Question

A straight line $\overline{r}=\overline{a}+\lambda \overline{b}$ meets the plane $\overline{r}.\overline{n}=0$ at a point $p$. The position vector of $p$ is

• A. $\overline{a}+\left(\frac{\overline{a}.\overline{n}}{\overline{b}.\overline{n}}\right)\overline{b}$
• B. $\overline{a}-(\overline{b}.\overline{n})\overline{b}$
• C. $\overline{a}-\left(\frac{\overline{a}.\overline{n}}{\overline{b}.\overline{n}}\right)\overline{b}$
• D. $\overline{a}+(\overline{b}.\overline{n})\overline{b}$

Answer 1

The correct option is C $\overline{a}-\left(\frac{\overline{a}.\overline{n}}{\overline{b}.\overline{n}}\right)\overline{b}$

$\to$ Intersection of $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ and $\overrightarrow{r}.\overrightarrow{n}=0$ is $(\overrightarrow{a}+\lambda \overrightarrow{b}),\overrightarrow{n}=0$

$\Rightarrow \overrightarrow{a}.\overrightarrow{n}+\lambda \overrightarrow{b}.\overrightarrow{n}=0\Rightarrow \lambda =-\frac{\overrightarrow{a}.\overrightarrow{n}}{\overrightarrow{b}.\overrightarrow{n}}$

Putting $\lambda$ in $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$

$\overrightarrow{p}=\overrightarrow{a}-\left(\frac{\overrightarrow{a}.\overrightarrow{n}}{\overrightarrow{b}.\overrightarrow{n}}\right)\overrightarrow{b}\Rightarrow \left(c\right)$