Question

The shortest distance between the lines $x = y + 2 = 6z - 6 $ and $x + 1 = 2y = - 12z$ 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,
$\vec{a}_{1}=-2 \hat{j}+\hat{k}, b_{1}=6 \hat{i}+6 \hat{j}+\hat{k}, \vec{a}_{2}=-\hat{i}, \vec{b}_{2}=12 \hat{i}+6 \hat{j}-\hat{k}$
$\vec{ b }_{1} \times \vec{ b }_{2}=\begin{vmatrix}\hat{i} & \hat{j} & k \\ 6 & 6 & 1 \\ 12 & 6 & -1\end{vmatrix}=-12 \hat{1}+18 \hat{\jmath}-36 \hat{k}$
Shortest distance $=\frac{\left|\left(\vec{a_{2}}-\vec{a_{1}}\right) \cdot\left(\vec{b_{1}} \times \vec{b_{2}}\right)\right|}{\left|\vec{b_{1}} \times \vec{b_{2}}\right|}$
$=\frac{|(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(-12 i +18 \hat{j}-36 \hat{k})|}{\sqrt{(-12)^{2}+(18)^{2}+(-36)^{2}}}$
$=\frac{|+12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2$