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)

$14$

Answer 2

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, $d=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right)\cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$ $=\left|\frac{\left(7\hat{i}+38\hat{j}-5\hat{k}\right)\cdot\left(24\hat{i}+36\hat{j}+72\hat{k}\right)}{\sqrt{\left(24\right)^{2}+\left(36\right)^{2}+\left(72\right)^{2}}}\right|$ $=\left|\frac{168+1368-360}{\sqrt{576+1296+5184}}\right|=\left|\frac{1176}{\sqrt{7056}}\right|$ $=\frac{1176}{84}=14$