Question

If \(\vec{a}\) and \(\vec{b}\) are unit vectors, then the greatest value of
\(\sqrt{3}\,|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|\)
is _______.

Hint:

Use \(|\vec a+\vec b|^2+|\vec a-\vec b|^2=4\) for unit vectors and maximise the linear combination (Cauchy-Schwarz).

Answer 1

Correct Answer: 4

Let \(x=|\vec a+\vec b|\), \(y=|\vec a-\vec b|\). By the parallelogram law, \(x^2+y^2=2(|\vec a|^2+|\vec b|^2)=4\).
We need the maximum of \(\sqrt{3}x+y\) subject to \(x^2+y^2=4\).
By Cauchy-Schwarz: \(\sqrt{3}x+y\le\sqrt{(3+1)(x^2+y^2)}=\sqrt{4\cdot4}=4\).
Equality holds when \(\frac{x}{\sqrt{3}}=\frac{y}{1}\), i.e. \(x=\sqrt{3},\ y=1\), which is attainable (angle between the vectors \(60^\circ\): \(|\vec a+\vec b|=\sqrt{3}\), \(|\vec a-\vec b|=1\)).
Greatest value \(=4\).
Answer: 4