Question: Find the angle between the diagonal of a cube and one of its edges.
I know how to do this question but in particular I do not understand why $\vec b=<1,0,0>$ rather than $<0,0,0>$
For this question, I assumed that the cube has side lengths of 1 and is placed such that its left corner is at the origin $<0,0,0>$. So the vectors $\vec b$ and $\vec a$ have components $<0,0,0>$ and $<1,1,1>$ respectively, where $\vec b$ is the bottom left corner of the cube at the origin and $\vec a$ is the position vector from the origin to the top right corner of the cube.
So the angle between the diagonal and one of its edges is
$$θ=\cos^{-1}\frac{\vec a.\vec b}{|\vec a||\vec b|}=\cos^{-1}\frac{<1,1,1>.<0,0,0>}{√3}=\cos^{-1}\frac{0}{√3}=90°$$
But this answer is wrong as $\vec b=<1,0,0>$.
So my question is why is $\vec b=<1,0,0>$ and not $<0,0,0>$?
Best Answer
$(0,0,0)$ is the vector from $(0,0,0)$ to $(0,0,0)$ so it has length 0 and in particular cannot possibly represent an edge of the cube.
By the symmetry of the cube we can assume that without loss of generality it is given by the standard cube $[0,1]^3$. Moreover it does not matter which diagonal and which edge we choose to compare, as long as they share a point. Hence we may take the diagonal $(1,1,1)$ and any of the edges $(1,0,0),$ $(0,1,0)$ or $(0,0,1)$. Indeed we find that $$\langle (1,0,0),(1,1,1)\rangle = ... = \langle (0,0,1),(1,1,1) \rangle = 1$$ Thus we can compute $$\vartheta = \cos^{-1} \dfrac{\langle (0,1,0),(1,1,1) \rangle}{\sqrt 3} = \cos^{-1} \dfrac{1}{\sqrt 3}$$