The shortest distance between the lines
Question:

The shortest distance between the lines $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$

and $x+y+z+1=0,2 x-y+z+3=0$ is :

 

  1. 1

  2. $\frac{1}{\sqrt{3}}$

  3. $\frac{1}{\sqrt{2}}$

  4. $\frac{1}{2}$


Correct Option: , 2

Solution:

For line of intersection of planes $x+y+z+1=0$ and

$2 x-y+z+3=0$

$\vec{b}_{2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 1\end{array}\right|=2 \hat{i}+\hat{j}-3 \hat{k}$

Put $y=0$, we get $x=-2$ and $z=1$

$L_{2}: \bar{r}=(-2 \hat{i}+\hat{k})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and

$L_{1}: \bar{r}=(\hat{i}-\hat{j})+\mu(-\hat{j}+\hat{k})$ (Given)

Now, $\bar{b}_{1} \times \bar{b}_{2}=-2[\hat{i}+\hat{j}+\hat{k}]$ and $\bar{a}_{2}-\bar{a}_{1}=-3 \hat{i}+\hat{j}+\hat{k}$

$\therefore$ Shortest distance $=\frac{1}{\sqrt{3}}$

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