Question

Find the coordinates of the foot of the perpendicular drawn from the $2\hat{i}-\hat{j}+5\hat{k}$ point to the line $\overline{r}=\left(11\hat{i}-2\hat{j}-8\hat{k}\right)+\lambda \left(10\hat{i}-4\hat{j}-11\hat{k}\right)$

Answer 1

Let the foot of perpendicular be $B(x_1,y_1,z_1)$ and the given point be $A(2,-1,5)$

So, putting $\overrightarrow{r}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$ in the equation of line we get

$\left(x_1\hat{i}+y_1\hat{j}+z_1\hat{k}\right)=\left(11\hat{i}-2\hat{j}-8\hat{k}\right)+\lambda \left(10\hat{i}-4\hat{j}-11\hat{k}\right)$

$\begin{array}{c}\Rightarrow x_1\hat{i}+y_1\hat{j}+z_1\hat{k}=\left(11+10\lambda \right)\hat{i}-\left(2+4\lambda \right)\hat{j}-\left(8+11\lambda \right)\hat{k}\\ \Rightarrow x_1=11+10\lambda \\ y_1=-\left(2+4\lambda \right)\\ z_1=-\left(8+11\lambda \right)\end{array}$

Direction Ratio of line $AB$ are $x_1-2,y_1+1,z_1-5$

since, line $AB$ is $\perp$ to the given line

Therefore,

$\begin{array}{c}10\left(x_1-2\right)-4\left(y_1+1\right)-11\left(z_1-5\right)=0\\ \Rightarrow 10\left(11+10\lambda -2\right)-4\left(-2-4\lambda +1\right)-11\left(-8-11\lambda -5\right)=0\\ \Rightarrow 10\left(10\lambda +9\right)-4\left(-4\lambda -1\right)-11\left(-11\lambda -13\right)=0\\ \Rightarrow 100\lambda +90+16\lambda +4+121\lambda +143=0\\ \Rightarrow 237\lambda +237=0\\ \Rightarrow 237\lambda =-237\\ \Rightarrow \lambda =-1\end{array}$

So,

$\begin{array}{c}{x}_1=11+10\times -1\\ =11-10\\ =1\\ y_1=-\left(2-4\right)\\ =2\\ z_1=-\left(8-11\right)\\ =3\end{array}$

Hence, coordinates of the foot of the perpendicular from $2\hat{i}-\hat{j}+5\hat{k}$ to the given line are $\left(1,2,3\right)$.