Find the cross product AxB

**Answer 1**

Here is how I do it. Cross out the first row and column and perform your calculations then repeat for the other 2 rows I get

(6*4 – 3*-9)+(6*4-(-4*-9)+(6*3-(-4)*6)

Its just a process that you need to memorize.

For the first set of parenthesis I got

j k

6 -9

3 4

second set I got

i k

6 -9

-4 4

3rd set I got

i j k

6 6

-4 3

please note that you also get rid of the letters I just left them so you could see the columns

**Answer 2**

OK Let me ask you this. Have you been studying matrix theory? Because that is where this ‘table’ method is coming from. Anyway, let me show you the basic form.

Suppose we have two vectors a and b which, in component form, are given by…

a = a1i + a2j + a3k

b = b1i + b2j + b3k

Then a X b is given by

i j k

a1 a2 a3

b1 b2 b3

= [(a2b3) – (b2a3)]i + [(a2b3) – (b1a3)]j + [(a1b2) – (b1a2)]k

To find the i-component of the product, cover the i-column, and the i-j-k row. You are left with

a2 a3

b2 b3

Now you multiply them in a X-pattern [ since we are finding the cross product, its a good way to remember what to do.]

a2b3 b2a3

You then subtract the results

a2b3 – b2a3

This gives you the magnitude of the i-component aka the x-component of the cross product, which you will recall is a vector.

To find the j-component, cover the j-column and the i-j-k row to leave

a1 a3

b1 b3

Again multiply them in an X-pattern

a1b3 b1a3

Subtracting gives

(a1b3 – b1a3)j

which is the j-component aka the y-component of the resultant vector.

Finally, the k-component

a1 a2

b1 b2

(a1b2 – b1a2)k

That’s the basic theory. So lets see what we can do with your problem.

i j k

6 6 -9

-4 3 4

i-component

[(6 . 4) – (3 . -9)]i = [(24 – (-27)]i = 51i

j-component

[(6 . 4) – (-4 . -9)]j = [(24) – (36)]j = – 12j

k-component

[(6 . 3) – (-4 . 6)]k = [(18 + 24)]k = 42k

So

A X B = 51i – 12j + 42k

I hope this helps. As always, be careful with the negative signs and make liberal use of brackets to avoid silly mistakes.