Cross Product Properties Question

[Sho Kano]

Homework Statement

A\cdot B\times C\quad =\quad 2\\ (2A+B)\quad \cdot \quad [(A-C)\quad \times \quad (2B+C)]\quad =\quad ?

Homework Equations

Various cross product and dot product properties

The Attempt at a Solution

I've only managed to get so far, don't really know what to do next
A\cdot B\times C\quad =\quad 2\\ (2A+B)\quad \cdot \quad [(A-C)\quad \times \quad (2B+C)]\\ \\ =(2A+B)\quad \cdot \quad [(A-C)\times 2B\quad +\quad (A-C)\times C]\\ =(2A+B)\quad \cdot \quad [A\times 2B\quad -\quad C\times 2B\quad +\quad A\times C]

 

[haruspex]

Homework Statement

A\cdot B\times C\quad =\quad 2\\ (2A+B)\quad \cdot \quad [(A-C)\quad \times \quad (2B+C)]\quad =\quad ?

Homework Equations

Various cross product and dot product properties

The Attempt at a Solution

I've only managed to get so far, don't really know what to do next
A\cdot B\times C\quad =\quad 2\\ (2A+B)\quad \cdot \quad [(A-C)\quad \times \quad (2B+C)]\\ \\ =(2A+B)\quad \cdot \quad [(A-C)\times 2B\quad +\quad (A-C)\times C]\\ =(2A+B)\quad \cdot \quad [A\times 2B\quad -\quad C\times 2B\quad +\quad A\times C]

The scalar triple product has a very useful cyclic property. X.(YxZ)=Y.(ZxX)=Z.(XxY). Switching the cyclic order swaps the sign.

 

[Sho Kano]

The scalar triple product has a very useful cyclic property. X.(YxZ)=Y.(ZxX)=Z.(XxY). Switching the cyclic order swaps the sign.

I tried that already, the problem is I end up getting this mess after distributing
2A\cdot A\times 2B\quad -\quad 2A\cdot C\times 2B\quad +\quad 2A\cdot A\times C\quad +\quad B\cdot A\times 2B\quad -\quad B\cdot C\times 2B\quad +\quad B\cdot A\times C
and there's no way of using the triple scalar product to simplify that, other than the second term

 

[haruspex]

I tried that already, the problem is I end up getting this mess after distributing
2A\cdot A\times 2B\quad -\quad 2A\cdot C\times 2B\quad +\quad 2A\cdot A\times C\quad +\quad B\cdot A\times 2B\quad -\quad B\cdot C\times 2B\quad +\quad B\cdot A\times C
and there's no way of using the triple scalar product to simplify that, other than the second term

The first, third, fourth and fifth terms simplify so much using that hint that they disappear immediately. Post an attempt at using it on the first term.

 

[Sho Kano]

The first, third, fourth and fifth terms simplify so much using that hint that they disappear immediately. Post an attempt at using it on the first term.

OH they simplify to 0. I was too focused on matching the given information with the terms.
2A\cdot A\times B\\ =\quad 2[A\cdot A\times B]\\ =\quad 2[A\times A\cdot B]\\ =\quad 0
So we are left with
-2A\cdot C\times 2B\quad +\quad B\cdot A\times C\\ =\quad -2A\cdot 2[C\times B]\quad +\quad B\cdot A\times C\\ =\quad -4[A\cdot C\times B]\quad +\quad B\cdot A\times C\\ =\quad -4[A\times C\cdot B]\quad +\quad A\times C\cdot B\\ =\quad -3[A\times C\cdot B]\\ =\quad -3[A\cdot C\times B]\\ =\quad -3[-(A\cdot B\times C)]\\ =\quad -3(-2)\\ =\quad 6

 

[haruspex]

OH they simplify to 0. I was too focused on matching the given information with the terms.
2A\cdot A\times B\\ =\quad 2[A\cdot A\times B]\\ =\quad 2[A\times A\cdot B]\\ =\quad 0
So we are left with
-2A\cdot C\times 2B\quad +\quad B\cdot A\times C\\ =\quad -2A\cdot 2[C\times B]\quad +\quad B\cdot A\times C\\ =\quad -4[A\cdot C\times B]\quad +\quad B\cdot A\times C\\ =\quad -4[A\times C\cdot B]\quad +\quad A\times C\cdot B\\ =\quad -3[A\times C\cdot B]\\ =\quad -3[A\cdot C\times B]\\ =\quad -3[-(A\cdot B\times C)]\\ =\quad -3(-2)\\ =\quad 6

Looks right.

 

[Sho Kano]

Looks right.

Thanks, it turned out to be so simple!