Question

Prove that `veca . [(vecb + vecc) xx (veca + 3vecb + 4vecc)] = [(veca, vecb, vecc)]`

Answer 1

We are given the expression:

`veca * [(vecb + vecc) xx (veca + 3vecb + 4vecc)]`

We need to show:

`veca * [(vecb + vecc) xx (veca + 3vecb + 4vecc)] = [veca, vecb, vecc]`

where `[veca, vecb, vecc]` represents the scalar triple product:

`[veca, vecb, vecc] = veca * (vecb xx vecc)`

Let’s expand the vector triple product using distributive property:

`veca * [(vecb + vecc) xx (veca + 3vecb + 4vecc)]`

Break it down:

`veca * [vecb xx veca + vecb xx 3vecb + vecb xx 4vecc + vecc xx veca + vecc xx 3vecb + vecc xx 4vecc]`

Now simplify each cross product:

`vecb xx veca = -(veca xx vecb)`

`vecb xx vecb = vec0`

`vecb xx vecc` remains as is

`vecc xx veca = -(veca xx vecc)`

`vecc xx vecb = -(vecb xx vecc)`

`vecc xx vecc = vec0`

Substitute and simplify:

`veca * [-veca xx vecb + 0 + 4vecb xx vecc - veca xx vecc - 3vecb xx vecc + 0]`

Combine like terms:

`veca * [-veca xx vecb - veca xx vecc + vecb xx vecc]`

Now apply the scalar triple product identity:

`veca * (veca xx vecb) = 0`   ...(Since it’s orthogonal)

`veca * (veca xx vecc) = 0`

So, `veca * [-veca xx vecb - veca xx vecc + vecb xx vecc] = veca * (vecb xx vecc) = [veca, vecb, vecc]`

`veca * [(vecb + vecc) xx (veca + 3vecb + 4vecc)] = [veca, vecb, vecc]`