In ABC, |BC|=3, |CA|=5, |BA|=7.

Question

In $	riangle ABC$, $|\vec{BC}|=3$, $|\vec{CA}|=5$, $|\vec{BA}|=7$. The projection of $\vec{BA}$ on $\vec{BC}$ is:

Accepted Answer

By the law of cosines, $\cos(\angle ABC) = \dfrac{BA^2+BC^2-CA^2}{2\cdot BA\cdot BC} = \dfrac{49+9-25}{2\cdot7\cdot3} = \dfrac{33}{42}=\dfrac{11}{14}$.

$$\text{Projection of }\vec{BA}\text{ on }\vec{BC} = |BA|\cos(\angle ABC) = 7\cdot\frac{11}{14} = \frac{11}2$$

Solutions · 1