Question

Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the line
$\overline{r}=2\overline{i}-\overline{j}+2\overline{k}+\overline{\lambda }(3\overline{i}+4\overline{j}+2\overline{k})$ and the plane $\overline{r}.(\overline{i}-\overline{j}+\overline{k})=5$

Answer 1

$\left[\right(2\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+4\hat{j}+2\hat{k}\left)\right].(\hat{i}-\hat{j}+\hat{k})=5$

$\left[\right(2\hat{i}-\hat{j}+2\hat{k}+3\lambda \hat{i}+4\lambda \hat{j}+2\lambda \hat{k}\left]\right(\hat{i}-\hat{j}+\hat{k})=5$

$\left[\right(2+3\lambda )\hat{i}+(-1+4\lambda )\hat{j}+(2+2\lambda \left)\hat{k}\right](\hat{i}-\hat{j}+\hat{k})=5$

$(2+3\lambda )+(-1+4\lambda )(-1)+(2+2\lambda )L=5$

$\lambda +5=5$

$\lambda =0$

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

Let the point of intersection be $(x,y,z)$

$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$x\hat{i}+y\hat{j}+z\hat{k}=2\hat{i}-\hat{j}+2\hat{k}$

$x=2$ $y=-1$ $z=2$

Point of intersection $=(2,-1,2)$

$(2,-1,2)$ $(-1,-5,-10)$

$=13$