lines $x = y+2 = 6z-6$ and $x + 1 = 2y = -12\,z$ is

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

Answer 1

Answer: (B)

The lines are $\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}$
and $\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}$
Here, $\vec{a}_{1}=-2\hat{j}+\hat{k}$,
$\vec{b}_{1}=6\hat{i}+6\hat{j}+\hat{k}$,
$\vec{a}_{2}=-\hat{i}$,
$\vec{b}_{2}=12\hat{i}+6\hat{j}-\hat{k}$
$\vec{b}_{1}\times\vec{b}_{2}=-12\hat{i}+18\hat{j}-36\hat{k}$
Shortest distance
$=\frac{\left|\left(\vec{a}_{2}-\vec{a}_{1}\right)\cdot\left(\vec{b}_{1}\times\vec{b}_{2}\right)\right|}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}=\frac{\left|\left(-\hat{i}+2\hat{j}-\hat{k}\right)\cdot\left(-12\hat{i}+18\hat{j}-36\hat{k}\right)\right|}{\sqrt{\left(-12\right)^{2}+\left(18\right)^{2}+\left(-36\right)^{2}}}$
$=\frac{\left|12+36+36\right|}{\sqrt{1764}}=\frac{84}{42}=2$