A) 48
B) 16
C) 4
D) 2
Correct Answer: B
Solution :
[b] We have, \[\text{\vec{a}}\,\text{.\vec{b}}=|\vec{a}||\vec{b}|cos\frac{\pi }{6}\]\[=4\times 2\times \frac{\sqrt{3}}{2}=4\sqrt{3}.\] \[\text{Now, }{{\left( \text{\vec{a}}\times \text{\vec{b}} \right)}^{2}}+{{\left( \text{\vec{a}}\,\text{.}\,\text{\vec{b}} \right)}^{2}}={{\text{a}}^{2}}{{\text{b}}^{2}};\]\[\Rightarrow {{\left( \text{\vec{a}}\times \text{\vec{b}} \right)}^{2}}+48=16\times 4\Rightarrow {{\left( \text{\vec{a}}\times \text{\vec{b}} \right)}^{2}}=16\]
You need to login to perform this action.
You will be redirected in
3 sec
