Two vectors \( \vec{a} \) and \( \vec{b} \) are inclined at angle \( \theta=60^\circ \) such that \( |\vec{a}|=1,\ |\vec{b}|=2 \) , then \( \left|(\vec{a}+3\vec{b})\times(3\vec{a}-\vec{b})\right|^2 \) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 300
Since \( \vec{a}\times\vec{a}=\vec{b}\times\vec{b}=\vec{0} \) and \( \vec{b}\times\vec{a}=-\vec{a}\times\vec{b} \), the cross product expands as
\( \displaystyle (\vec{a}+3\vec{b})\times(3\vec{a}-\vec{b})=3\,\vec{a}\times\vec{a}-\vec{a}\times\vec{b}+9\,\vec{b}\times\vec{a}-3\,\vec{b}\times\vec{b} \)
\( \displaystyle =-\vec{a}\times\vec{b}+9\,\vec{b}\times\vec{a} \)
\( \displaystyle =-10\,(\vec{a}\times\vec{b}) \)
\( \displaystyle |\vec{a}\times\vec{b}|=|\vec{a}|\,|\vec{b}|\sin 60^\circ \)
\( \displaystyle =1\times 2\times\frac{\sqrt{3}}{2} \)
\( \displaystyle =\sqrt{3} \)
\( \displaystyle \left|(\vec{a}+3\vec{b})\times(3\vec{a}-\vec{b})\right|^2=100\,|\vec{a}\times\vec{b}|^2 \)
\( \displaystyle =100\times 3 \)
\( \displaystyle =300 \)