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Vector: Outer Product and Volume
Motivation:
Find the volume spanned by the three vectors as below:
And from previous study, we can find that the area of parallelogram produced from and is
2 OAB =
For the height of the quadrangular prism, it is
h = the projection length of on normal vector of and
Let be the unit vector on the direction of normal vector of and . Then h =
With height and the area of the bottom parallelogram,
Volume of prism = h = ( )
Now, we put the whole thing into a 3-dimensional coordinate system. Let = (a1, a2, a3 ) = (b1, b2, b3 ) = (c1, c2, c3 )
=
From this, we hope that we can define a vector from and such that
1. the vector is perpendicular to and ; 2. the magnitude of the vector is equal to the area of parallelogram spanned by and
That is how we define the outer product of and .
Definition ( Outer Product ) Let , , be three unit vectors along the positive directions of x-axis, y-axis, and z-axis of the 3 dimensional coordinate system. If we have two vectors
= a1 + a2 + a3 = b1 + b2 + b3
then we define the outer product of and as
= + +
In the form of the determinant, it can be represented as
=
If you are not familiar with determinant, just treat is as an equivalent representation of the one above.
Please notice that outer product is not commutative; in other words,
Theorem ( Volume of Quadrangular Prism )
The volume of the quadrangular prism shown below is
V = | ( ) |
In the form of determinant, it is
V = | | |