Question

Find the vector and Cartesian equations of the line through the point $(1,2,-4)$ and perpendicular to the two lines
$\overrightarrow{r}=(8\overrightarrow{i}-19\overrightarrow{j}+10\hat{k})+\lambda (3\overrightarrow{i}-16\hat{j}+7\hat{k})$ and $\overrightarrow{r}=(15\hat{i}+29\hat{j}+5\hat{k})+\mu (3\hat{i}+8\hat{j}-5\hat{k})$

Answer 1

Since the line is perpendicular to the given lines, it would be parallel to the vector, which will be obtained by the cross product of the given vectors.

$(3\hat{i}-16\hat{j}+7\hat{k})$ $\times$ $(3\hat{i}$ + $8\hat{j}$ - $5\hat{k})$

$=0+24\hat{k}+15\hat{j}+48\hat{k}-0+80\hat{i}+21\hat{j}-56\hat{i}+0$

$=24\hat{i}+36\hat{j}+72\hat{k}$

Can also be written as $2\hat{i}+3\hat{j}+6\hat{k}$ since we only need the direction.

Since, the line passes through the point $(1,2,-4)$, the vector equation of the line becomes $\overrightarrow{r}=(\hat{i}-2\hat{j}+4\hat{k})+\alpha (2\hat{i}+3\hat{j}+6\hat{k})$

In Cartesian form, $x=1+2\alpha ,y=-2+3\alpha ,z=4+6\alpha$