In a triangle $$ABC$$, if $$|\overrightarrow{BC}| = 3$$, $$|\overrightarrow{CA}| = 5$$ and $$|\overrightarrow{BA}| = 7$$, then the projection of the vector $$\overrightarrow{BA}$$ on $$\overrightarrow{BC}$$ is equal to:

We are given a triangle $$ABC$$ with $$|\overrightarrow{BC}| = 3$$, $$|\overrightarrow{CA}| = 5$$, and $$|\overrightarrow{BA}| = 7$$. We need to find the projection of $$\overrightarrow{BA}$$ onto $$\overrightarrow{BC}$$.

The projection of $$\overrightarrow{BA}$$ on $$\overrightarrow{BC}$$ is given by $$\frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BC}|}$$. To find the dot product, we use the cosine rule at vertex $$B$$.

By the cosine rule applied to side $$CA$$ (opposite to angle $$B$$): $$|CA|^2 = |BA|^2 + |BC|^2 - 2|BA||BC|\cos B$$ $$25 = 49 + 9 - 2(7)(3)\cos B$$ $$25 = 58 - 42\cos B$$ $$\cos B = \frac{33}{42} = \frac{11}{14}$$

The dot product $$\overrightarrow{BA} \cdot \overrightarrow{BC} = |\overrightarrow{BA}||\overrightarrow{BC}|\cos B = 7 \times 3 \times \frac{11}{14} = \frac{231}{14} = \frac{33}{2}$$.

The projection of $$\overrightarrow{BA}$$ on $$\overrightarrow{BC}$$ is: $$\frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BC}|} = \frac{33/2}{3} = \frac{33}{6} = \frac{11}{2}$$