A) \[4\sqrt{2}\]
B) \[5\sqrt{2}\]
C) \[6\sqrt{2}\]
D) \[3\sqrt{2}\]
Correct Answer: B
Solution :
Since, \[\vec{a},\,\,\vec{b},\,\,\vec{c}\]are perpendicular to \[\vec{b}+\,\vec{c},\,\,\,\,\vec{c}+\vec{a}\]and \[\vec{a}+\vec{b}.\] \[\therefore \] \[\vec{a}.(\vec{b}\,+\,\vec{c})=0,\,\,\vec{b}.(\vec{c}+\vec{a})=0\] and \[\vec{c}.(\vec{a}+\vec{b})=0\] Now, \[{{\left| \vec{a}+\vec{b}+\vec{c} \right|}^{2}}=(\vec{a}+\vec{b}+\vec{c}).(\vec{a}+\vec{b}+\vec{c})\] \[=|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+|\vec{c}{{|}^{2}}+\vec{a}.(\vec{b}+\vec{c})\] \[+\vec{b}.(\vec{a}+\vec{c})+\vec{c}.(\vec{a}+\vec{b})\] \[=9+16+25=50\] \[\Rightarrow \] \[|\vec{a}+\vec{b}+\vec{c}|=\sqrt{50}=5\sqrt{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
