Vector Algebra Question 224
Question: If in a right angled triangle ABC, the hypotenuse $ AB=p, $ then $ \overrightarrow{AB}.\overrightarrow{AC}+\overrightarrow{BC},.\overrightarrow{BA}+\overrightarrow{CA}.\overrightarrow{CB} $ is equal to
Options:
A) $ 2p^{2} $
B) $ \frac{p^{2}}{2} $
C) $ p^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- We have $ \overrightarrow{AB},.,\overrightarrow{AC}+\overrightarrow{BC},.,\overrightarrow{BA}+\overrightarrow{CA},.,\overrightarrow{CB} $ $ (AB),(AC)\cos \theta +(BC),(BA)\cos (90^{o}-\theta )+0 $ = $ AB(AC\cos \theta +BC\sin \theta )=AB( \frac{{{(AC)}^{2}}}{AB}+\frac{{{(BC)}^{2}}}{AB} ) $ $ =AC^{2}+BC^{2}=AB^{2}=p^{2} $ .
BETA
