Saturday

February 13, 2016
Posted by **Mathwise** on Wednesday, November 4, 2009 at 6:33am.

find: Ax(BxC)

pls help with solution.

- Vector & Geometry -
**Reiny**, Wednesday, November 4, 2009 at 7:44amA=[1,1,0]

B=[2,-3,1]

C=[4,0,-3]

Let's do BxC

I use a very simple algorith to form the cross-product of two vectors.

- write the two vectors above each other

2 -3 1

4 0 -3

- for the first number, with your pinkie or with a pencil, block off the first column and find the right cross-product of the remaining square matrix, that is, (-3)(-3) - (0)(1) = 9

-for the second number, with your pinkie or withe a pencil, block off the middle column and find the**negative**right cross-product of the remaining matrix, that is

-( (2)(-3) - (1)(4) ) = 10

- and finally for the third number, block off the third column and find the right cross-product of the remaining square matrix, that is, (2)(0) - (-3)(4) = 12

So BxC = [9,10,12]

( I always check by taking the dot product of this with the two original vectors, you should get zero)

Now repeat by taking [1,1,0]x[9,10,12]

I got [12, -12, 1} or 12i - 12j + k