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Prove Property 6 of cross products, that is,

$$ a \times (b \times c) = (a \cdot c) b - (a \cdot b) c $$

$a \times(b \times c)=(a \cdot c) b-(a \cdot b) c$

Vectors

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Welcome back to another cross product problem. We're going to try a really messy proof that a cross be crossed see is equal to a dot C, times b minus a dot B times C. I started by writing up the definition of the cross see just using our regular cross product method. Uh So we're going to focus on now, is writing out in full a cross be crusty. So we're looking at the vector A which we'll just call x a y az cross, this entire thing, It's gonna be a bit of a mess and I'm not going to be showing all of my work, but let's do our best here. So if you want to write this out and fall, you're welcome to work through it. But first we ignore our first column of the matrix, we're gonna look at a Y times all of this will be B x c, y minus b, y, c x minus hazy times all of this. And I'm going to be working that negative sign in. So that will be B z c x minus B x c Z. And all of this is multiplied by i minus. Then we'll be covering up our second column, we'll be looking at A X times the last component. Again we be X. See why minus B, Y c X and then minus hey, Z times the first calm little B b y, Sesay minus B Z. See why all times J. And then lastly when we ignore the third column, that will be A X Times This Middle one. Again, working in that negative sign E c c x minus B E x c Z minus A.Y. Times 1st 1 B Y c Z minus busy. See why all times K. Now all of this is going to get really messy. I know it's already pretty messy but all we're going to do is ignore the second two terms. For now, we're just going to focus on what happens in the first component of this vector. What we're gonna do is we're going to rewrite it, let's pretend to be X. And then what's left over is we have a Y C. Y. And we have a B. X. Here leaving us with an eggs Sesay. And then let's pull out a C. X. Leaving us with that's going to be a Y. B. Y. All time sex A Y. B, Y plus A Z. B. Z. All right. And we can verify, you can double check my work that all of this is equal to all of this. Now we're going to do something a little strange. We're going to add, let's make sure I'm adding the right thing here. We're going to add A X B E X C X. And then we'll subtract a X b x c x. And because we're adding and subtracting the exact same thing effectively, we're just adding zero to this whole thing. So this is completely valid. But by doing this we can now write this first component as B x times will be X times we have an A X E X right here and we have a X c X plus a Y, c, Y plus as z minus C X times. And again, we've got a minus c X right here. So again, we're looking at A X B X plus A, Y, B, Y plus a Z, B Z. And this should look pretty familiar at this point. The stuff in the parentheses is exactly the same because this is just the dot product between a and see double check that for yourself if you want. But we have B x times a dot c minus c x times a day. Dot b. And we're now 1/3 of the way done the problem. Now, what I've done is gone and worked ahead For the 2nd 2 components of this sector. We noticed we did the same thing, rewriting it and then adding and subtracting the same thing, Rewriting the third component and then adding and subtracting the same thing. And by doing this we can rewrite the J component as B y times a dot c minus c y times a day be. And the K component as busy times a dot c minus. Easy times a dot be. And so now we have our new vector is all of this times I plus all of this time's jay plus all of this times K. And we put that all together in a single vector. All we're left with is a dot c, times E I sat down. We're left with hey dot C, times B minus, and then once again we have a dot be times see? And that all together minus Hey dot be times See. And remember the original claim was that this was a cross be cross see. So this is one way of proving this identity. It's terribly messy. This is an awful proof, but it's a proof that it does work. Thanks for watching.