I'm working through a proof on the equivalence between the vector component formula and the sin formula for the cross product of two vectors, $a$ and $b$.
One point in the proof involves finding the area of the parallelogram of $a$ and $b$, angle between them $\theta$, when it is projected onto the $xy$ plane. Call the area of our original parallelogram, $Q$, and the area of the projected one, $P$.
After determining that the angle which the cross product is rotated away from the z-axis ($\alpha$) is the same angle the plane of the parallelogram is rotated away from the xy plane, the presenter states without justification that the area of the projected parallelogram ($P$) is equal to: $Q \cos(\alpha)$.
This is holding me up since I cannot justify this myself, and in my attempt to do so I come up with a different result for $P$: by identifying the side lengths for P as the adjacent sides of right triangle with hypotenuse lengths $a$ and $b$, the expression I would get to describe the area $P$ would be: $$ a \cos(\alpha) \cdot b \cos(\alpha) \cdot \sin(\theta) = Q \cos^2(\alpha) $$ For reference the proof is described here (timestamped to point of interest): https://youtu.be/cXKDJ7_rmyM?t=4603
He seems to make an error by writing what should be cosine down as sine (unless I am grossly mistaken), but looking past that I cannot justify or discover any geometric method to find $Q$ as anything other than what she above. Is he justified in saying the area should be $Q \cos(\alpha)$, or am I correct with the above formulation?