Question

Find the equation of the ccontaining rhe lines $\overrightarrow{r}=i+2\hat{j}-\hat{k}+\lambda (2\hat{i}-\hat{j}+\hat{k})$ and $\hat{r}=\hat{i}+2\hat{j}-\hat{k}+\mu (\hat{i}-\hat{j}+2\hat{k})$. Also find the distance of this plane from origin.

Answer 1

Given lines,

$\begin{array}{c}\overline{r}=\hat{i}+2\hat{j}-\hat{k}+\lambda (2\hat{i}-\hat{j}+\hat{k})-----\left(1\right)\\ \overline{r}=\hat{i}+2\hat{j}-\hat{k}+\mu (\hat{i}-\hat{j}+2\hat{k})-----\left(2\right)\end{array}$

we find the equation of plane containing lines form (1) & (2)

and the distance from origin to this plane from (1 )&(2), contain point with

Parallel to vector is :

$\begin{array}{c}\overline{b}=2\hat{i}-\hat{j}+\hat{k}\\ \overline{c}=\hat{i}-\hat{j}+2\hat{k}\end{array}$

position Vector is :

$\overline{OA}=\hat{i}+2\hat{j}-\hat{k}$

then required equation of plan is :

$\begin{array}{c}\left[\overline{r}-\overline{OA}\overline{b}\overline{c}\right]=0\\ \left(\begin{array}{ccc}x-1& y-2& z+1\\ 2& -1& a_1\\ 1& -1& 2\end{array}\right)=0\\ (x-1)(-1)-(y-2)\left(3\right)+(z+1)(-1)=0\\ -x+1-3y+6-z=0\\ -x-z-3y+6=0\\ or,x+z+3y-6=0\end{array}$

So, distance from origin to this plane=$=\frac{\left|-6\right|}{\sqrt{1+9+1}}=\frac{6}{\sqrt{11}}$ units.