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# Prove the property of cross products (Theorem 11).Property 2: $(ca) \times b = c (a \times b) = a \times (cb)$

## $$\mathbf{a} \times(c \mathbf{b})$$

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

Welcome back to another cross product problem today will be trying to prove some cross products properties like a constant times an arbitrary vector A crossed with the vector B is equal to that same constant and a cross P. And that should also be equal to a cross with that same constant times B. Let's see if we can make this work. The first thing we'll do is just calculate the cross product between A and B, which I've already set up in our matrix here so that we can use these steps outlined in the textbook in order to calculate the cross product first, we will ignore the first column And look at a two times B three -3 times be too like I said, that's a oops A two, I'm just be three minus a three times B to all times I minus. Then we'll ignore the second column And look at a one B 3 -3 B one. That's a one, B three minus a three. Be one All Times J. And lastly we'll ignore the third column And look at a one B 2 -11. That's a one. Me too minus a two B one. Okay. If we multiply this whole thing by sea and we're left with the vector. See times A two B three minus a. Three B two comma. See are negative C. Actually we can incorporate the negative science Oc times A three B one minus a one B three. Since it's negative, we just brought that in and then the last coordinate almost have enough room there. Well just Bc times A one beat you minus a two B one. However, if we want to look at sea A crossed with B and we can just bring that see directly in to our matrix here and then do the same thing. In that case our new cross product would be ignore the first column And we have see a two times B three Minour CA three mm B two. Let's see. Yeah, thank you. B three minus C A three B two. I will notice. We've got to see we can bring out front minus. Then we ignore our second column. See a one B 3 -3A 3 B one. See A one B three minus C A three B one jay. Don't Ignore that 2nd bracket there. Okay. There we are. And lastly plus then we ignore our third column And look at CA one B 2 minus C A to B. One. See A one B two minus C A two. The one. Okay. And sure enough if we write this as a single vector, make sure we've got enough room here. That gives us see times A two B 3 minus a three B two, same as before. Uh huh. Then we're going to bring the minus sign in. We're looking at sea times a three B one minus a one B. Three same as before. And then I last coordinate See a 1 B2 -A. to b. one also matches up. If we wanted to do this one more time, we could try and show that a cross C times B is equal to the same thing. And if we perform the exact same steps, let's fly through this here, we'll ignore our first column. And then we're looking at he too CB three minus a three CB two minus ignore the second column. And then we're looking at a one CB three minus a three CB one jay plus. And then we ignore the third column. 81 CB two minus a two CB one. Okay. And if we bring out all of those seas again, we're looking at sea times a two B three. Mine is a three B 2. Bring out the negative sign. Bring out the sea. We're left with C A three B 1 minus a one B three. And lastly, see times A one B two minus a two B one. And sure enough, that's exactly the same as what we got before. Therefore, see a cross B is equal to see times a crispy is equal to a grass E. B. Or scaling any one of the vectors in a cross product is equal to scale like any other of the vectors in a cross product. Thanks for watching.

Coker College

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp