Determine the moment produced by force F about the diagonal AF of the rectangular block. Express the result as a Cartesian vector.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
Solution:
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We will use the unit vector AF and the position vector AB to find the moment produced by F about the diagonal AF. Note that you can also use a position vector from F to B (try it yourself).
Our position vector, r_{AB} is:
r_{AB}=\left\{0i+3j+0k\right\} m
Let us now calculate the unit vector AF. To do so, we need to find a position vector from A to F.
r_{AF}=\left\{3i+3j-1.5k\right\}
The magnitude of this position vector is:
magnitude of r_{AF}\,=\,4.5
We can now calculate the unit vector :(don’t remember?)
Now that we have both a position vector and a unit vector, we can find the moment about the diagonal AF. This can be found by:
M_{AF}=\begin{bmatrix}\dfrac{3}{4.5}&\dfrac{3}{4.5}&\dfrac{1.5}{4.5}\\0&3&0\\-6&3&10\end{bmatrix}
(Remember, force F is already given to us as a Cartesian vector in the question. Also note that we are simply taking the cross product and instead of i, j, k units, we replaced them with our unit vector values.)
M_{AF}=14 N•M