Question

If in a right angle triangle $ACB$, the hypotenuse $AB=p,$ then $\overrightarrow{AB}\cdot \overrightarrow{AC}+\overrightarrow{BC}\cdot \overrightarrow{BA}+\overrightarrow{CA}\cdot \overrightarrow{CB}$ is equal to

• A. $2p^2$
• B. $\frac{p^2}{2}$
• C. $p^2$
• D. None of these

Answer 1

Answer: (C)

$p^2$

In $\Delta ABC$

Hypotenuse AB=P

$\begin{array}{c}\sin B=\frac{AC}{AB}\\ \sin {90}^{\circ }=\frac{AC}{AB}\\ 1=\frac{AC}{P}\\ AC=P,\\ \cos B=\frac{BC}{AB}\\ \cos {90}^{\circ }=\frac{BC}{AB}\\ 0=\frac{BC}{P}\\ 0=BC,\end{array}$

Thus,

$\begin{array}{c}=\overrightarrow{AB}\cdot \overrightarrow{AC}+\overrightarrow{BC}\cdot \overrightarrow{BA}+\overrightarrow{CA}\cdot \overrightarrow{CB}\\ =P\cdot P+0\cdot P+P\cdot 0\\ =P^2\end{array}$

Hence, Option $C$ is the correct answer.