Question

Find the angle (in degrees) between the longest edge and the longest diagonal of a 2 by 5 by 6 rectangular box.

Answer 1

Task. Find the angle (in degrees) between the longest edge and the longest diagonal of a 2 by 5 by 6 rectangular box.

Solution. Choose coordinats $(x,y,z)$ so that one of the vertics of the box has coordinates $O=(0,0,0)$, and other vertices adjacent to $O$ are

$A=(2,0,0),\qquad B=(0,5,0),\qquad C=(0,0,6).$

Let $D=(2,5,6)$ be the verex opposite to $O$. Then the longest edge is $OC$, and the longest diagonal is $OD$.

So we have to find the angle $\alpha$ between the vectors

$\vec{OC}=(0,0,6)\qquad\text{and}\qquad\vec{OD}=(2,5,6).$

Notice that their scalar product is

$\vec{OC}\cdot\vec{OD}=0*2+0*5+6*6=36.$

On the other hand, this their scalar product can be computed as follows:

$\vec{OC}\cdot\vec{OD}=|OC|\cdot|OD|\cdot\cos\alpha,$

whence

$\cos\alpha=\frac{\vec{OC}\cdot\vec{OD}}{|OC|\cdot|OD|}.$

Since

$|OC|=\sqrt{0^{2}+0^{2}+6^{2}}=\sqrt{36}=6,$

$|OD|=\sqrt{2^{2}+5^{2}+6^{2}}=\sqrt{65},$

we obtain that

$\cos\alpha=\frac{\vec{OC}\cdot\vec{OD}}{|OC|\cdot|OD|}=\frac{36}{6*\sqrt{65}}=\frac{6}{\sqrt{65}}\approx 0.74421.$

Hence

$\alpha=\arccos 0.74421=41.9{}^{\circ}.$

Answer. $41.9{}^{\circ}$.