1.Vectors a⃗\vec{a}a and b⃗\vec{b}b make an angle θ=2π3\theta = \dfrac{2\pi}{3}θ=32π. If ∣a⃗∣=1|\vec{a}| = 1∣a∣=1, ∣b⃗∣=2|\vec{b}| = 2∣b∣=2 then {(a⃗+3b⃗)×(3a⃗−b⃗)}2=\{(\vec{a} + 3\vec{b}) \times (3\vec{a} - \vec{b})\}^2 ={(a+3b)×(3a−b)}2=a.225225225b.250250250c.275275275d.300300300Login to continueOnly logged in users canattempt or see the solution.
