Vector Algebra Question 321
Question: If a, b, c are unit vectors such that $ \mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}, $ then $ \mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a}= $
[MP PET 1988; Karnataka CET 2000; UPSEAT 2003, 04]
Options:
A) 1
B) 3
C) ? 3/2
D) 3/2
Show Answer
Answer:
Correct Answer: C
Solution:
- Squaring $ (\mathbf{a}+\mathbf{b}+\mathbf{c})=\mathbf{0}, $ we get $ {{\mathbf{a}}^{2}}+{{\mathbf{b}}^{2}}+{{\mathbf{c}}^{2}}+2\mathbf{a}.\mathbf{b}+2\mathbf{b}.\mathbf{c}+2\mathbf{c}.\mathbf{a}=0 $
Þ $ |\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+|\mathbf{c}{{|}^{2}}+2(\mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a})=0 $
Þ $ 2(\mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a})=-3 $
$ \Rightarrow \mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a}=-\frac{3}{2} $ .
BETA
