Home  18.013A  Chapter 3  Section 3.3 


Express the square of the area of a parallelogram with sides $\stackrel{\u27f6}{v}$ and $\stackrel{\u27f6}{w}$ in terms of dot products.
Solution:
The area of a parallelogram is the length of its base multiplied by its altitude. If we consider the vector
$\stackrel{\u27f6}{v}$
as its base, then its altitude will be the length of
$\stackrel{\u27f6}{w}$
multiplied by the sine of the angle between
$\stackrel{\u27f6}{v}$
and
$\stackrel{\u27f6}{w}$
. The square of this area will then be the square of the length of
$\stackrel{\u27f6}{v}$
multiplied by the square of the length of
$\stackrel{\u27f6}{w}$
, multiplied by by the square of the sine of the angle between them.
To get this in the form of dot products we use the Pythagorian theorem to write the square of the sine of this angle as 1 minus the square of its cosine.
The result then consists of two terms:
The first is just the product of the squares of the sides, which is $(\stackrel{\u27f6}{v},\stackrel{\u27f6}{v})*(\stackrel{\u27f6}{w},\stackrel{\u27f6}{w})$ .
The second which must be subtracted, can be identified as $(\stackrel{\u27f6}{v},\stackrel{\u27f6}{w})*(\stackrel{\u27f6}{w},\stackrel{\u27f6}{v})$ , and our answer is
AREA of PARALLELOGRAM SQUARED $=(\stackrel{\u27f6}{v},\stackrel{\u27f6}{v})*(\stackrel{\u27f6}{w},\stackrel{\u27f6}{w})(\stackrel{\u27f6}{v},\stackrel{\u27f6}{w})*(\stackrel{\u27f6}{w},\stackrel{\u27f6}{v})$
