A cube is placed so that one corner is at the origin and three edges are along the x-, y-, and z-axes of a coordinate system (Fig. P1.80). Use vectors to compute (a) the angle between the edge along the z-axis (line ab) and the diagonal from the origin to the opposite corner (line ad), and (b) the angle between line ac (the diagonal of a face) and line ad.

The given data can be expressed below as:

This dot product is described as the algebraic operation which mainly takes the “two equal-length” number and also returns a single piece of the number.

Dividing the dot product of the line with the multiplication of the lines provides the angle between the edges, lines, and the diagonals.

a) The free body diagram of the cube is presented above in Fig. P1.80.

Here, if we assume that the sides of the cube are a then the length of the diagonal will be $\sqrt{2}a$using the Pythagorean theorem, and the angle between the cube are respectively.

As the sides of the cube are a, hence, from the first figure, it can be identified that

role="math" localid="1655790038397" $$\begin{gathered} a\overrightarrow{b}.a\overrightarrow{d}=\left|a\overrightarrow{b}\right|\left|a\overrightarrow{d}\right|\cos \theta \\ \cos \theta =\frac{a^2}{\sqrt{3}{a}^2} \\ \theta ={\cos }^{-1}\frac{1}{\sqrt{3}} \end{gathered}$$

Thus, the angle between the edge and the diagonal is ${\cos }^{-1}\frac{1}{\sqrt{3}}.$

b) a) the free body diagram of the cube can be expressed as:

pic

Here, the sides of the cube are a, $\sqrt{3}a$ and $\sqrt{2}a$are the angle between the cube are $\theta$and $\alpha$respectively.

As the sides of the cube are a, hence, from the first figure, it can be identified that

$a\hat{c}=d\hat{i}+d\hat{k}anda\hat{d}=d\hat{i}+d\hat{j}+d\hat{k}.$

From the rule of the dot product of the vectors, the angle between the edge and the diagonals that can be expressed as:

$$\begin{gathered} a\overrightarrow{c}.a\overrightarrow{d}=\left|a\overrightarrow{c}\right|\left|a\overrightarrow{d}\right|\cos \theta \\ \cos \theta =\frac{\sqrt{2}{a}^2}{\sqrt{3}{a}^2} \\ \theta ={\cos }^{-1}\frac{\sqrt{2}}{\sqrt{3}} \end{gathered}$$

Thus, the angle between the lines is${\cos }^{-1}\left[\frac{\sqrt{2}}{\sqrt{3}}\right].$