Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Area of parallelogram spanned by two vectors

Posted by peeterjoot on August 18, 2009

Parallelogram Area

Parallelogram Area

As depicted in figure, one can see that the area of a parallelogram spanned by two vectors is computed from the base times height. In the figure \mathbf{u} was picked as the base, with length \Vert \mathbf{u} \Vert. Designating the second vector \mathbf{v}, we want the component of \mathbf{v} perpendicular to \hat{\mathbf{u}} for the height. An orthogonal decomposition of \mathbf{v} into directions parallel and perpendicular to \hat{\mathbf{u}} can be performed in two ways.

\begin{aligned}\mathbf{v} &= \mathbf{v} \hat{\mathbf{u}} \hat{\mathbf{u}} = (\mathbf{v} \cdot \hat{\mathbf{u}}) \hat{\mathbf{u}} + (\mathbf{v} \wedge \hat{\mathbf{u}}) \hat{\mathbf{u}} \\     &= \hat{\mathbf{u}} \hat{\mathbf{u}} \mathbf{v} = \hat{\mathbf{u}} (\hat{\mathbf{u}} \cdot \mathbf{v}) + \hat{\mathbf{u}} (\hat{\mathbf{u}} \wedge \mathbf{v}) \end{aligned}

The height is the length of the perpendicular component expressed using the wedge as either \hat{\mathbf{u}} (\hat{\mathbf{u}} \wedge \mathbf{v}) or (\mathbf{v} \wedge \hat{\mathbf{u}}) \hat{\mathbf{u}}.

Multiplying base times height we have the parallelogram area

\begin{aligned}A(\mathbf{u},\mathbf{v}) &= \Vert \mathbf{u} \Vert \Vert \hat{\mathbf{u}} ( \hat{\mathbf{u}} \wedge \mathbf{v} ) \Vert \\ &= \Vert \hat{\mathbf{u}} ( \mathbf{u} \wedge \mathbf{v} ) \Vert \end{aligned}

Since the squared length of an Euclidean vector is the geometric square of that vector, we can compute the squared area of this parallogram by squaring this single scaled vector

\begin{aligned}A^2 &= (\hat{\mathbf{u}} ( \mathbf{u} \wedge \mathbf{v} ) )^2 \end{aligned}

Utilizing both encodings of the perpendicular to \hat{\mathbf{u}} component of \mathbf{v} computed above we have for the squared area

\begin{aligned}A^2&= (\hat{\mathbf{u}}( \mathbf{u} \wedge {\mathbf{v}} ) )^2 \\ &= (( \mathbf{v} \wedge {\mathbf{u}} ) \hat{\mathbf{u}}) (\hat{\mathbf{u}} ( {\mathbf{u}} \wedge \mathbf{v} )) \\ &= ( \mathbf{v} \wedge \mathbf{u} ) ( \mathbf{u} \wedge \mathbf{v} ) \end{aligned}

Since \mathbf{u} \wedge \mathbf{v} = -\mathbf{v} \wedge \mathbf{u}, we have finally

\begin{aligned}A^2 = -( \mathbf{u} \wedge \mathbf{v} )^2 \end{aligned}

There are a few things of note here. One is that the parallelogram area can easily be expressed in terms of the square of a bivector. Another is that the square of a bivector has the same property as a purely imaginary number, a negative square.

It can also be noted that a vector lying completely within a plane anticommutes with the bivector for that plane. More generally components of vectors that lie within a plane commute with the bivector for that plane while the perpendicular components of that vector commute. These commutation or anticommutation properties depend both on the vector and the grade of the object that one attempts to commute it with (these properties lie behind the generalized definitions of the dot and wedge product to be seen later).


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