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If ∣a⃗∣=3∣b⃗∣|\vec{a}| = 3|\vec{b}|, ∣b⃗∣=2|\vec{b}| = 2 and angle between a⃗\vec{a} and b⃗\vec{b} is 60∘60^\circ, then ∣a⃗−b⃗∣|\vec{a} - \vec{b}| is equal to:

Solution

✅ Correct Option: 2

∣a⃗∣=6|\vec{a}| = 6, ∣b⃗∣=2|\vec{b}| = 2, a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡60∘=6⋅2⋅1/2=6\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos 60^\circ = 6 \cdot 2 \cdot 1/2 = 6. ∣a⃗−b⃗∣2=∣a⃗∣2+∣b⃗∣2−2a⃗⋅b⃗=36+4−12=28|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 \vec{a}\cdot\vec{b} = 36 + 4 - 12 = 28. So ∣a⃗−b⃗∣=28=27|\vec{a} - \vec{b}| = \sqrt{28} = 2\sqrt{7}.

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