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Using cosine law prove that u v = U,V1 + uzVz + Uzlz forthe vectors in 3- space:...

Question

Using cosine law prove that u v = U,V1 + uzVz + Uzlz forthe vectors in 3- space:

Using cosine law prove that u v = U,V1 + uzVz + Uzlz forthe vectors in 3- space:



Answers

Prove: For any vectors $u, v \in \mathbf{C}^{n}$ and any scalar $z \in \mathbf{C},$ (i) $u \cdot v=\overline{v \cdot u},$ (ii) $(z u) \cdot v=z(u \cdot v)$ (iii) $u \cdot(z v)=\bar{z}(u \cdot v)$.

Okay. We are given our vector U v n w. And we are going to show that that the triple scalar product, the dot product between you and the cross product of BMW is given by this equation here. Okay, so first of all, let's just consider when we would be finding the determinant of the matrix. See here on the right, what we would do, we would take that you value and we multiply it by v squared time Visa, two times W c three minus WC two Times B three. Right, Well, if that was an I instead we would be doing the exact same thing. But we'd say that we were in the I direction And then later we would multiply by you of one. The same thing would happen for our U two or Jay Component. Um you two would get that same multiple. Um except on the left. If we would just say that it was in the J and then later we would take our use of two and multiply it by that term so you can see how these are similar. Um You know the right side gets there directly. Um The left side you would first find your vector with your I. J. And K. Components and then you would start multiplying your you one times your eye component, Peace, you two times that J component and so on and so forth. But the two sides would be equivalent.

Hello, everybody. In this video, I'm going to be showing you had a solve exercise 74 Chapter 13 Section four of calculus early. Transcendental. Now, in this problem, they want us to prove that for four vectors U V, W and X. The quantity u cross v dotted with W cross X is equal to the quantity you dot w times x dot v minus you dot x times VW. And while this seems like a really complicated expression to prove we can actually do this pretty easily, using the results from some previous now the first step we want to take create a vector C or that I can call c equal toe w cross X, and this will help make the ensuing steps a little easier to see. With that, this expression on the left here turns into U cross V started with C and now, for the next step we want to do is use the result from exercise 72. So if you don't understand this next step, I would go back and try and work out that problem for yourself. But with this results were led to write this left expression here as you started with V Cross E and now we can substitute C back in to get you started. The gross W. Ross. Now we'll move down here. But for this step, notice that we can use a result from exercise 73 to break this triple cross product down into a couple of terms. That will help us. So let's get the U run here and you is now dotted with dot product of the don't x times a vector w it subtracted from that we have we w tons of vector x And again, if you don't understand the step, I encourage you to revisit exercise 73 verify this result for yourself. But now notice that these two dot products are scale er's so using the distributive property of the dot product as well as its linearity in each term, we can pull these dot products in front and don't you with the remaining vector. And so what this turns into is the quantity you don't w from dotting you with the vector w here and then keeping the scaler the that was originally in front of W. And then similarly But we can subject from this is you dotted with X and then keeping the scaler D w and this is identical. So what we were trying to prove it only looks a little bit different, but we can use the symmetry of the dot product to verify that this is the exact same as above we have. You don't w as above times Vida X, which is the same as xtalpi it subtracted from that we have you don't X as is here times VW, which is also here. So really just this term to be flipped, But still it remains equal and we arrive at the result that we wanted. And so that's how you solve exercise 74.

Hello, everybody. In this video, I'm gonna be showing you how to solve Exercise 73 chapter 13 Section four of calculus early. Transcendental. Now, in this problem, they want us to prove that with the three vectors u V and W that the quantity you cross v cross w equal to you dot w times v minus U v w Now to begin the starvation I've gone ahead and previously written of the across W in a component form, you can derive this using the determinant method is described in here in 0.6. But I thought it would be helpful to write them here for now just so that we can immediately use it in this problem. And to make things even easier, I'm gonna go ahead and call this victor. See? So now if we want to compute this cross product, you cross with this vector what we want to dio Let's create a similar three by three matrix and find its determinant where this matrix is i, j and K, the standard unit vectors in the top row. The second row has two components of U U one U two and u three and he stirred ERA has the components of that cross product that we've labeled C see one, You 83. What? You're just these components that I already had written down. Now, this is a pretty long derivation, especially if we want to do all three components. So what I'm gonna dio is for the remainder of the video I'm gonna sold for the first component Only The reason I'm doing this is that the second and third components follow very naturally from the same process of calculating just the 1st 1 So here I'll show you the calculation of the first component of this quantity, and you can go ahead and take almost all the same steps to calculate the other two. No, the first component of this vector is going to be Hi, I'm the coefficient of I is going to be you too. I'm C three minus you three. I'm see too. Now let's go ahead and sub n. C two and C three from the quantities we have up here. First term is you two times 1/3 component which is V one W two minus V two w one. It's attracted from this. We have you three times. A second component, his V three w one minus V one will be you three. So now let's distribute these you components into these two quantities in the first term, we have you to the one w two minus you too. You too. W one minus you. Three The three w one plus you three one w three. And now at this point, what we want to do perform a little bit of a trick What we want to dio It's ad you won the one w one has determined here and before you move on, we'll go ahead and add these other two positive terms to that. You too. He won w two plus you three one w three And then from that we want to subtract you one view one and w one So that this expression remains the same is what we had before subtracting the other negative quantities. We also have YouTube v two w one minus you three. The three do you want. So now let's focus on the positive and negative branches separately. Notice that in the positive branch of this expression, each term has the same factor of the one in it. so we can go ahead and pull this out. This turns into V one times you one w one plus you to W two lets you three you three Now for the negative bridge. Similarly, we can pull out a factor of W one for each term. So we subtract w one times u one v one. Yes, you too. You too. Plus you three, three. But no notice. What's in these two, uh, quantities in parentheses In the positive bridge. This is just the dot product of U W So we can go ahead and say that this positive branch is equal to you. Don't w. Times v one. Well, in this negative branch concentric dot product of U and V multiplied by the first component W. And so here we have the first component of the answer notice that this represents the first component of the answer as our answer simply has the full vector V in the full vector w in the quantity. So if you do this exact same process using the other two components instead of just this 1st 1 you will again find that, for example, with the second component that will be equal to you dot w tons V two minus c dot v times w two and the same will go for the third component. But all three will have the same form and therefore it will equal this right hand side that we're so again with this, I encourage you to go ahead and sulfur the other two. But with this procedure, that is how you solve exercise 73 overall.

Okay. This question wants us to prove this constant property for the DOT product. So basically, it says if we have a constant, it doesn't matter where that goes in the dot product. It's not gonna affect the answer if we do it in Vector One vector to or just pull it out and do it at the very end. So for this one, I think it's easy to use the component definition. So will start from the fact that you got the is just equal to the ex components multiplied together plus the why components multiplied together. Plus the Z components multiplied together. And then we see that just adding a constant on the outside. Well, that just attach is a constant teach expression. And let me add the constants here on sex. Okay, and now we can see that we just have this product here, and it doesn't matter where that see comes from when it's getting multiplied, because we could group the sea in the u X or we could group the sea and the Vieques so the C could come either from the V vector or the U vector. But I think it's easiest just to explicitly show that all three of these evaluate to the same thing. So see you started with the is equal to see you x v X plus See you Why v y plus cuz vz And then finally you dotted with CV is equal to you x times cvx plus you y times c v y plus u z plus you ve time CVC And we can see that all three of these products are equal to each other because it doesn't matter what order we multiply things in.


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