Question

A line passing through the point $A$ with position vector $\overrightarrow{a}=4\hat{i}+2\hat{j}+2\hat{k}$ is parallel to the vector $\overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}$. Find the length of the perpendicular drawn on this line from a point $P$ with position vector $\overrightarrow{r_1}=\hat{i}+2\hat{j}+3\hat{k}$.

Answer 1

Let $\overrightarrow{r}=a+\lambda \overrightarrow{b}$

$=4\hat{i}+2\hat{j}+2\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$

$=(4+2\lambda )\hat{i}+(2+3\lambda )\hat{j}+(2+6\lambda )\hat{k}$

Let $L$ be the foot of the perpendicular

$\therefore$ Position vector of $L$ is:

$(2\lambda +4)\hat{i}+(3\lambda +2)\hat{j}+(6\lambda +2)\hat{k}$

$\overrightarrow{PL}=(2\lambda +4-1)\hat{i}+(3\lambda +2-2)\hat{j}+(6\lambda +2-3)\hat{k}$

$=(2\lambda +3)\hat{i}+3\lambda \hat{j}+(6\lambda -1)\hat{k}$

$\overrightarrow{PL}.\overrightarrow{b}=2(2\lambda +3)+3\left(3\lambda \right)+6(6\lambda -1)=0$

$\Rightarrow 4\lambda +6+9\lambda +36\lambda -6=0$

$\Rightarrow 49\lambda =0$

$\Rightarrow \lambda =0$

Thus, $\overrightarrow{PL}=3\hat{i}-\hat{k}$

Length of the perpendicular,

$\left|\overrightarrow{PL}\right|=\sqrt{(3{)}^2+(-1{)}^2}=\sqrt{10}$ units