Find the distance of the point from the point of intersection of the line <img alt="\vec{r}= 2\hat{i}-j+2\hat{k}

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

Answers (1)

Equation of line is
$\vec{r}= 2\hat{i}-\hat{j}+2\hat{k}+\lambda \left ( 3\hat{i}+4\hat{j}+2\hat{k} \right )---(i)$
coordinates of any point on this line are
$\left ( 2+3\lambda \right )\hat{i}+\left ( -1+4\lambda \right )\hat{j}+\left ( 2+2\lambda \right )\hat{k}$
Equation of plane is
$\vec{r}\cdot \left ( \hat{i}-\hat{j}+\hat{k} \right )= 5---(ii)$
Since the point on line lies on the plane
$\vec{r}\cdot \left ( \hat{i}-\hat{j}+\hat{k} \right )= 5$
$\left [ \left ( 2+3\lambda \right ) \hat{i}+\left ( 2+4\lambda \right )\hat{j}+\left ( 2+2\lambda \right )\right ]\left ( \hat{i} -\hat{j}+\hat{k}\right )= 5$
$\Rightarrow \left ( 2+3\lambda \right )-\left ( -1+4\lambda \right )+\left ( 2+2\lambda \right )= 5$
$\Rightarrow \lambda +5= 5$
$\Rightarrow \lambda = 0$
So equation of line is
$\vec{r}= \left ( 2\hat{i}-\hat{j}+2\hat{k} \right )+0\left ( 3\hat{i}+4\hat{j}+2\hat{k} \right )$
$\Rightarrow \vec{r}= 2\hat{i}-\hat{j}+2\hat{k}---(iii)$
let point of intersection be (x,y,z)
$\therefore \vec{r}= x\hat{i}+y\hat{j}+z\hat{k}---(iv)$
From equations (iii) and (iv)
$x= 2,y= -1,z= 2$
$\therefore$ Point of intersection is (2,-1,2)
Distance between points (2,-1,2) and (-1,-5,-10) is
$= \sqrt{\left ( -1-2 \right )^{2}+\left ( -5+1 \right )^{2}+\left ( -10-2 \right )^{2}}$
$= \sqrt{\left ( -3 \right )^{2}+\left ( -4 \right )^{2}+\left ( -12\right )^{2}}$
$=\sqrt{9+16+144}= \sqrt{169}= 13\, units$