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If a⃗\vec a and b⃗\vec b are two unit vectors and ∣a⃗−b⃗∣=3|\vec a-\vec b|=\sqrt3, then the value of ∣a⃗+b⃗∣|\vec a+\vec b| is:

Solution

✅ Correct Option: 1

For unit vectors, ∣a⃗−b⃗∣2=2−2a⃗⋅b⃗=3|\vec a-\vec b|^2=2-2\vec a\cdot\vec b=3, so a⃗⋅b⃗=−12\vec a\cdot\vec b=-\frac{1}{2}.

Then ∣a⃗+b⃗∣2=2+2a⃗⋅b⃗=1|\vec a+\vec b|^2=2+2\vec a\cdot\vec b=1, giving ∣a⃗+b⃗∣=1|\vec a+\vec b|=1.

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