Question

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

Answer 1

$\overrightarrow{r}=(2\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+4\hat{j}+2\hat{k})$

equation of plane is $r.(\hat{i}-\hat{j}+\hat{k})=5$

To find point in intersection of line and plane putting values of $\overrightarrow{r}$ from equation of line into equations of plane

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

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

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

$\left[\right(2+3\lambda )\times 1+(-1+4\lambda )\times (-1)+(2+2\lambda )\times 1=5$

$2+3\lambda +1-4\lambda +2+2\lambda =5$

$\lambda +5=5\Rightarrow \lambda =0$

So, the equation of line is

$\overrightarrow{r}=(2\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+4\hat{j}+2\hat{k})$

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

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

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

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

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