# Math Help - Verctors in R3: cross product

1. ## Verctors in R3: cross product

I have this problem:

d,e,f are vectors in R3

$(d\times f)\bullet e$

I don't need to hear the solution, I just need some pointers about where to start. I mean, I did the cross multiplication and now I have letters everywhere.

Is there a cross product rule I'm missing? or a dot product rule?

2. Originally Posted by Freaky-Person
I have this problem:
d,e,f are vectors in R3
$(d\times f)\bullet e$
I don't need to hear the solution, I just need some pointers about where to start. I mean, I did the cross multiplication and now I have letters everywhere.
Is there a cross product rule I'm missing? or a dot product rule?
First you find $g=d\times f$.
Then you find $g\cdot e$.

3. using this, should I still have a couple of lines full of letters? Cause that's what I'm getting to >.>

4. Originally Posted by Freaky-Person
using this, should I still have a couple of lines full of letters? Cause that's what I'm getting to >.>
$\left( {d_2 f_3 - d_3 f_2 } \right)e_1 - \left( {d_1 f_3 - d_3 f_1 } \right)e_2 + \left( {d_1 f_2 - d_2 f_1 } \right)e_3$

5. ooooh i,j,k disappear! And oh! Wow! I totally get it now! ha...

Thank you!

6. Originally Posted by Plato
$\left( {d_2 f_3 - d_3 f_2 } \right)e_1 - \left( {d_1 f_3 - d_3 f_1 } \right)e_2 + \left( {d_1 f_2 - d_2 f_1 } \right)e_3$
waaaait why is it $\left( {d_2 f_3 - d_3 f_2 } \right)e_1 - \left( {d_1 f_3 - d_3 f_1 } \right)e_2$? That minus, is it a typo?

7. Originally Posted by Freaky-Person
waaaait why is it $\left( {d_2 f_3 - d_3 f_2 } \right)e_1 - \left( {d_1 f_3 - d_3 f_1 } \right)e_2$? That minus, is it a typo?
NO!
$\left| {\begin{array}{ccc}
i & j & k \\
{d_1 } & {d_2 } & {d_3 } \\
{f_1 } & {f_2 } & {f_3 } \\

\end{array} } \right| = \left( {d_2 f_3 - d_3 f_2 } \right)i - \left( {d_1 f_3 - d_3 f_1 } \right)j + \left( {d_1 f_2 - d_2 f_1 } \right)k$

8. my book and wikipedia say otherwise.

Wikipedia:
This equation is the sum of nine simple cross products. After all the multiplication is carried out using the basic cross product relationships between i, j, and k defined above,

a × b = a1b1(0) + a1b2(k) + a1b3(−j) + a2b1(−k) + a2b2(0) + a2b3(i) + a3b1(j) + a3b2(−i) + a3b3(0).

This equation can be factored to form

a × b = (a2b3 − a3b2) i + (a3b1 − a1b3) j + (a1b2 − a2b1) k = (a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1).
Oh wait, I see it. I go: 23, 31, 12. You go: 23, 13, 12.

9. Your book is not wrong. I am not wrong.
You are wrong for not reading carefully.
$- \left( {d_1 f_3 - d_3 f_1 } \right) = \left( {d_3 f_1 - d_1 f_3 } \right)$