Vector Algebra Question 346
Question: $ \mathbf{a},,\mathbf{b} $ and c are three vectors with magnitude $ |\mathbf{a}|,=4, $ $ |\mathbf{b}|,=4, $ $ |\mathbf{c}|,=2 $ and such that $ \mathbf{a} $ is perpendicular to $ (\mathbf{b}+\mathbf{c}),,\mathbf{b} $ is perpendicular to $ (\mathbf{c}+\mathbf{a}) $ and $ \mathbf{c} $ is perpendicular to $ (\mathbf{a}+\mathbf{b}). $ It follows that $ |\mathbf{a}+\mathbf{b}+\mathbf{c}| $ is equal to
[UPSEAT 2004]
Options:
A) 9
B) 6
C) 5
D) 4
Show Answer
Answer:
Correct Answer: B
Solution:
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Here |a|=4; $ |\mathbf{b}|=4;\,|\mathbf{c}|=2 $ and $ \mathbf{a}.(\mathbf{b}+\mathbf{c})=0\Rightarrow \mathbf{a}.\mathbf{b}+\mathbf{a}\mathbf{.c}=0 $ .....(i) $ \mathbf{b}.(\mathbf{c}+\mathbf{a})=0\Rightarrow \mathbf{b}\mathbf{.c}+\mathbf{b}\mathbf{.a}=0 $ .....(ii) $ \mathbf{c}.(\mathbf{a}+\mathbf{b})=0\Rightarrow \mathbf{c}\mathbf{.a}+\mathbf{c}\mathbf{.b}=0 $ .....(iii) Adding (i), (ii) and (iii), we get, $ 2[\mathbf{a}\mathbf{.b}+\mathbf{b}\mathbf{.c}+\mathbf{c}\mathbf{.a}]=0 $ \ $ |\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{|\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+|\mathbf{c}{{|}^{2}}+2(\mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a})} $ $ =\sqrt{|\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+|\mathbf{c}{{|}^{2}}} $ = $ \sqrt{16+16+4} $
Þ $ |\mathbf{a}+\mathbf{b}+\mathbf{c}|=6 $ .
BETA
