A cube is 3 cm on a side, with one corner at the origin. What is the unit vector pointing from the origin to the diagonally opposite corner at location$<3,3,3>$cm? What is the angle from this diagonal to one of the adjacent edges of the cube?

Side of cube = 3cm

Diagonally opposite corner at

To find the vector that points from the origin to the opposite corner of the cubesubtract the initial location from the final location

cm

The magnitude of this vector is

$\left|\stackrel{\rightharpoonup }{r}\right|=\sqrt{{\left(3\right)}^2+{\left(3\right)}^2+{\left(3\right)}^2}=5.196$cm

The unit vector for that vector is

Recall that the unit vector could be given in terms of the direction cosines

which means:

therefore

$\cos \left({\theta }_x\right),\cos \left({\theta }_y\right),\cos \left({\theta }_z\right)=0.577$

So, the angle from the diagonal vector to one of the adjacent edges is ( $n.b.$the three adjacent edges are on the$x,y,$and $z‐$axis since the side that the origin

$$\begin{gathered} \theta ={\cos }^{-1}\left(0577\right) \\ \theta =54.76° \end{gathered}$$

Where,${\theta }_x={\theta }_y={\theta }_z=\theta$

Therefore, the required vector isand the angle from the diagonal to the adjacent edges of the cube is${\theta }_x={\theta }_y={\theta }_z=\theta =54.76°$.