For the 1D case on this page we calculated the full table for the outer product for null basis, but here the full table is a bit big so I have put it at the bottom of this page, here we just have the entries for cross multiplying two vectors:

a^b |
b.n_{0} |
b.n_{∞} |
b.n_{1} |
b.n_{2} |

a.n_{0} |
0 |
n_{0∞ }-1/2 |
n_{01} |
n_{02} |

a.n_{∞} |
-n_{0∞}+ 1/2 |
0 | n_{∞1} |
n_{∞2} |

a.n_{1} |
-n_{01} |
-n_{∞1} |
0 | n_{12} |

a.n_{2} |
-n_{02} |
-n_{∞2} |
-n_{12} |
0 |

Each term in this table is calculated from:

n_{0}= (e_{1} + e_{2})/2

n_{∞}= (e_{1} - e_{2})/2

n_{1}= e_{3}

n_{2}= e_{4}

So we just multiply out each term by converting to 'e' basis, doing inner product, then converting back to 'n' basis.

## The outer product of two null vectors

Lets take the point 'p' (x1,y1) in Euclidean space, this gives,

p=(-1,x1²+y1²,x1,y1)

and we want to take the outer product with q:

q=(-1,x2²+y2²,x2,y2)

So, multiplying out the terms using the above table, the outer product is the multivector:

scalar = (x1²+y1² -x2²- y2²)/2

n_{0∞} = x2²+y2² - x1²-y1²

n_{01} = x1 - x2

n_{02} = y1 - y2

n_{∞1} = x1²x2 - x2²x1 = x1x2(x1 - x2)

n_{∞y} = y1²y2 - y2²y1 = y1y2(y1 - y2)

n_{12} = x1 y2 - x2 y1

## Meet

In the above example if x1=x2 and y1=y2 then:

scalar = 0

n_{0∞} = 0

n_{01} = 0

n_{02} = 0

n_{∞1} = 0

n_{∞y} = 0

n_{12} = 0

So this gives us a way to test if x1 and x2 represent the same point since, if they do then,

p^q=0