Discussion Forum

The equation of line passing through (3,-1,2) and perpendicular to the lines

 $\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2\hat{i}-2\hat{j}+2\hat{k})$

 and $\overrightarrow{r}=(2\hat{i}+\hat{j}-3\hat{k})+\mu(\hat{i}-2\hat{j}+2\hat{k})$  is 

Answer:

Explanation:

 Direction ratio of line perpendicular to the lines  $\overrightarrow{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$

 and  $\overrightarrow{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$ is $\alpha(\overrightarrow{b_{1}}\times\overrightarrow{b_{2}})$

 $\therefore$    Direction ratio of line perpendicular to the lines

   $\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2\hat{i}-2\hat{j}+2\hat{k})$

and 

  $\overrightarrow{r}=(2\hat{i}+\hat{j}-3\hat{k})+\mu(\hat{i}-2\hat{j}+2\hat{k})$  is  

                                        $\alpha\begin{bmatrix}\hat{i} & \hat{j}&\hat{k} \\2 & -2&1 \\1 &-2 &2 \end{bmatrix}$

 = $\alpha [{(-4+2)\hat{i}-(4-1)\hat{j}+(-4+2)\hat{k}]}$

    = $\alpha [{-2\hat{i}-3\hat{j}-2\hat{k}]}$

  Now, equation of line passing through (3,-1,2) and parallel to ${-2\hat{i}-3\hat{j}-2\hat{k}}$

 $\overrightarrow{r}=3\hat{i}-\hat{j}-2\hat{k}+\beta(2\hat{i}+3\hat{j}+2\hat{k})$

 Hence, cartesian form of the above equation is 

$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$