Question

Find the foot of the perpendicular drawn from the point $\hat{i}+6\hat{j}+3\hat{k}$ to the line $\overrightarrow{r}=\hat{j}+2\hat{k}+\lambda \left(\hat{i}+2\hat{j}+3\hat{k}\right).$ Also, find the length of the perpendicular

Answer 1

Let L be the foot of the perpendicular drawn from the point P ($\hat{i}+6\hat{j}+3\hat{k}$) to the line $\overrightarrow{r}=\hat{j}+2\hat{k}+\lambda \left(\hat{i}+2\hat{j}+3\hat{k}\right).$

Let the position vector L be
$\overrightarrow{r}=\hat{j}+2\hat{k}+\lambda \left(\hat{i}+2\hat{j}+3\hat{k}\right)=\lambda \hat{i}+\left(1+2\lambda \right)\hat{j}+\left(2+3\lambda \right)\hat{k}$ ...(1)


Now,
$$\begin{gathered} \overrightarrow{PL}=\mathrm{Position}\mathrm{vector}\mathrm{of}L-\mathrm{Position}\mathrm{vector}\mathrm{of}P \\ \Rightarrow \overrightarrow{PL}=\left\{\lambda \hat{i}+\left(1+2\lambda \right)\hat{j}+\left(2+3\lambda \right)\hat{k}\right\}-\left(\hat{i}+6\hat{j}+3\hat{k}\right) \\ \Rightarrow \overrightarrow{PL}=\left(\lambda -1\right)\hat{i}+\left(2\lambda -5\right)\hat{j}+\left(3\lambda -1\right)\hat{k}...\left(2\right) \end{gathered}$$


Since $\overrightarrow{PL}$ is perpendicular to the given line, which is parallel to $\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}$, we have

$$\begin{gathered} \overrightarrow{PL}.\overrightarrow{b}=0 \\ \Rightarrow \left\{\left(\lambda -1\right)\hat{i}+\left(2\lambda -5\right)\hat{j}+\left(3\lambda -1\right)\hat{k}\right\}.\left(\hat{i}+2\hat{j}+3\hat{k}\right)=0 \\ \Rightarrow 1\left(\lambda -1\right)+2\left(2\lambda -5\right)+3\left(3\lambda -1\right)=0 \\ \Rightarrow \lambda =1 \end{gathered}$$

Substituting $\lambda =1$ in (1), we get the position vector of L as $\hat{i}+3\hat{j}+5\hat{k}$.

Substituting $\lambda =1$ in (2), we get
$\overrightarrow{PL}=-3\hat{j}+2\hat{k}$
$$\begin{gathered} =\sqrt{{\left(-3\right)}^2+2^2} \\ =\sqrt{13} \end{gathered}$$

Hence, the length of the perpendicular from point P on PL is $\sqrt{13}\mathrm{units}$.