Question

The shortest distance between the lines $\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}$ and $\frac{x-10}{3}=\frac{y-30}{8}=\frac{4-z}{-5}$ is

• A. $\frac{234}{7}units$
• B. $\frac{288}{21}units$
• C. $\frac{221}{3}units$
• D. $\frac{234}{21}units$

Answer 1

Answer: (B)

$\frac{288}{21}units$

Given lines are $\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}$ and $\frac{x-10}{3}=\frac{y-30}{8}=\frac{4-z}{5}=\frac{z-4}{-5}$
The vector form of given lines are
$r=7\hat{i}-4\hat{j}+6\hat{k}+\lambda (3\hat{i}-16\hat{j}+7\hat{k})$
and $r=10\hat{i}+30\hat{j}+4\hat{k}+\mu (3\hat{i}+8\hat{j}-5\hat{k})$
On comparing these equations with
$r=a_1+\lambda b_1$ and $r=a_2+\mu b_2$, we get
$\overrightarrow{a_1}=7\hat{i}-4\hat{j}+6\hat{k}$
$\overrightarrow{a_2}=10\hat{i}+30\hat{j}+4\hat{k}$
$\overrightarrow{b_1}=3\hat{i}-16\hat{j}+7\hat{k}$
and $\overrightarrow{b_1}=3\hat{i}+8\hat{j}-5\hat{k}$
Shortest distance $=\left|\frac{({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)\cdot ({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)}{|b_1\times b_2|}\right|$
$=\left|\frac{(3\hat{i}+34\hat{j}-2\hat{k})\cdot (24\hat{i}+36\hat{j}+72\hat{k})}{84}\right|$
$=\left|\frac{72+1224-144}{84}\right|=\left|\frac{1152}{84}\right|=\frac{288}{21}units$

Hence, option B is correct.