11b.2. The cross product, normal vectors, areas

The cross product

The cross product (sometimes also called the vector product ) is defined as follows: (a1a2a3)×(b1b2b3)=(a2b3b2a3(a1b3b1a3)a1b2b1a2) It is a bit unwieldy: we should practice the computation until we can reproduce it accurately.

Normal vectors and areas

There are 2 main reasons for introducing the cross product. The first is that the result of the cross product, n=a×b is a vector that is perpendicular to both a and b. This is sometimes called the normal vector and will be used extensively in chapter 11c when we work with planes in 3 dimensions.

The second is that the cross product leads to a formula very similar to the dot product formula we encountered in the previous section:

where θ is the angle between vectors a and b. Notice the two differences between this and the dot product formula: first we take the magnitude of a×b on the LHS of the equation, and second we have a sine instead of a cosine.

The derivation of the formula is a bit out of our scope so we will take the formula as is. Hopefully the formula doesn't sound too foreign to us: we have seen it in the calculation of areas of triangles and parallelograms in previous study. This gives us two formulas:

Solution to examples

1. Points A and B have coordinates A(1,2,3) and B(5,0,1).
Find a vector n, such that n is perpendicular to OA and OB.

Solution: n=(123)×(501)=(21610)
Remark: Any parallel vector will work as an answer.


2. Find the area of triangle OAB.

Solution: Area of triangle OAB=12|a×b|=12|2i16j10k|= 310 units2.