Show that the position vectors $ a⃗ _1 , a⃗ _2, $ and $ a⃗ _3 $ are the edges of a cube, and determine the cube’s volume.
$ a⃗ _1 = \begin{bmatrix}
10 \\
-5 \\
10
\end{bmatrix} $
$ a⃗ _2 = \begin{bmatrix}
-11 \\
-2 \\
10
\end{bmatrix} $
$ a⃗ _3 = \begin{bmatrix}
-2 \\
-14 \\
-5
\end{bmatrix} $
I searched the web for some time and could not find any correlation between vectors and a cube. I can imagine that there should be some result after pointing out their directions like of they are really the edges or not.
Best Answer
First of all I think $$ a_1 = \begin{bmatrix} 10 \\ -5 \\ 10 \end{bmatrix} $$ since this makes the computations correct.
Method 1
For it to be a cube we need two things
$$\lVert a_1 \rVert = \lVert a_2 \rVert = \lVert a_3 \rVert \tag{1}$$
$$ a_1 \perp a_2 \ \land a_2 \perp a_3 \ \land a_1 \perp a_3 \tag{2}$$
Lets check $(1)$ first,
$$ \lVert a_1 \rVert = \sqrt{10^2 + 5^2 + 10^2} = \sqrt{100 + 25 + 100} = 15$$
$$ \lVert a_2 \rVert = \sqrt{(-11)^2 + (-2)^2 + 10^2} = \sqrt{121 + 4 + 100} = 15$$
$$ \lVert a_3 \rVert = \sqrt{(-2)^2 + (-14)^2 + (-5)^2} = \sqrt{4 + 196 + 25} = 15$$
Weyhoo!
Now we can check $(2)$, i.e. whether $a_i \perp a_j$ by computed
$$ a_i \cdot a_j = \lVert a_i \rVert \rVert a_j \rVert \cos \theta = 0$$ because this means the angle $\theta$ between them is 90 degrees.
$$ a_1 \cdot a_2 = -110 + 10 + 100 = 0 $$ $$ a_2 \cdot a_3 = 22 + 28 - 50 = 0 $$ $$ a_1 \cdot a_3 = -20 + 70 - 5 = 0 $$
Now since we know it is a cube, we simply compute $V = 15^3 = 3375$.
Method 2
A smarter but slightly more advanced method is to compute the volume $V$ of a prism as the vector triple product between the three vectors.
$$ V = \det \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \det \begin{bmatrix} 10 & -5 & 10 \\ -11 & -2 & 10 \\ -2 & -14 & -5 \end{bmatrix} = 10 \cdot \left(10 + 140 \right) - (-5) \cdot \left( 55 + 20\right) + 10 \cdot \left(154 -4 \right) = 1500 + 375 + 1500 = 3375 $$ Now we use the fact that we know a prism attains its highest possible volume when the sides if and only if the sides orthogonal, and that this volume is then equal to the length of the vectors multiplied. Because we computed the volume of the prism to be exactly equal to that of the sides multiplied, we know it must be a cube.