Question

If $A=(-1,2,-3)$ , $B=(-16,6,4)$, $C=(1,-1,3)$ and $D=(4,9,7)$, then the shortest distance between the lines $\overrightarrow{AB},\overrightarrow{CD}$ is

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is C $5$

Shortest Distance is given by

$=\frac{\left|\right(\overrightarrow{a}-\overrightarrow{c}\left)\right(\overrightarrow{b}\times \overrightarrow{d}\left)\right)|}{|\overrightarrow{b}\times \overrightarrow{d}|}$

Line $AB=(-1,2,-3)+5(-15,4,7)$
Line $CD=(1,-1,3)+t(3,10,4)$

$\overline{b}\times \overline{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -15& 4& 7\\ 3& 10& 4\end{array}\right|=-54\hat{i}+81\hat{j}-162\hat{k}$

$\overrightarrow{a}-\overrightarrow{c}=(-2,3,-6)$
Putting values in equation$=\frac{\left|\right(\overrightarrow{a}-\overrightarrow{c}\left)\right(\overrightarrow{b}\times \overrightarrow{d}\left)\right)|}{|\overrightarrow{b}\times \overrightarrow{d}|}$

$=\frac{6+13-5+54}{\sqrt{9+(4.5{)}^2+81}}$

$=5$