In a triangle $$ABC$$, if $$|\overrightarrow{BC}| = 8$$, $$|\overrightarrow{CA}| = 7$$, $$|\overrightarrow{AB}| = 10$$, then the projection of the vector $$\overrightarrow{AB}$$ on $$\overrightarrow{AC}$$ is equal to :

In triangle $$ABC$$, we have $$|\overrightarrow{BC}| = a = 8$$, $$|\overrightarrow{CA}| = b = 7$$, and $$|\overrightarrow{AB}| = c = 10$$. The projection of $$\overrightarrow{AB}$$ on $$\overrightarrow{AC}$$ is $$\frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|}$$.

Using the cosine rule, $$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{49 + 100 - 64}{2 \cdot 7 \cdot 10} = \frac{85}{140} = \frac{17}{28}$$.

Now $$\overrightarrow{AB} \cdot \overrightarrow{AC} = |\overrightarrow{AB}||\overrightarrow{AC}|\cos A = 10 \cdot 7 \cdot \frac{17}{28} = \frac{1190}{28} = \frac{85}{2}$$.

The projection is $$\frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|} = \frac{85/2}{7} = \frac{85}{14}$$.