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Two vectors a⃗\vec{a} and b⃗\vec{b} are inclined at angle θ=60°\theta=60° such that ∣a⃗∣=1|\vec{a}|=1, ∣b⃗∣=2|\vec{b}|=2, then ∣(a⃗+3b⃗)×(3a⃗−b⃗)∣2\left|(\vec{a}+3\vec{b})\times(3\vec{a}-\vec{b})\right|^2 is equal to

Solution

✅ Correct Option: 1

Expanding, the cross products of a vector with itself vanish, leaving −a⃗×b⃗+9(b⃗×a⃗)⋅(−1)-\vec a\times\vec b+9(\vec b\times\vec a)\cdot(-1) terms, i.e. −10(a⃗×b⃗)-10(\vec a\times\vec b). Since ∣a⃗×b⃗∣=1⋅2sin⁡60°=3|\vec a\times\vec b|=1\cdot2\sin60°=\sqrt3, the magnitude is 10310\sqrt3 and its square is 300300.

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