Kinematics Question 601
Question: If $ |\vec{a}|=4,|\vec{b}|=2 $ and the angle between $ \vec{a} $ and $ \vec{b} $ it’s $ \pi /6 $ then $ {{(\overrightarrow{a}\times \overrightarrow{b})}^{2}} $ is equal to
Options:
A) 48
B)16
C) 4
D)2
Show Answer
Answer:
Correct Answer: B
Solution:
[b] We have, $ \vec{a} {.\vec{b}}=|\vec{a}||\vec{b}|cos\frac{\pi }{6} $
$ =4\times 2\times \frac{\sqrt{3}}{2}=4\sqrt{3}. $
Now $ {{( \vec{a}\times \vec{b} )}^{2}}+{{( \vec{a}\text{.}\vec{b} )}^{2}}={{a^{2}}}{{b^{2}}}; $
$ \Rightarrow {{( \vec{a}\times \vec{b} )}^{2}}+48=16\times 4\Rightarrow {{( \vec{a}\times \vec{b} )}^{2}}=16 $
