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 $\overline{b}$ lf $PA=6$, then the position vector of $P$ is

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

Answer 1

The correct option is B $\overline{a}+\frac{6}{|\overline{b}|}\overline{b}$

$\overrightarrow{PA}=k\overrightarrow{b}$
$\left|\overrightarrow{PA}\right|=6$
$k\left|\overrightarrow{b}\right|=6$
$k=\frac{6}{\left|\overrightarrow{b}\right|}$
$\overrightarrow{AP}=\frac{6}{\left|\overrightarrow{b}\right|}\overrightarrow{b}$
$-\overrightarrow{a}+\overrightarrow{p}=\frac{6\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$
$\overrightarrow{a}+\frac{6\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\overrightarrow{p}$