Area of projected parallelogram onto a plane. 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.

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:
$$\require{cancel}
\begin{align}
\left( u - ( u \cdot  n)\, n\right) \times \left( v - ( v \cdot  n)\, n\right) &=  u \times  v- ( v \cdot  n) u \times  n - ( u \cdot  n)  n \times  v + \cancel{( u \cdot  n)( v \cdot  n)  n \times  n} \\[5px]
 &=  u \times  v - \big(( v \cdot  n) u - ( u \cdot  n) v\big) \times  n \\[5px]
 &=  u \times  v - \big( n \times \left( u \times  v\right)\big) \times  n \\[5px]
 &= \cancel{ u \times  v} + \left( n \cdot ( u \times  v)\right)  n - \cancel{\left( n \cdot  n\right) u \times  v} \\[5px]
 &= \left(( u \times  v) \cdot  n\right)  n
\end{align}
$$
