Magnitude of Difference of Two Equal Vectors

Let the angle between $$\vec a$$ and $$\vec b$$ be $$\theta$$

Since $$|\vec a|=|\vec b|=1,$$

we have

$$|\vec a+\vec b| = \sqrt{|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b}$$

$$=\sqrt{2+2\cos\theta}$$

$$=2\cos\frac\theta2$$

Similarly,

$$|\vec a-\vec b| =\sqrt{2-2\cos\theta}$$

$$=2\sin\frac\theta2$$

Therefore,

$$4|\vec a+\vec b|+|\vec a-\vec b|$$

$$=8\cos\frac\theta2+2\sin\frac\theta2$$

Let

$$x=\frac\theta2$$

Then

$$x\in[0,\pi/2]$$

and expression becomes

$$8\cos x+2\sin x$$

Its maximum value is

$$\sqrt{8^2+2^2}$$

$$=\sqrt{68}$$

$$=2\sqrt{17}$$

Its minimum value in $$[0,\pi/2]$$ is $$2$$

Hence range is $$[2,2\sqrt{17}]$$

Now,

$$2\sqrt{17}\approx8.24$$

Therefore integers in the range are $$2,3,4,5,6,7,8$$

Total integers $$=7$$