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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 $<0,0,0>$?

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  • $\begingroup$ The vector $\langle 0, 0, 0 \rangle$ has length $0$. How can it be a side of a cube then? $\endgroup$ Jun 8, 2020 at 6:40
  • $\begingroup$ There are multiple diagonals.... this one case is the “easiest” to consider if you are coordinatizing the cube. You could also have used sides $\langle 0, 1, 0 \rangle$ or $\langle 0, 0, 1 \rangle$ $\endgroup$ Jun 8, 2020 at 6:46
  • $\begingroup$ So any initial point of the diagonal I choose, the length would still be 1? in other words, the diagonal on the flat side from one corner to other would be the same as in diagonal of the interior from origin to the corner of the cube? $\endgroup$
    – user314
    Jun 8, 2020 at 6:50
  • $\begingroup$ The answer is 55°. But if I choose the two components <1,1,1> and <1,0,0> and find the angle between those two vectors I get 35°. How would I know which components to choose if the answer varies depending on the chosen components? $\endgroup$
    – user314
    Jun 8, 2020 at 6:58
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    $\begingroup$ The components of a vector are the differences between the coordinates of its endpoint and startpoint. Vector $\langle 0,0,0\rangle$ is NOT a diagonal of your cube. $\endgroup$ Jun 8, 2020 at 7:08

1 Answer 1

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$(0,0,0)$ 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 $$\vartheta = \cos^{-1} \dfrac{\langle (0,1,0),(1,1,1) \rangle}{\sqrt 3} = \cos^{-1} \dfrac{1}{\sqrt 3}$$

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  • $\begingroup$ Why can't I choose the two position vectors to be $\vec a = <1,1,1>$ and $\vec b = <1,1,0>$? If I now find the angle between $\vec a$ and $\vec b$, the angle I get is 35° $<=>$ $cos^{-1}$(2/√6). Your answer is right, but if I choose $\vec b = <1,1,0>$, it is a corner of the cube, but will give a different answer $\endgroup$
    – user314
    Jun 8, 2020 at 7:16
  • $\begingroup$ But how is $(1,1,0)$ an edge of the cube $[0,1]^3$? Did you try drawing this cube? For any corner of a cube there are exactly three adjacent edges. In our case we chose precisely that cube, which edges adjacent to the corner $(0,0,0)$ are the standard basis vectors $(1,0,0),...,(0,0,1)$... $\endgroup$ Jun 8, 2020 at 7:22
  • $\begingroup$ <1,1,0> is not an edge of the cube. You need one of the vectors to be an edge... $\endgroup$ Jun 8, 2020 at 7:23
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    $\begingroup$ You really need to draw the cube... if the diagonal is <1,1,1>, then <0,1,1> can not be an edge... your three edges would be <1,0,0>, <0,1,0>, and <0,0,1> $\endgroup$ Jun 8, 2020 at 7:35
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    $\begingroup$ I think I know now. So to make sure, the only diagonal is <1,1,1> because the initial point of this vector is at the origin and the terminal point is at <1,1,1>. And the edges are <1,0,0>, <0,1,0>, <0,0,1>, and I can choose any one of three these edges along with <1,1,1> as they give the same angle ? $\endgroup$
    – user314
    Jun 8, 2020 at 8:06

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