Question

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

• A. $\frac{25}{4}$
• B. $\frac{85}{14}$
• C. $\frac{127}{20}$
• D. $\frac{115}{16}$

Answer 1

The correct option is B

$\frac{85}{14}$

Explanation for the correct option:

Finding projection of $\overrightarrow{AB}$ on $\overrightarrow{AC}$:

Given: In the triangle $ABC$, $\overrightarrow{BC}=8$, $\overrightarrow{CA}=7$, and $\overrightarrow{AB}=10$.

The triangle $ABC$ is as shown,

Let projection of $\overrightarrow{AB}$ on $\overrightarrow{AC}$ be $\overrightarrow{AC}=\left|AN\right|$.

We know that,

$$\begin{gathered} \left|\overrightarrow{AN}\right|=\left|\overrightarrow{AB}\right|\times \cos \theta \\ \left|\overrightarrow{AN}\right|=\left|\overrightarrow{AB}\right|\times \left(\frac{b^2+c^2-a^2}{2bc}\right) \end{gathered}$$.

Now substituting $a$ as $8$, $b$ as $7$, $c$ as $10$, and $\left|\overrightarrow{AB}\right|$ as $10$ we get

$\begin{array}{rcl}\left|\overrightarrow{AN}\right|& =& 10\times \left(\frac{7^2+{10}^2-8^2}{2\times 7\times 10}\right)\\ & =& 10\times \left(\frac{49+100-64}{140}\right)\\ & =& 10\times \left(\frac{149-64}{140}\right)\\ & =& 10\times \left(\frac{85}{140}\right)\\ & =& \frac{85}{14}\end{array}$

Therefore, the projection of $\overrightarrow{AB}$ on $\overrightarrow{AC}$ is $\frac{85}{14}$.

Hence, option (B) is the correct answer.