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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.

ArsenBerk
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Erald Sn
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1 Answers1

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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:

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

dxiv
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  • Hey, great answer. How can i prove this though:

    " 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 $, v,$"

     – Erald Sn Dec 25 '17 at 10:01
  • @EraldSn Every vector $u$ can be written in a unique way as the sum of a vector orthogonal to a given plane and a vector contained in that plane. 1. $,u \cdot n = |u| , |n| \cos \theta = |u| \cos \theta,$ so multiplying by $n$ gives a vector along the normal $n$ of magnitude $|u| \cos \theta$ i.e. the projection of $u$ onto $n,$. 2. $,\big(u - ( u \cdot n)\big), \cdot n$ $= u \cdot n - (u \cdot n)(n \cdot n)$ $= u \cdot n - u \cdot n = 0,$, so $,u - ( u \cdot n), n,$ is orthogonal to $n,$. – dxiv Dec 25 '17 at 17:42