# Cross product!

1. Oct 18, 2009

### Noxide

Why is the cross product defined this way

(blah)i + (blah)j + (blah)k = u

and used this way

(blah)i - (blah)j + (blah)k = u

2. Oct 18, 2009

Well; simply make the second blah in the first line a -blah,- it remains a " blah" doesn't it? :)

The " trick" you mention makes use of a determinant to calculate aXb .

3. Oct 18, 2009

### HallsofIvy

Making the second coeffient "-" makes it fit a nice mnemonic:
expanding a determinant along the top row
$$\left|\begin{array}{ccc}a & b & c \\ d & e & f\\ g & h & i\end{array}\right|= a\left|\begin{array}{cc}e & f \\ h & i\end{array}\right|- b\left|\begin{array}{cc}d & f \\ g & i\end{array}\right|+ c\left|\begin{array}{cc}d& e \\ g & h\end{array}\right|$$

If $\vec{u}= a\vec{i}+ b\vec{j}+ c\vec{k}$ and $\vec{v}= x\vec{i}+ y\vec{j}+ z\vec{k}$, then writing $\ve{u}\times \vec{v}= (bz-cy)\vec{i}+ (az- cx)\vec{j}+ (ay- bx)\vec{k}$ as $(bz-cy)\vec{i}- (cx- az)\vec{j}+ (au= bx-)\vec{k}$ makes it clearer that we can think of it as
$$\vec{u}\times\vec{v}= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ a & b & c \\ x & y & z\end{array}\right|$$

4. Oct 18, 2009

### Noxide

Thanks ivy!

We basically got the formula thrown at us because it is useful when finding torques. I see that cross products are also going to be covered in our linear algebra class, but we have not yet covered determinants in the class so cross products are a little out of my reach in terms of gaining a full understanding. I think your explanation nudged me in the right direction.