Question

The shortest distance between the lines
$\vec{r}=\left(8+3\lambda \right)\hat{i}-\left(9+16\lambda \right)\hat{j}+\left(10+7\lambda \right)\hat{k}$ and
$\vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu \left(3\hat{i}-8\hat{j}-5\hat{k}\right)$ is

• A. $84$
• B. $14$
• C. $21$
• D. $16$

Answer 1

Answer: (B)

Given, equation of lines are
$\vec{r}=\left(8+3\lambda \right)\hat{i}-\left(9+16\lambda \right)\hat{j}+\left(10+7\lambda \right)\hat{k}$
i.e., $\vec{r}=8\hat{i}-9\hat{j}+10\hat{k}+\lambda \left(3\hat{i}-16\hat{j}+7\hat{k}\right)$
and $\vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu \left(3\hat{i}+8\hat{j}-5\hat{k}\right)$
Here, ${\vec{a}}_1=8\hat{i}-9\hat{j}+10\hat{k}$,
${\vec{b}}_1=3\hat{i}-16\hat{j}+7\hat{k}$
${\vec{a}}_2=15\hat{i}+29\hat{j}+5\hat{k}$,
${\vec{b}}_2=3\hat{i}+8\hat{j}-5\hat{k}$
${\vec{a}}_2-{\vec{a}}_1=\left(15\hat{i}+29\hat{j}+5\hat{k}\right)-\left(8\hat{i}-9\hat{j}+10\hat{k}\right)$
$=7\hat{i}+38\hat{j}-5\hat{k}$
${\vec{b}}_1\times {\vec{b}}_2=\hat{i}\left(80-56\right)-\hat{j}\left(-15-21\right)+\hat{k}\left(24+48\right)$
$=\left(24\hat{i}+36\hat{j}+72\hat{k}\right)$
$\therefore$ Shortest distance, $d=\left|\frac{\left({\vec{a}}_2-{\vec{a}}_1\right)\cdot \left({\vec{b}}_1\times {\vec{b}}_2\right)}{|{\vec{b}}_1\times {\vec{b}}_2|}\right|$
$=\left|\frac{\left(7\hat{i}+38\hat{j}-5\hat{k}\right)\cdot \left(24\hat{i}+36\hat{j}+72\hat{k}\right)}{\sqrt{{\left(24\right)}^2+{\left(36\right)}^2+{\left(72\right)}^2}}\right|$

$=\left|\frac{168+1368-360}{\sqrt{576+1296+5184}}\right|=\left|\frac{1176}{\sqrt{7056}}\right|$

$=\frac{1176}{84}=14$