In $\mathrm{△}\mathrm{ABC}\text{, if }\left|\overline{\mathrm{BC}}\right|=3,\left|\overline{\mathrm{CA}}\right|=5\text{ and }\left|\overline{\mathrm{BA}}\right|=7$, then the projection of the vector $\overline{BA}$ on $\overline{BC}$ is equal to

Given, $\mathrm{△}\mathrm{ABC},\left|\mathrm{BC}\right|=3,\left|\mathrm{CA}\right|=5,\left|\mathrm{BA}\right|=7$

Now, projection of BA on $\mathrm{BC}=\left|\mathrm{BA}\right|\cos \mathrm{\angle }\mathrm{ABC}$

$\begin{array}{c}\mathrm{Now}, \cos \mathrm{\angle }\mathrm{ABC}=\frac{(7{)}^2+(3{)}^2-(5{)}^2}{2\cdot 7\cdot 3}\\ =\frac{49+9-25}{42}=\frac{33}{42}=\frac{11}{14}\end{array}$

$\begin{array}{c}\therefore \text{ Projection }=\left|\mathrm{BA}\right|\cdot \frac{11}{14}\\ =7\times \frac{11}{14}=\frac{11}{2}\end{array}$