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?


Your result is wrong because the equations:

  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.


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