Area of parallelogram and volume of parallelpiped using cross-product

Question

I just started my vector calculus course a couple of weeks ago and I understand that the area of a parallelogram is Area = |A x B|, where A and B are position vectors (base and a side of the parallelogram).

What I need help with is visualization. I know that A x B creates a third vector orthogonal to A and B, how is its magnitude equivalent to the parallelogram's area?

Same goes for the parallelpiped's volume V = A • (B x C) where A, B, and C are vectors.

Where do these equations derive from and how do I visualize the process?

Please bear with me because I only understand things once I can visualize them.

All the best!

Answer 1

Consider the picture below of a triangle with sides a and b. The angle between a and b is $\theta$ and the coordinates of the vectors with these sides are shown.

The area is

1. $\frac{1}{2}bh$ = $\frac{1}{2}basin(\theta)$

2. The area of the rectangle minus the area of the red, blue and green triangles. This is the modulus of $\frac{1}{2}(x_1y_2-y_1x_2)$.

Multiply by 2 to get the area of the corresponding parallelogram. This second form is precisely the modulus of the cross product of vectors a and b. A second form for this modulus is $absin(\theta)$. Exactly the same relationship holds in 3D when you work in the plane defined by the 3 triangle vertices.

The volume of a parallelpipid is the base area multiplied by the height. If a and b are in the horizontal plane and c is a third vector lying above the plane then its vertical component is $c cos(\alpha)$, where $\alpha$ is the angle it makes with the horizontal. This is simply c.k, where k is the unit vector in the vertical direction. The volume of the parallelpipid is then $c.k(x_1y_2-y_1x_2) = c.(a\times{b})$

Answer 2

The cross product $a\times b$ of two vectors in ${\mathbb R}^3$ can be defined in various ways. Of course one has to prove that all these definitions mean the same vector. However $a\times b$ was defined in your case, in the end you should be able to work freely with all descriptions of $a\times b$, depending on the problem at hand.

Assume that two vectors $a$, $b\in{\mathbb R}^3$ are given.

(I) Let $a=(a_1,a_2,a_3)$, $\>b=(b_1,b_2,b_3)$. Then $$a\times b:=(a_2b_3-a_3b_2 ,a_3b_1-a_1b_3 ,a_1b_2-a_2b_1)\ .$$ (II) If $a$ and $b$ are linearly dependent then $a\times b:=0$. Otherwise $a$ and $b$ span a two-dimensional subspace $U\subset{\mathbb R}^3$. Choose a unit vector $n\in U^\perp$ such that the triple $(a,b,n)$ is "positively oriented". Then $$a\times b:= {\rm area}(P)\ n\ ,$$ where $$P:=\{u\>a+v\>b\>|\>(u,v)\in[0,1]^2\}$$ is the parallelogram spanned by $a$ and $b$. One has ${\rm area}(P)=|a|\>|b|\>\sin\gamma$, where $\gamma$ is the angle enclosed by $a$ and $b$.

(III) Given a variable vector $x$ the (signed) volume $\epsilon(a,b,x)$ of the parallelotope spanned by $a$, $b$, and $x$ is a linear function of $x$. Therefore there is a certain vector $r\in{\mathbb R}^3$ with $$\epsilon(a,b,x)=r\cdot x\qquad\forall \ x\in{\mathbb R}^3\ .$$ This vector $r$ depends in a skew-bilinear way on $a$ and $b$. It is called the cross-product of $a$ and $b$, and is denoted by $a\times b$.