A) \[5\sqrt{2}\]
B) 50
C) \[10\sqrt{2}\]
D) 10
E) 20
Correct Answer: D
Solution :
\[\because \]\[|\overrightarrow{a}+\overrightarrow{b}|=6\] \[\Rightarrow \]\[|\overrightarrow{a}{{|}^{2}}+|\overrightarrow{b}{{|}^{2}}+2\overrightarrow{a}.\overrightarrow{b}=36\] ...(i) Similarly, \[|\overrightarrow{b}{{|}^{2}}+|\overrightarrow{c}{{|}^{2}}+2\overrightarrow{b}.\overrightarrow{c}=64\] ...(ii) and \[|\overrightarrow{c}{{|}^{2}}+|\overrightarrow{a}{{|}^{2}}+2\overrightarrow{c}.\overrightarrow{a}=100\] ...(iii) On adding Eqs. (i), (ii) and (iii), we get \[|\overrightarrow{a}{{|}^{2}}+|\overrightarrow{b}{{|}^{2}}+|\overrightarrow{c}{{|}^{2}}+(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=100\] \[|\overrightarrow{a}{{|}^{2}}+|\overrightarrow{b}{{|}^{2}}+|\overrightarrow{c}{{|}^{2}}=100\] ...(iv) \[(\because \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=0)\] Now \[|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{{|}^{2}}=|\overrightarrow{a}{{|}^{2}}+|\overrightarrow{b}|+|\overrightarrow{c}{{|}^{2}}\] \[+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})\] \[\Rightarrow \] \[|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{{|}^{2}}=100\] \[\Rightarrow \] \[|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=10\]
You need to login to perform this action.
You will be redirected in
3 sec
