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Given a cross product:

$\vec{u} \times \vec{v} = \left< -1, 1, -3 \right>$

I'm trying to find:

$(\vec{u} - 3\vec{v}) \times (\vec{u} + 2\vec{v})$

as a vector.

Clearly there are some properties of cross-products that I'm not aware of that would help solve this, but I can't for the life of me find them.

I do know the following rules:

$(xa) \times b = x(a \times b) = a \times (xb)$

and

$a \times (b + c) = a \times b + a \times c$

(where $a$, $b$, and $c$ are vectors and $x$ is a scalar)

But I have no idea how/if these rules apply to the above.

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Let me start if off for you:

$$\begin{align}(\vec{u} - 3\vec{v}) \times (\vec{u} + 2\vec{v}) &= (\vec{u} - 3\vec{v})\times u + (\vec{u} - 3\vec{v})\times(2\vec{v})\\&=(\vec{u} - 3\vec{v})\times \vec u + 2\cdot((\vec{u} - 3\vec{v})\times \vec v)\end{align}$$

In the first line, I used the second rule you wrote, and in the second, I used the first rule you wrote.

Now, continue doing that, and use two extra rules:

  • $(\vec a + \vec b)\times \vec c = $ something
  • $\vec a \times \vec a =$ something.
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  • $\begingroup$ Thank you so much! (u - 3) x u + 2((u - 3v) x v) -> (u x u + -3v x u) + 2(u x v + -3u x u) -> [u x u and -3u x u both equal 0] -3v x u + 2u x v -> 3u x v + 2u x v [we need u x v because that's what we know] -> Then it's just simple vector math: <-3, 3, -9> + <-2, 2, -6> -> -5, 5, -15 $\endgroup$ – TranquilMarmot Oct 4 '18 at 15:35

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