If I want to set the cross product of a and b to zero, what do I do next?

Question

a x b = 0
what I have after the crossing is a bunch of numbers in i j and k like this.
(x^2+xy)i - (x+y)j + (x^2+y^2)k = 0
what do I do next to plug in values for x and y?

okay, actually, this is the real equation
(8-4xy-y^2-y)i - (4x+y)j + (4x^2 +5xy+x+y+y^2-6)k =0
how do I plug in x and y values that work for this equation?

Answer 1

In the first case, x = y = 0

x+y = 0 => x = -y. But x^2 + y^2 = 0 too, so they have to be both 0. And x^2 + xy = 0 if you plug x = y = 0 in this equation.

In the second one:

If (8-4xy-y^2-y)i - (4x+y)j + (4x^2 +5xy+x+y+y^2-6)k =0, then I would begin using the fact that 4x + y must be 0.

y = -4x

Lets now plug this in the first equations:

8-4xy-y^2-y = 8 - 4x (-4x) - (-4x)^2 - (-4x) = 0

8 + 16 x^2 - 16 x^2 + 4x = 0. So, x = -2 and y = 8

Lets now see if these values accomplish the last equation:

4x^2 +5xy+x+y+y^2-6 = 4* (-2)^2 + 5 (-2)(8) + (8) + (8)^2 - 6 =

= 16 - 80 + 8+ 64 - 6 = -86 + 86 = 0. Ok

All is fine.

Hope this helps.

Ana

Answer 2

If (x^2+xy)i - (x+y)j + (x^2+y^2)k is the result of your cross product, and you want it to be the zero vector, than the only way is to set x=0 and y=0, because that's the only way that x^2+y^2=0.

If I have misintepreted your question, please clarify.

Answer 3

Make one of the a or b a linear multiple of the other.


Doug