In Fig. , a cube of edge length a sits with one corner at the origin of an xyz coordinate system. A body diagonal is a line that extends from one corner to another through the center. In unit-vector notation, what is the body diagonal that extends from the corner at (a) coordinates (0,0,0), (b) coordinates (a,0,0), (c) coordinates (0,a,0), and (d) coordinates (a,a,0)? (e) Determine the angles that the body diagonals make with the adjacent edges. (f) Determine the length of the body diagonals in terms of a.

Use the vector subtraction to find the diagonals. Use the trigonometry to find the angle between the body diagonal and adjacent edges. To find the diagonals, use the following equation.

$\stackrel{\rightharpoonup }{a}-\overrightarrow{b}=\left(a_x-b_x\right)\hat{i}+\left(a_y-b_y\right)\hat{j}+\left(a_z-b_z\right)\hat{k}$ (i)

The find the angle, use the following equation,

$\tan \theta =\frac{a_y}{a_x}$(ii)

It is given that length of the edge of cube is a , the origin has coordinates (0,0,0) and the point that diagonally opposite to origin has coordinates (a,a,a) .

The point opposite to origin has coordinates (a,a,a) and has position vector $a\hat{i}+a\hat{j}+a\hat{k}$.

Thus, using the equation (i), the unit vector for the diagonal that extends from coordinates (0,0,0) is $a\hat{i}+a\hat{j}+a\hat{k}$.

The point (a,0,0) corresponds to vector $a\hat{i}$and the diagonally opposite point has coordinates (0,a,a) and has position vector $a\hat{j}+a\hat{k}$.

So, using the equation (i), the vector along the line is role="math" localid="1656302367453" $-a\hat{i}+a\hat{j}+a\hat{k}$.

Therefore, the unit vector for the diagonal that extends from coordinates (a,0,0) is $-a\hat{i}+a\hat{j}+a\hat{k}$.

The point (0,a,0) corresponds to vector $a\hat{j}$and the diagonally opposite point has coordinates (a,0,a) and has position vector $a\hat{i}+a\hat{k}$.

Therefore, using the equation (i) vector along the line is $a\hat{i}-a\hat{j}+a\hat{k}$.

Thus, the unit vector for the diagonal that extends from coordinates (0,a,0) is $a\hat{i}-a\hat{j}+a\hat{k}$.

The point $\left(a,a,0\right)$is corresponds to vector $a\hat{i}+a\hat{j}$and the diagonally opposite point has coordinates (0,0,a) and has position vector $a\hat{k}$.

So, using equation (i), the vector along the line is $-a\hat{i}-a\hat{j}+a\hat{k}$.

Thus, the unit vector for the diagonal that extends from coordinates (a,a,0) is $-a\hat{i}-a\hat{j}+a\hat{k}$.

The vector $a\hat{i}$is parallel to X axis and vector $a\hat{j}+a\hat{k}$is perpendicular to X axis. Using this orientation, find angle between the axis and perpendicular vector. The magnitude of $a\hat{i}$is a and magnitude of $a\hat{j}+a\hat{k}\mathrm{is}a\sqrt{2}$.

From the equation (ii),

$$\begin{gathered} \tan \theta =\frac{a}{a\sqrt{2}} \\ =\frac{1}{\sqrt{2}} \\ \tan \theta =\frac{1}{\sqrt{2}} \\ \theta =54.7° \end{gathered}$$

Thus, the angle between the body diagonal and adjacent side is $54.7°$.

Body diagonal has position vector $a\hat{i}+a\hat{j}+a\hat{k}$

So, its length is $\sqrt{a^2+a^2+a^2}=a\sqrt{3}$

Therefore, the length of the body diagonal is $a\sqrt{3}$.