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Prove the property of cross products (Theorem 11).

Property 4: $ (a + b) \times c = a \times c + b \times c $

$$

\mathbf{a} \times \mathbf{c}+\mathbf{b} \times \mathbf{c}

$$

Vectors

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Welcome back to another cross product problem where we're trying to calculate the cross product of A plus B cross C. And show that that distributes to across C plus B. Cross C. One of the ways we can prove this is take three arbitrary vectors A. B and C. And plugged them into our matrix. So into our second rover matrix, we'll plug in the vector A plus B. Giving us a one plus B. One, A two plus beat you and a three plus B. Three. And in the 3rd row we'll just put c. Giving AC one C two in C. Three. Then we can use the method outlined in the book in order to calculate this cross product. Remember we ignore the first row and then we'll look at a two plus B two times C three minus a three plus B, three times C. Two. Oh, which I will write as C three times a. Two plus B two minus c, two times a. Three plus B. Three. All of this times I now we'll subtract remember. Big nor the second column. Look at a one plus B one time C three minus a three plus B. Three M. C one. And again we'll write that as C three times A one plus B one minus see one times A three plus B. Three. You're welcome to expand any of this if you want, but it's not necessary at the step. And then lastly we'll ignore the third column. We'll get a one plus B one time C two minus A to place B two times C one. And if we rewrite that, I guess that's c. two Times A. one plus b. one minus. See one times A two plus B two all times. Okay, so that's our cross product A plus B crossed with C. If we compare this to a cross C. Plus secrecy, we can just use the definitions and if now we add these together. A cross C plus B. Grassy. We can simplify this a little bit. Let's use our vector notation here. We've got a C. Three in both of these. So we can write this at C three times A two Plus B two minus. And then we've got to see two in both of these. So we're multiplying by a three minus B three. That's our first coordinate for a second. We'll subtract C three times A one plus B one plus since we're subtracting a negative, That'll give us C2 times. Uh Sorry that will give us C1 minus. Let's just right at C1 times. There we go. A. Three plus B. Three. Just expanding these terms and our final coordinate. We've got to see too times A one plus B one minus. See one times A two. Hey to plus B two. And if we compare that to what we got originally, we'll see that we have the exact same thing just written in i j k notation instead of are vector notation here. Since we did this for arbitrary vectors, A, B and C, we have proven that cross products distribute across edition. Thanks for watching.