Question

Let $A\left(\overrightarrow{a}\right),$ $B\left(\overrightarrow{b}\right),$ $C\left(\overrightarrow{c}\right)$ and $D\left(\overrightarrow{d}\right)$ be four points such that $\overrightarrow{a}=-2\hat{i}+4\hat{j}+3\hat{k},$ $\overrightarrow{b}=2\hat{i}-8\hat{j},$ $\overrightarrow{c}=\hat{i}-3\hat{j}+5\hat{k},$ $\overrightarrow{d}=4\hat{i}+\hat{j}-7\hat{k}.$ If $m$ is the shortest distance between the lines $AB$ and $CD,$ then which of the following is (are) CORRECT?

• A. $m=0,$ hence $AB$ and $CD$ intersect
• B. $m=\frac{\left[\overrightarrow{AB}\overrightarrow{CD}\overrightarrow{BD}\right]}{|\overrightarrow{AB}\times \overrightarrow{CD}|}$
• C. $AB$ and $CD$ are skew lines and $m=\frac{23}{13}$
• D. $m=\frac{\left[\overrightarrow{AB}\overrightarrow{CD}\overrightarrow{AC}\right]}{|\overrightarrow{AB}\times \overrightarrow{CD}|}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $m=\frac{\left[\overrightarrow{AB}\overrightarrow{CD}\overrightarrow{BD}\right]}{|\overrightarrow{AB}\times \overrightarrow{CD}|}$
C $AB$ and $CD$ are skew lines and $m=\frac{23}{13}$
D $m=\frac{\left[\overrightarrow{AB}\overrightarrow{CD}\overrightarrow{AC}\right]}{|\overrightarrow{AB}\times \overrightarrow{CD}|}$

Given, $\overrightarrow{a}=-2\hat{i}+4\hat{j}+3\hat{k};\overrightarrow{b}=2\hat{i}-8\hat{j}$ $\overrightarrow{c}=\hat{i}-3\hat{j}+5\hat{k};\overrightarrow{d}=4\hat{i}+\hat{j}-7\hat{k}$ $\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}=4\hat{i}-12\hat{j}-3\hat{k}$ $\overrightarrow{CD}=\overrightarrow{d}-\overrightarrow{c}=3\hat{i}+4\hat{j}-12\hat{k}$ $\overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a}=3\hat{i}-7\hat{j}+2\hat{k}$ $\overrightarrow{BD}=\overrightarrow{d}-\overrightarrow{b}=2\hat{i}+9\hat{j}-7\hat{k}$

By definition,
$m=\frac{(\overrightarrow{AB}\times \overrightarrow{CD})\cdot \overrightarrow{AC}}{|\overrightarrow{AB}\times \overrightarrow{CD}|}\cdots \left(1\right)$
$=\frac{(\overrightarrow{AB}\times \overrightarrow{CD})\cdot \overrightarrow{BD}}{|\overrightarrow{AB}\times \overrightarrow{CD}|}\cdots \left(2\right)$

$\overrightarrow{AB}\times \overrightarrow{CD}=13(12\hat{i}+3\hat{j}+4\hat{k})$ $\therefore |\overrightarrow{AB}\times \overrightarrow{CD}|=13\sqrt{144+9+16}=169$
$\therefore m=\frac{13(12\hat{i}+3\hat{j}+4\hat{k})}{169}\cdot (3\hat{i}-7\hat{j}+2\hat{k})$ [ Using $\left(1\right)$]
$=\frac{13}{169}[36-21+8]$
$=\frac{23}{13}$

Also, $m=\frac{13(12\hat{i}+3\hat{j}+4\hat{k})\cdot (2\hat{i}+9\hat{j}-7\hat{k})}{169}$ [ Using $\left(2\right)$ ]
$=\frac{13}{169}[24+27-28]$
$=\frac{23}{13}$