Question

$P$ is a point on the line through the point $A$ whose position vector is $\overline{a}$ and the line is parallel to the vector $\overline{b}$. If $PA=6$, then the position vector of $P$ is

• A. $\overline{a}+6\overline{b}$
• B. $\overline{a}\pm \frac{6}{\left|\overline{b}\right|}\overline{b}$
• C. $\overline{a}-6\overline{b}$
• D. $\overline{b}+\frac{6}{\left|\overline{a}\right|}\overline{a}$

Answer 1

Answer: (B)

$\overline{a}\pm \frac{6}{\left|\overline{b}\right|}\overline{b}$

Given

$|\overrightarrow{r}-\overrightarrow{a}|=6$......(i)

$\overrightarrow{r}-\overrightarrow{a}=\lambda \overrightarrow{b}$.......(ii)

$|\overrightarrow{r}-\overrightarrow{a}{|}^2={\lambda }^2|\overrightarrow{b}{|}^2$

$\Rightarrow \frac{36}{{\lambda }^2}=|\overrightarrow{b}{|}^2$

$\therefore |\overrightarrow{b}|=\pm \frac{6}{\lambda }$

$\therefore \overrightarrow{r}=\overrightarrow{a}\pm \frac{6\overrightarrow{b}}{|b|}$

$\therefore \overrightarrow{r}=\overrightarrow{a}\pm \frac{6\overrightarrow{b}}{|\overrightarrow{b}|}$