Question

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

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

Answer 1

The correct option is B

$a+\left(\frac{6}{\left|b\right|}\right)b$

Explanation for the correct option:

To find the position vector of $P$:

Let the position vector of $P$ be $\overrightarrow{r}$.

Given,

$\left|PA\right|=6\to \left(i\right)$

That the line $PA$ is parallel to the vector $\overrightarrow{b}$

$\Rightarrow PA=k\overrightarrow{b}\to \left(ii\right)$

Since $a$ be the position vector of a point $A$,

Then,

$PA=\overrightarrow{r}–\overrightarrow{a}\to \left(iii\right)$

Equating equation $\left(ii\right)$ and $\left(iii\right)$

$$\begin{gathered} \Rightarrow \overrightarrow{r}–\overrightarrow{a}=k\overrightarrow{b} \\ \Rightarrow \overrightarrow{r}=\overrightarrow{a}+k\overrightarrow{b}\to \left(iv\right) \end{gathered}$$

Solve the equation $\left(ii\right)$ and find the value of $k$

$$\begin{gathered} \Rightarrow PA=k\overrightarrow{b} \\ \Rightarrow \left|PA\right|=\left|k\overrightarrow{b}\right|\left[\text{taking modulus on both sides}\right] \\ \Rightarrow \left|PA\right|=k\left|\overrightarrow{b}\right|\left[\because \left|k\right|=k;\text{where}k\text{is a scalar}\right] \\ \Rightarrow 6=k\left|\overrightarrow{b}\right|\left[\because \left|PA\right|=6\right] \\ \Rightarrow k=\frac{6}{\left|\overrightarrow{b}\right|} \end{gathered}$$

Substitute $k=\frac{6}{\left|\overrightarrow{b}\right|}$ in equation $\left(iv\right)$

$\overrightarrow{r}=\overrightarrow{a}+\left(\frac{6}{\left|\overrightarrow{b}\right|}\right)\overrightarrow{b}$

Hence, option (B) is the correct answer.