Find the Unit Vector and Position Vector on a Given Line

Expert intro: This is a standard vector-scaling problem: first normalize the direction, then use the required length to build the new vector on the same line.

Detailed walkthrough

Given

$\overrightarrow{OA}=4i+3j.$

a) Unit vector in the direction of $\overrightarrow{OA}$

First find the magnitude:

$|\overrightarrow{OA}|=\sqrt{4^2+3^2}=\sqrt{16+9}=5$

So the unit vector is

$\frac{1}{5}(4i+3j)=\frac45 i+\frac35 j.$

b) Find $\overrightarrow{OC}$

Since $C$ lies on $OA$, $\overrightarrow{OC}$ has the same direction as $\overrightarrow{OA}$. Its magnitude is

$|\overrightarrow{OC}|=\frac{16}{5}.$

Hence

$\overrightarrow{OC}=\frac{16}{5}\left(\frac45 i+\frac35 j\right).$

Multiply through:

$\overrightarrow{OC}=\frac{64}{25}i+\frac{48}{25}j.$

So,

$\boxed{\text{a) } \frac45 i+\frac35 j \quad\text{and}\quad \text{b) } \frac{64}{25}i+\frac{48}{25}j}$

💡 Pitfall guide

The main trap is mixing up the direction vector with the unit vector. You must divide by the magnitude $5$ before scaling to length $16/5$. If you skip that normalization, the vector length will come out wrong.

🔄 Real-world variant

If $C$ were on the opposite side of $O$ along the same line, the unit vector would be the negative of the one above, and $\overrightarrow{OC}$ would be the same length but with opposite sign. The method is identical; only the direction changes.

🔍 Related terms

unit vector, magnitude, scalar multiple