Question

Find the angles between (a) the space diagonals of a cube; (b) a space diagonal and an edge; (c) a space diagonal and a diagonal of a face.

Short answer

(a) The angles between the space diagonals of a cube is$70.5°$ .

(b) The angle between a space diagonal and an edge is$54.7°$ .

(c) The angle between a space diagonal and a diagonal of a face is$35.5°$ .

Step-by-step solution

Concept of the cube:

A cube with unit side length and one side at the origin. Since the cube is symmetric, use only one diagonal, one side, and one diagonal on the face.

Let the side be$\mathbf{A}\mathbf{=}\mathbf{i}$ the face diagonal be$\mathbf{}\mathbf{B}\mathbf{=}\mathbf{i}\mathbf{+}\mathbf{j}$and the diagonal be$\mathbf{}\mathbf{C}\mathbf{=}\mathbf{i}\mathbf{+}\mathbf{j}\mathbf{+}\mathbf{k}$ .$\mathbf{}\mathbf{}\mathbf{A}\mathbf{=}\sqrt{{\mathbf{1}}^{\mathbf{1}}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{B}\mathbf{=}\sqrt{{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{1}}^{\mathbf{2}}}\mathbf{=}\sqrt{\mathbf{2}}$and $\mathbf{C}\mathbf{=}\sqrt{{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{1}}^{\mathbf{2}}}\mathbf{=}\sqrt{\mathbf{3}}\mathbf{.}$

(a) The angles between the space diagonals of a cube:

Another diagonal would point from$(0,1,0)$to$(1,0,1)$, so let this diagonal be $(\mathrm{i}+\mathrm{k})=-\mathrm{i}+\mathrm{j}-\mathrm{k}$

$$\begin{gathered} D=\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{3} \end{gathered}$$

Let the angle between the vectors C and D be ${\theta }_1$.

role="math" localid="1658984390608" $$\begin{gathered} \mathrm{C}.\mathrm{D}={\mathrm{CDcos\theta }}_1\to {\mathrm{cos\theta }}_1 \\ =\frac{\mathrm{C}.\mathrm{D}}{\mathrm{CD}} \\ {\mathrm{cos\theta }}_1=\frac{\mathrm{C}.\mathrm{D}}{\mathrm{CD}} \\ =\frac{\left(1\right)\left(-1\right)+\left(1\right)\left(1\right)+\left(1\right)\left(-1\right)}{\sqrt{3}\times \sqrt{3}} \\ =\frac{-1}{3} \\ {\mathrm{\theta }}_1={\cos }^{-1}\left(-\frac{1}{3}\right) \\ =70.5^{\circ } \end{gathered}$$

(b) The angle between a space diagonal and an edge:

Let the angle between the vectors C and A be ${\theta }_2$.

role="math" localid="1658984500259" $$\begin{gathered} \mathrm{C}.A=CA{\mathrm{cos\theta }}_2\to {\mathrm{cos\theta }}_2 \\ =\frac{\mathrm{C}.A}{CA} \\ {\mathrm{cos\theta }}_2=\frac{\mathrm{C}.A}{CA} \\ =\frac{\left(1\right)\left(1\right)+\left(1\right)\left(0\right)+\left(1\right)\left(0\right)}{\sqrt{3}\times 1} \\ =\frac{1}{\sqrt{3}} \\ {\mathrm{\theta }}_2={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ =54.7^{\circ } \end{gathered}$$

(c) The angle between a space diagonal and a diagonal of a face:

Let the angle between the vectors C and B be ${\theta }_3$.

$$\begin{gathered} \mathrm{C}.B=CB{\mathrm{cos\theta }}_3\to {\mathrm{cos\theta }}_3 \\ =\frac{\mathrm{C}.B}{CB} \\ {\mathrm{cos\theta }}_3=\frac{\mathrm{C}.B}{CB} \\ =\frac{\left(1\right)\left(1\right)+\left(1\right)\left(1\right)+\left(1\right)\left(0\right)}{\sqrt{3}\times \sqrt{2}} \\ =\frac{2}{\sqrt{6}} \\ =\sqrt{\frac{2}{3}} \\ {\mathrm{\theta }}_3={\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right) \\ =35.3^{\circ } \end{gathered}$$