Angle of the body diagonals

Using vector dot product calculate the angle of the body diagonals of the cube.

Correct result:

A =  70.5288 °

Solution:

$$\begin{gathered} D_1=(1,1,1) \\ {D}_2=(1,1,-1) \\ {d}_1=\mathrm{\mid }{D}_1\mathrm{\mid } \\ {d}_1=\sqrt{1^2+1^2+1^2}=\sqrt{3}\doteq 1.7321 \\ {d}_2=\mathrm{\mid }{D}_2\mathrm{\mid } \\ {d}_2=\sqrt{1^2+1^2+(-1{)}^2}=\sqrt{3}\doteq 1.7321 \\ {D}_1\cdot D_2=d_1{d}_2\cos A \\ c=\cos A={\displaystyle \frac{D_1\cdot D_2}{d_1{d}_2}} \\ c={\displaystyle \frac{1\cdot 1+1\cdot 1+1\cdot (-1)}{d_1\cdot d_2}}={\displaystyle \frac{1\cdot 1+1\cdot 1+1\cdot (-1)}{1.7321\cdot 1.7321}}={\displaystyle \frac{1}{3}}\doteq 0.3333 \\ {A}_0=\arccos (c)=\arccos (0.3333)\doteq 1.231\text{ rad } \\ A=A_0\to {}^{\circ }=A_0\cdot {\displaystyle \frac{180}{\pi }}{}^{\circ }=1.23095941734\cdot {\displaystyle \frac{180}{\pi }}{}^{\circ }=70.529{}^{\circ }=70.5288^{\circ }=70^{\circ }31^{\mathrm{\prime }}44\mathrm{"} \end{gathered}$$

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