A) 3
B) \[\sqrt{3}\]
C) \[(\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{3}})}/3\]
D) 1
Correct Answer: B
Solution :
Since \[\hat{a},\text{ }\hat{b}\] and \[\hat{c}\] are mutually perpendicular to each other, then \[\hat{a}-\hat{b}=\hat{b}\,.\,\,\hat{c}=\hat{c}\,.\,\,\hat{a}=0\] \[{{\left| \hat{a}+\hat{b}+\hat{c} \right|}^{2}}=(\hat{a}+\hat{b}+\hat{c}).(\hat{a}+\hat{b}+\hat{c})\] \[[\because \,\,|\hat{a}{{|}^{2}}=\hat{a}\,\,.\,\,\hat{a}]\] \[=\,\,\,|\hat{a}{{|}^{2}}+|\hat{b}{{|}^{2}}+|\hat{c}{{|}^{2}}\] \[+\,\,2\,\,(\hat{a}\,\,.\,\,\hat{b}+\hat{b}\,\,.\,\,\hat{c}+\hat{c}\,.\,\,\hat{a})\] \[=1+1+1+2\,\,(0+0+0)\] \[[\ \because \,\,\hat{a}\,\,.\,\,\hat{b}=\hat{b}\,\,.\,\,\hat{c}=\hat{c}\,\,.\,\,\,\hat{a}=0]\] \[\Rightarrow \,\,\,\,|\hat{a}+\hat{b}+\hat{c}{{|}^{2}}=3\] \[\therefore \,\,\,\,|\hat{a}+\hat{b}+\hat{c}|=\sqrt{3}\]
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