Let \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a}|=1 \), \( |\vec{b}|=2 \) and angle between \( \vec{a} \) and \( \vec{b} \) is \( 120^{\circ} \), then the value of \( \left|(\vec{a}+3\vec{b})\times(3\vec{a}-\vec{b})\right|^{2} \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 300
Expanding the cross product, with \( \vec{a}\times\vec{a}=\vec{b}\times\vec{b}=\vec{0} \) and \( \vec{b}\times\vec{a}=-\vec{a}\times\vec{b} \):
\( \displaystyle (\vec{a}+3\vec{b})\times(3\vec{a}-\vec{b}) \)
\( \displaystyle =-\vec{a}\times\vec{b}+9(\vec{b}\times\vec{a}) \)
\( \displaystyle =-10(\vec{a}\times\vec{b}) \)
The magnitude of \( \vec{a}\times\vec{b} \) is
\( \displaystyle |\vec{a}||\vec{b}|\sin 120^{\circ}=1\cdot 2\cdot\frac{\sqrt{3}}{2} \)
\( \displaystyle =\sqrt{3} \)
Therefore the required value is \( 10^{2}\times 3=300 \).