Vector Algebra Question 190
Question: If $ |\mathbf{a}|=3,,|\mathbf{b}|=1,|\mathbf{c}|=4 $ and $ \mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}, $ then $ \mathbf{a},.,\mathbf{b}+\mathbf{b},.,\mathbf{c}+\mathbf{c},.,\mathbf{a}= $
[MP PET 1995; RPET 2000]
Options:
A) ? 13
B) ? 10
C) 13
D) 10
Show Answer
Answer:
Correct Answer: A
Solution:
- $ {{(\mathbf{a}+\mathbf{b}+\mathbf{c})}^{2}}=0 $
Þ $ |\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 $ 9+1+16+2(\mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a})=0 $
Þ $ \mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a}=-\frac{26}{2}=-13. $
BETA
