As given by the question
There are given that the vector:
[tex]\vec{v}=\vec{2i}+\vec{3j}[/tex]
Now,
From the formula to find the unit vector in same direction is:
[tex]\vec{u}=\frac{\vec{v}}{\lvert\vec{v}\rvert}[/tex]
Then,
[tex]\begin{gathered} \vec{u}=\frac{\vec{v}}{\lvert\vec{v}\rvert} \\ \vec{u}=\frac{\vec{2i}+\vec{3j}}{\lvert\vec{2i}+\vec{3j}\rvert} \\ \vec{u}=\frac{\vec{2i}+\vec{3j}}{\lvert\sqrt[]{2^2+3^2}\rvert} \end{gathered}[/tex]
Then,
[tex]\begin{gathered} \vec{u}=\frac{\vec{2i}+\vec{3j}}{\sqrt[]{2^2+3^2}} \\ \vec{u}=\frac{\vec{2i}+\vec{3j}}{\sqrt[]{4+9}} \\ \vec{u}=\frac{\vec{2i}+\vec{3j}}{\sqrt[]{13}} \end{gathered}[/tex]
Then,
Rationalize the denominator:
So,
[tex]\begin{gathered} \vec{u}=\frac{\vec{2i}+\vec{3j}}{\sqrt[]{13}} \\ \vec{u}=\frac{\vec{2i}+\vec{3j}}{\sqrt[]{13}}\times\frac{\sqrt[]{13}}{\sqrt[]{13}} \\ \vec{u}=\frac{\vec{\sqrt[]{13}(2i}+\vec{3j})}{13} \\ \vec{u}=\frac{2\sqrt[]{13}}{13}i+\frac{3\sqrt[]{13}}{13}j \end{gathered}[/tex]
Hence, the unit vector is shown below:
[tex]\vec{u}=\frac{2\sqrt[]{13}}{13}i+\frac{3\sqrt[]{13}}{13}j[/tex]
