Question

If the straight line $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ meets the plane $\overrightarrow{r}.\overrightarrow{n}=0$ in $P$. The position vector of $P$ is

• A. $\overrightarrow{a}+\frac{\overrightarrow{a}.\overrightarrow{n}}{\overrightarrow{b}.\overrightarrow{n}}\overrightarrow{b}$
• B. $\overrightarrow{b}+\frac{\overrightarrow{a}.\overrightarrow{n}}{\overrightarrow{b}.\overrightarrow{n}}\overrightarrow{a}$
• C. $\overrightarrow{a}-\frac{\overrightarrow{a}.\overrightarrow{n}}{\overrightarrow{b}.\overrightarrow{n}}\overrightarrow{b}$
• D. None of these

Answer 1

Answer: (C)

$\overrightarrow{a}-\frac{\overrightarrow{a}.\overrightarrow{n}}{\overrightarrow{b}.\overrightarrow{n}}\overrightarrow{b}$

Straight line $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ meets the plane $\overrightarrow{r}.\overrightarrow{n}=0$ in $P$

$P$ satisfy $(\overrightarrow{a}+\lambda \overrightarrow{b}).\overrightarrow{n}=0$

Solving above equation, we get

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

Thus, the position vector of $P$ is $\overrightarrow{a}-\frac{\overrightarrow{a}.\overrightarrow{n}}{\overrightarrow{b}.\overrightarrow{n}}\overrightarrow{b}$.