Question

How do you find the cross product `<0,1,2>times<1,1,4>` and verify that the resulting vectors are perpendicular to the given vectors?

Answer 1

To take the cross product, for some

`veca = << a_1,a_2,a_3 >>`

and

`vecb = << b_1,b_2,b_3 >>,`

you can use this formula:

`\mathbf(veca xx vecb = << a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 >>)`

If you notice, the subscripts are cyclic. They go `23, 31, 12`, and you subtract the "mirror-image" subscripts.

So, the cross product is:

`color(blue)(veca xx vecb)`

`= << 1*4 - 2*1, 2*1 - 0*4, 0*1 - 1*1 >>`

`= color(blue)(<< 2,2,-1 >>)`

An easy way to verify the perpendicularity of `vecaxxvecb` is by performing the dot product with `veca` and then with `vecb`. If they are perpendicular, and neither of them is `vec0` (which would defeat the purpose), then the dot product should be `0`.

(It doesn't matter which direction you do it either, because dot products are commutative.)

`color(blue)((vecaxxvecb)cdotveca) = vecacdot(vecaxxvecb)`

`= << 2,2,-1 >>cdot<< 0,1,2 >>`

`= 2*0 + 2*1 + (-1)*2`

`= color(blue)(0)`

`color(blue)((vecaxxvecb)cdotvecb) = vecbcdot(vecaxxvecb)`

`= << 2,2,-1 >>cdot<< 1,1,4 >>`

`= 2*1 + 2*1 + (-1)*4`

`= color(blue)(0)`

And to further prove this, here is another formula you should know (modified to substitute `vecaxxvecb` for one of the vectors):

`\mathbf((vecaxxvecb)cdotveca = ||(vecaxxvecb)||cdot||veca||costheta)`
`\mathbf((vecaxxvecb)cdotvecb = ||(vecaxxvecb)||cdot||vecb||costheta)`

Since both results were `0`, and since neither vector was `vec0` (hence their norms are nonzero), it must be that `costheta = 0`. Therefore, `theta = 90^@` or `270^@`, and either way, CW or CCW, that indicates perpendicularity.

Hence, `\mathbf(vecaxxvecb)` is perpendicular to `\mathbf(veca)` and is also perpendicular to `\mathbf(vecb)`.