If $a$ is a vector that makes equal angles with ${\mathbf i},{\mathbf j},{\mathbf k}$ and has magnitude $3$, then find the angle of $a$ with either of these unit vectors?
Wouldn't the answer simply be $120$ degrees? However, the answer is given us $\cos^{-1} (1/\sqrt{3})$.
Let $a=(x,y,z)$ We know that $a\cdot (1,0,0)=a\cdot (0,1,0)=a\cdot (0,0,1)$ and so $x=y=z$ so $a=\lambda(1,1,1)$ for some $\lambda\in\mathbb{R}$. Given that $\|a\|=3$ then we have $\sqrt{3\lambda^2}=3$ so $\lambda=\sqrt{3}$.
So, $a\cdot (1,0,0)=\sqrt{3}$ and using the dot product formula, this means that $$\sqrt{3}=\|a\| \|(1,0,0)\|\cos\theta=(3)(1)\cos\theta$$ so $\cos\theta=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$. Finally then $\theta=\cos^{-1}\frac{1}{\sqrt{3}}$.