Say you have a parallelogram which is defined by the to vectors: $\vec u$, $\vec v$. Prove that the area of its projection on a plane with a perpendicular vector $\vec n$ (where $|\vec n|=1$) is: $E=|(\vec u \times \vec v)\ \vec n|$. Now I know that the area of the original parallelogram is: $|\vec u \times \vec v|$, but i can't relate this with the other area, or find it from scratch.
2026-04-09 16:58:54.1775753934
Area of projected parallelogram onto a plane.
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The geometric intuition is that the projected area is equal to the original area multiplied by $\,\cos \theta\,$ where $\,\theta\,$ is the angle between the planes. But $\,\vec u \times \vec v\,$ is a vector along the normal to the plane spanned by$\,(\vec u, \vec v)\,$, so the angle between $\,\vec u \times \vec v\,$ and $\,\vec n\,$ is precisely the angle between the two planes. The dot product of $\,\vec u \times \vec v\,$ with unit vector $\,\vec n\,$ then introduces the projection factor of $\,\cos \theta\,$.
Outline of an algebraic proof (where $\,\vec \cdot \,$ arrows are omitted, and $\, a \cdot b\,$ is the dot product):
the projection of $ u$ onto the normal $\, n\,$ is $\,( u \cdot n)\, n\,$, so the projection onto the given plane orthogonal to $\, n\,$ is $\, u - ( u \cdot n)\, n\,$, and the same goes for $\, v\,$
the projected parallelogram is the parallelogram formed by the projections of the two original vectors, so its area is the magnitude of $\,\left( u - ( u \cdot n)\, n\right) \times \left( v - ( v \cdot n)\, n\right)\,$
the latter simplifies, using the triple product identity $\, a \times ( b \times c)=(a \cdot c)b - (a\cdot b)c\,$, to:
$$\require{cancel} \begin{align} \left( u - ( u \cdot n)\, n\right) \times \left( v - ( v \cdot n)\, n\right) &= u \times v- ( v \cdot n) u \times n - ( u \cdot n) n \times v + \cancel{( u \cdot n)( v \cdot n) n \times n} \\[5px] &= u \times v - \big(( v \cdot n) u - ( u \cdot n) v\big) \times n \\[5px] &= u \times v - \big( n \times \left( u \times v\right)\big) \times n \\[5px] &= \cancel{ u \times v} + \left( n \cdot ( u \times v)\right) n - \cancel{\left( n \cdot n\right) u \times v} \\[5px] &= \left(( u \times v) \cdot n\right) n \end{align} $$