Q. Find the equation of a straight line in the plane $\overrightarrow{r} \bullet \overrightarrow{n} = d$ which is parallel to $r = \overrightarrow{a} + \lambda\ \overrightarrow{b}$ and passes through the foot of the perpendicular drawn from point $P\left( \overrightarrow{a} \right)$ to $\overrightarrow{r} \bullet \overrightarrow{n} = d$ (where $\overrightarrow{n} \bullet \overrightarrow{b} = 0$)
A) $\overrightarrow{r} = \overrightarrow{a} + \left( \frac{d - \overrightarrow{a} \bullet \overrightarrow{n}}{n^{2}} \right)\overrightarrow{n} + \lambda\ \overrightarrow{b}$
B) $\overrightarrow{r} = \overrightarrow{a} + \left( \frac{d - \overrightarrow{a} \bullet \overrightarrow{n}}{n} \right)\overrightarrow{n} + \lambda\ \overrightarrow{b}$
C) $\overrightarrow{r} = \overrightarrow{a} + \left( \frac{\overrightarrow{a} \bullet \overrightarrow{n} - d}{n^{2}} \right)\overrightarrow{n} + \lambda\ \overrightarrow{b}$
D) $\overrightarrow{r} = \overrightarrow{a} + \left( \frac{\overrightarrow{a} \bullet \overrightarrow{n} - d}{n} \right)\overrightarrow{n} + \lambda\ \overrightarrow{b}$
Explanation
(a)
Foot of the perpendicular drawn from point $A\left( \overrightarrow{a} \right)\$on the plane $\overrightarrow{r}.\overrightarrow{n} = d$ is $\overrightarrow{a} + \left( \frac{d - \overrightarrow{a}.\overrightarrow{n}}{{|\overrightarrow{n}|}^{2}} \right)\overrightarrow{n}$
Therefore, equation of the line parallel to $\overrightarrow{r} = \overrightarrow{a} + \lambda\overrightarrow{b}$ in the plane $\overrightarrow{r}.\overrightarrow{n} = d$ is given by
$\overrightarrow{r} = \overrightarrow{a} + \left( \frac{d - \overrightarrow{a}.\overrightarrow{n}}{{|\overrightarrow{n}|}^{2}} \right)\overrightarrow{n} + \lambda\overrightarrow{b}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.