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, $\quad a_1=\hat{i}+\hat{j}, a_2=2 \hat{i}-3 \hat{k}$
$b=\hat{i}-2 \hat{j}+3 \hat{k}$
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}$
$b \times\left( a _2- a _1\right)=\begin{vmatrix} \hat{i} & \hat{ j } & \hat{ k } \\ 1 & -2 & 3 \\ 1 & -1 & -3\end{vmatrix}$
$ =\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}$
$\therefore \quad$ Distance $d=\left|\frac{9 \hat{i}+6 \hat{j}+\hat{k}}{\sqrt{14}}\right|$
$=\frac{1}{\sqrt{14}} \sqrt{(9)^2+(6)^2+(1)^2}$
$=\frac{1}{\sqrt{14}} \sqrt{81+36+1}$
$=\sqrt{\frac{118}{14}}$
$=\sqrt{\frac{59}{7}}$