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)

Given, lines are $\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}$
and $\frac{x-10}{3}=\frac{y-30}{8}=\frac{4-z}{5}$
The vector form of given equations 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
$a_1=7\hat{i}-4\hat{j}+6\hat{k}$
$a_2=10\hat{i}+30\hat{j}+4\hat{k}$
$b_1=3\hat{i}-16\hat{j}+7\hat{k}$
and $b_2=3\hat{i}+8\hat{j}-5\hat{k}$
$b_1\times b_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \mathbf{k}\\ 3& -16& 7\\ 3& 8& -5\end{array}\right|$
$=(80-56)\hat{i}+(21+15)\hat{j}+(24+48)\hat{k}$
$=24\hat{i}+36\hat{j}+72\hat{k}$
Then $\mid b_1\times b_2\mid =\sqrt{(24{)}^2+(36{)}^2+(72{)}^2}$
$=\sqrt{576+1296+5184}$
$=\sqrt{7056}=84$
$\therefore$ Shortest distance$=\left|\frac{\left(a_2-a_1\right)\cdot \left(b_1\times b_2\right)}{|b_1\times b_2|}\right|$
$=\left|\frac{(3\hat{i}+34\hat{j}-2\hat{k})\cdot (24\hat{i}+36\hat{j}+72\hat{k})}{84}\right|$
$=\left|\frac{72+1224-144}{84}\right|=\left|\frac{1152}{84}\right|=\frac{288}{21}$ units