# Determine the area of parrollelogram projected onto $xy$- plane, geometrically.

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?

1. $$a_{xy}= a\cos\alpha$$,
2. $$b_{xy}= b\cos\alpha$$,
3. $$\widehat{a_{xy},b_{xy}}=\theta$$,
do not hold in general, and never hold simultaneously provided that $$\alpha\ne0$$ and $$a\nparallel b$$.
On the other hand the relation $$A_{xy}=A\cos\alpha$$ holds for figures of any shape. Indeed, let $$\pi$$ be the line of intersection of two planes separated by angle $$\alpha$$. Consider a rectangle lying in one of the planes with side $$a$$ parallel and side $$b$$ perpendicular to $$\pi$$. Construct the projection of the rectangle on the other plane. The projection will be a rectangle with sides $$a'=a$$ and $$b'=b\cos\alpha$$. Therefore $$A'=A\cos\alpha$$. The same relation is valid for any figure, since it can be split in (infiniticimal) rectangles with sides directed as described above.