Question

If in a right angled triangle ABC, the hypotenuse $AB = p,$ then $\overset{\rightarrow}{AB}.\overset{\rightarrow}{AC} + \overset{\rightarrow}{BC}.\overset{\rightarrow}{BA} + \overset{\rightarrow}{CA}.\overset{\rightarrow}{CB}$ is equal to

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

Answer 1

Answer: (C)

$p^{2}$

Answer 2

We have $\overset{\rightarrow}{AB}.\overset{\rightarrow}{AC} + \overset{\rightarrow}{BC}.\overset{\rightarrow}{BA} + \overset{\rightarrow}{CA}.\overset{\rightarrow}{CB}$

$(AB)(AC)\cos\theta + (BC)(BA)\cos(90^{o} - \theta) + 0$

=$AB(AC\cos\theta + BC\sin\theta) = AB\left( \frac{(AC)^{2}}{AB} + \frac{(BC)^{2}}{AB} \right)$

$= AC^{2} + BC^{2} = AB^{2} = p^{2}$.