find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector

find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector $2\hat{i}+3\hat{j}+4\hat{k}$ to the plane $\vec{r}\cdot \left ( 2\hat{i}+\hat{j}+3\hat{k} \right )-26= 0.$ Also find image of P in the plane.

Answers (1)

Let L be the foot of the foot of the perpendicular drawn from $p\left ( 2\hat{i}+3\hat{j}+4\hat{k} \right )$ on the plane $\vec{r}\cdot \left ( 2\hat{i}+\hat{j}+3\hat{k} \right )-26= 0$ we know that the position vector of the foot of perpendicular from the point P with position vector $\vec{a}$ from the plane $\vec{r}\cdot \vec{n}= d$ is given by

$\overrightarrow{pn}= \frac{\left [ d-\left ( \vec{a}\cdot \vec{n} \right ) \right ]\vec{n}}{\left | \vec{n} \right |^{2}}$
$= \frac{\left [ 26-\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right ) \left ( 2\hat{i} +\hat{j}+3\hat{k} \right )\right ]}{\left ( \sqrt{4+1+9} \right )^{2}} \left ( 2\hat{i} +\hat{j}+3\hat{k} \right )$
$= \frac{\left [ 26-\left ( 4+3+12 \right ) \right ]\left ( 2\hat{i}+\hat{j}+3\hat{k} \right )}{14}= \frac{7}{14}\left ( 2\hat{i}+\hat{j}+3\hat{k} \right )$
$= \hat{i}+\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k}$ and the perpendicular distance is given by $\frac{\left | \vec{a}\cdot \vec{n}-d \right |}{\left | \vec{n} \right |}$
so required distance $= \frac{\left | \left ( 2\hat{i}+3\hat{j}+4\hat{k} \right ) \left ( 2\hat{i}+\hat{j}+3\hat{k} \right )-26\right |}{\sqrt{14}}$
$= \frac{7}{\sqrt{14}}$
Let Q be theimage of the point $P\left ( 2\hat{i}+3\hat{j}+4\hat{k} \right )$ to the plane $\vec{r}\cdot \left ( 2\hat{i}+\hat{j}+3\hat{k} \right )-26= 0.$ then PQ is normal to the plane.
$\therefore$ equation of line PQ is $\vec{r}= \left ( 2\hat{i}+3\hat{j}+4\hat{k} \right )+\lambda \left ( 2\hat{i}+\hat{j}+3\hat{k} \right )$
Since Q lies on line PQ so let the position vector of Q be $\left ( 2\hat{i}+3\hat{j}+4\hat{k} \right )+\lambda \left ( 2\hat{i}+\hat{j}+3\hat{k} \right )= \left ( 2+2\lambda \right )\hat{i}+\left ( 3+\lambda \right )\hat{j}+\left ( 4+3\lambda \right )\hat{k}$
Since, R is the mid point of PQ. therefore position vector of R is
$\frac{\left [ \left ( 2+2\lambda \right )\hat{i}+\left ( 3+\lambda \right )\hat{j} +\left ( 4+3\lambda \right )\hat{k}\right ]+\left ( 2\hat{i}+3\hat{j}+4\hat{k} \right )}{2}$
$= \left ( 2+\lambda \right )\hat{i}+\left ( 3+\frac{\lambda }{2} \right )\hat{j}+\left ( 4+3\frac{\lambda }{2} \right )\hat{k}$
since R lies on the plane $\vec{r}\cdot \left [ 2\hat{i}+\hat{j}+3\hat{k} \right ]-26= 0$
$\therefore \left \{ \left ( 2+\lambda \right )\hat{i} +\left ( 3+\frac{\lambda }{2} \right )\hat{j}+\left ( 4+3\frac{\lambda }{2} \right )\hat{k}\right \}$
$\left ( 2\hat{i}+\hat{j} +3\hat{k}\right )-26= 0$
$= \left ( 4+2\lambda \right )+3+\frac{\lambda }{2}+1 2 +\frac{9\lambda }{2}- 26=0$
$\Rightarrow \lambda = \frac{7}{7}= 1$
Thus the position vector of Q is $4\hat{i}+4\hat{j}+7\hat{k}$