Question

The distance between the lines given by $r=\hat{i}+\hat{j}+\lambda (\hat{i}-2\hat{j}+3\hat{k})$ and $r=(2\hat{i}-3\hat{k})$ $+\mu (\hat{i}-2\hat{j}+3\hat{k})$ is

• A. $\sqrt{\frac{59}{14}}$
• B. $\sqrt{\frac{59}{7}}$
• C. $\sqrt{\frac{118}{7}}$
• D. $\frac{\sqrt{59}}{7}$

Answer 1

Answer: (B)

Here, $a_1=\hat{i}+\hat{j},a_2=2\hat{i}-3\hat{k}$
$b=\hat{i}-2\hat{j}+3\hat{k}$
Disclarice od $=\left|\frac{b\times \left(a_2-a_1\right)}{|b|}\right|$
Now, $a_2-a_1=(2\hat{i}-3\hat{k})-(\hat{i}+\hat{j})$
$=\hat{i}-\hat{j}-3\hat{k}$
$b\times \left(a_2-a_1\right)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 3\\ 1& -1& -3\end{array}\right|$
$=\hat{i}(6+3)-\hat{j}(-3-3)+\hat{k}(-1+2)$
$=9\hat{i}+6\hat{j}+\hat{k}$
$|b|=\sqrt{1^2+(-2{)}^2+3^2}$
$=\sqrt{1+4+9}$
$=\sqrt{14}$
$\therefore$ Distance $d=\left|\frac{9\hat{i}+6\hat{j}+\hat{k}}{\sqrt{14}}\right|$
$=\frac{1}{\sqrt{14}}\sqrt{(9{)}^2+(6{)}^2+(1{)}^2}$
$=\frac{1}{\sqrt{14}}\sqrt{81+36+1}$
$=\sqrt{\frac{118}{14}}$
$=\sqrt{\frac{59}{7}}$