If a,b are unit vectors, the greatest value of 3∣a+b∣+∣a−b∣ is \_\_\_\_\_\_\_.
Answer: 4
Solutions · 1
Official solution
With θ the angle between a,b: ∣a+b∣=2cos2θ, ∣a−b∣=2sin2θ.
\[\sqrt{3}|\vec a+\vec b|+|\vec a-\vec b| = 2\sqrt{3}\cos\frac\theta2+2\sin\frac\theta2\]
Maximum value =(23)2+22=16=4.