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If ∣a⃗∣=2,∣b⃗∣=7|\vec a|=2, |\vec b|=7 and a⃗×b⃗=3i^+2j^+6k^\vec a\times\vec b=3\hat i+2\hat j+6\hat k, then angle between a⃗\vec a and b⃗\vec b is:

Solution

✅ Correct Option: 1

∣a⃗×b⃗∣=9+4+36=7|\vec a\times\vec b|=\sqrt{9+4+36}=7.

Since ∣a⃗∣∣b⃗∣sin⁡θ=14sin⁡θ=7|\vec a||\vec b|\sin\theta=14\sin\theta=7, we get sin⁡θ=12\sin\theta=\frac{1}{2}, so θ=π6\theta=\frac{\pi}{6}.

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