Question

In a triangle $ABC$, if $\left|\overrightarrow{AB}\right|=7$, $\left|\overrightarrow{BC}\right|=5$ and $\left|\overrightarrow{CA}\right|=3$, if the projection of $\overrightarrow{BC}$ on $\overrightarrow{CA}$ is equal to $\frac{n}{2}$, then the value of $n$ is

Answer 1

Finding the projection of $\overrightarrow{BC}$ on $\overrightarrow{CA}$ and the value of $n$

Given the sides of a triangle are,

$\left|\overrightarrow{AB}\right|=a=7$

$\left|\overrightarrow{BC}\right|=b=5$

$\left|\overrightarrow{CA}\right|=c=3$

The figure is as shown:

Projection of $\overrightarrow{BC}$ on $\overrightarrow{CA}$$=BC\cos \left(\theta \right)$, which implies that,

$$\begin{gathered} \left|CM\right|=\left|BC\right|\cos \left(\theta \right) \\ =\left|BC\right|\left|\frac{b^2+c^2-a^2}{2bc}\right| \\ =5\left|\frac{25+9-49}{30}\right| \\ =5\left|\frac{34-49}{30}\right| \\ =5\left|\frac{-15}{30}\right| \\ =\frac{5}{2} \end{gathered}$$

Comparing this with $\frac{n}{2}$, we get, $n=5$

Therefore, the value of $n$ is equal to $5$.