A) \[\sqrt{3}\]
B) 3
C) 1
D) 0
Correct Answer: A
Solution :
Three mutually perpendicular unit vectors \[=\mathbf{a}\], \[\mathbf{b}\] and \[\mathbf{c}\]. Therefore \[|\mathbf{a}|\,=\,|\mathbf{b}|\,=\,|\mathbf{c}|\,=1\] and \[\mathbf{a}.\mathbf{b}=\mathbf{b}.\mathbf{c}=\mathbf{c}.\mathbf{a}=0\]. We know that \[|\mathbf{a}+\mathbf{b}+\mathbf{c}{{|}^{2}}=(\mathbf{a}+\mathbf{b}+\mathbf{c})\,.\,(\mathbf{a}+\mathbf{b}+\mathbf{c})=\,\,|\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}\] \[+|\mathbf{c}{{|}^{2}}+2(\mathbf{a}\,.\,\mathbf{b}\,\,+\mathbf{b}\,.\,\mathbf{c}\,+\mathbf{c}\,.\,\mathbf{a})=1+1+1+0=3\] or \[|\mathbf{a}+\mathbf{b}+\mathbf{c}|\,=\sqrt{3}.\]
You need to login to perform this action.
You will be redirected in
3 sec
