Area of Projection of Parallelogram

Question

We are given a parallelogram $ABCD$ with $AB=\vec{u}$ and $AD=\vec{v}$. We know then that the area of $ABCD$ is given by $|\vec{u} \times \vec{v}|$. Show that the projection of $ABCD$ to a plane which has a vertical vector $\vec{n}$ with $|\vec{n}|=1$ has an area which is equal to $|(\vec{u} \times \vec{v})\dot{} \vec{n}|$ .

Answer 1

It seems reasonable to find the components of $\vec u$ and $\vec v$ on the plane with normal $\vec n$ and then take the cross product of those components. Here are two useful identities in analytic geometry which I will make use of them in the sequel

$$\vec v = {\vec v}_n + {\vec v}_t = (\vec v \cdot \vec n) \vec n - \vec n \times (\vec n \times \vec v)$$

$$\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\vec b - (\vec a \cdot \vec b)\vec c$$

And hence

$${\vec u}_t = - \vec n \times (\vec n \times \vec u) \\ {\vec v}_t = - \vec n \times (\vec n \times \vec v)$$

Now, we may evaluate the cross product of tangential components to get

$$\begin{align} {\vec u}_t \times {\vec v}_t &= [\vec n \times (\vec n \times \vec u)] \times [\vec n \times (\vec n \times \vec v)] \\ &= \{ [\vec n \times (\vec n \times \vec u)] \cdot (\vec n \times \vec v) \} \vec n - \{ [\vec n \times (\vec n \times \vec u)] \cdot \vec n \} (\vec n \times \vec v) \\ &= \{ [\vec n \times (\vec n \times \vec u)] \cdot (\vec n \times \vec v) \} \vec n - 0 (\vec n \times \vec v) \\ &= \{ [-\vec u + (\vec u \cdot \vec n) \vec n] \cdot (\vec n \times \vec v) \} \vec n \\ &= \{ -\vec u \cdot (\vec n \times \vec v) + (\vec u \cdot \vec n) \vec n \cdot (\vec n \times \vec v)\} \vec n \\ &= \{ -\vec u \cdot (\vec n \times \vec v) + 0\} \vec n \\ &= \{ (\vec u \times \vec v) \cdot \vec n \} \vec n \end{align}$$

and finally

$$ \boxed{ {\vec u}_t \times {\vec v}_t = [(\vec u \times \vec v) \cdot \vec n] \vec n}$$

where ${\vec u}_t$ and ${\vec v}_t$ are two sides of the projected parallelogram. I think that you can see the picture now! :)

Answer 2

By definition along the reference direction $z$ projection

$$ \dfrac{A_{\text{projected}}}{A _{\text{true}}} = \cos \theta = ( n \cdot z) / (|n||z|)$$

So to find projected area we have to take further dot product of $\vec{u} \times \vec{v}$ and $\vec{n}. $

which is a scalar triple product, in fact the triple product is Volume ( of the skewed parellelopiped),$ (u \times v \cdot z ) $, for which you have given unity as the normal vector.