Question

Find the angle between the body diagonals of a cube.

Short answer

The angle between the digonals is $70.{52}^{\circ }$.

Step-by-step solution

Explain the concept and draw the cube using given information

To find the angle between the diagonals, the vector diagonals must be evaluated and then the dot formula must be used.

The cube has 1 unit side and one of the corner is coinciding with the origin. using this information, the cube is drawn as follows:

Assume the vectors

The principle diagonals are $\overrightarrow{OM}$and $\overrightarrow{BA}$. The corrdinates of points O, M, A and B are $O=\left(0,0,0\right),M=\left(1,1,1\right),B=\left(0,1,0\right)$ and $A=\left(1,0,1\right)$.

Find the vectors $\overrightarrow{OM}$and $\overrightarrow{BA}$.

$$\begin{gathered} \overrightarrow{OM}=\left(1-0\right)i+\left(1-0\right)j+\left(1-0\right)k \\ =i+j+k \end{gathered}$$

Solve the vector $\overrightarrow{BA}$.

$$\begin{gathered} \overrightarrow{BA}=(1-0)i+(0-1)j+(1-0)k \\ =i-j+k \end{gathered}$$

Find the dot product between OM→ and AB→ .

The formula of the dot product of the vectors $\overrightarrow{OM}$and $\overrightarrow{BA}$is

$\overrightarrow{OM}\overrightarrow{BA}=\left|OM\right|\left|BA\right|\cos \theta ,\theta$is the angle between the vectos $\overrightarrow{OM}$and $\overrightarrow{BA}$.

Find the dot product of the vectors $\overrightarrow{OM}$and $\overrightarrow{BA}$.

$$\begin{gathered} \overrightarrow{OM}.\overrightarrow{BA}=\left|OM\right|\left|BA\right|\cos \theta \\ 2=3\cos \theta \\ \cos \theta =\frac{2}{3} \\ \theta =70.{52}^{\circ } \end{gathered}$$