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Let v, W1; and W2 be vectors in Rn_ Show that if U is orthogonal to both of W1 and w2 then U is orthogonal to every vector in span{W1, W2}. [20]Let u, €, be n...

Question

Let v, W1; and W2 be vectors in Rn_ Show that if U is orthogonal to both of W1 and w2 then U is orthogonal to every vector in span{W1, W2}. [20]Let u, €, be non-zero vectors in R3 Show that the vector w = Ilvll u + Ilull v bisects the (non-reflex) angle between u and U. [20]

Let v, W1; and W2 be vectors in Rn_ Show that if U is orthogonal to both of W1 and w2 then U is orthogonal to every vector in span{W1, W2}. [20] Let u, €, be non-zero vectors in R3 Show that the vector w = Ilvll u + Ilull v bisects the (non-reflex) angle between u and U. [20]



Answers

Let $\mathbf{u}$ and $\mathbf{v}$ be nonzero vectors in 2 - or 3 -space, and let $k=\|\mathbf{u}\|$ and $l=\|\mathbf{v}\| .$ Prove that the vector $\mathbf{w}=l \mathbf{u}+k \mathbf{v}$ bisects the angle between $\mathbf{u}$and $\mathbf{v}$.

So um We have the vectors V. W. And you and we had a. U. Is equal to the unit vector of B. Plus unit vector of W. And so having that, we need to show that the angle between you and me. They called to the angle between you and W. It's like. And the first thing I'm going to do is to show that the angle between you and me that is a call to the angle between you and um union victor bobby. And also we have that the angle between you and W. It's going to be equal to the angle between you and unit vector W. And that makes sense because the angle between you and W is the same as you and the unit vector of W. Because the unit vector W is the same direction. Just smaller. All right. So um first purpose now since the unit vector we and W are the same land. If I show that I knew that you need victory is equal to you. Dot product. You need victor. W. If I prove this that that means then that proves what we need to prove. So let's do so let's rewrite you as the you need vector formation of it. So, this is for the first one. And for you dot union vector W. We have this like getting inspector. We plus union victor. That will you multiplied by? Well, technically dot I forgot to put the dog product here. Adult. You need victor at W. All right. For the top one, you're gonna have unit vector V times T. V. Which is just one boss. You need victor. W thoughts. You need victor. Me. And for the bottom one, we're gonna have you need vector V. At times a unit vector W. Plus one. So, as you can see, these two are the same. Uh therefore the angle between U. And V. He recalled to angle between you and W. So we have a crew. That why that's the case. But let's look at it also geometrically, so geometrically, let's say this is me and this is w it really doesn't matter which ones they are. I just arbitrarily say they're these ones. And so what is the unit vector of B? Let's say this is the unit vector of we and this is a union victory of W. So they had the same length. Therefore for you, She's going to be the addition of the two. We can see that. Um This is gonna be you right here and as you can see the angle, the angles are the same. This angle between you and we looks about the same as a single between U N W. And that is the case because right here, if I circle it in red, this is going to be um like a kite, which means all the sides are the same length, which kind of proves that the angles are also the same, and so every time.

Hello there. So for this exercise we have this back to you and the soup spaced of you defined by the spine. Of these two vectors V one MBT. So technically these soup space W. Corresponds to a plane. We need to calculate the projection of this victory onto this place. Generically you can complete the projection of you into any said or subspace as the sun. All vectors in W. Of the England problem. The if you divided norm is part of the norm of the factors time to speak. So basically what I am saying here is the I am projecting you to each of the generators of the subspace. So in this case it means that the projection of you onto this of space of U. Is equal to the prediction of you On the Vector B. one. That this is fine by this formula. Plus the product of you, B to the responsible too projection of you wanted the vector. Uh Okay so we just need to calculate each of the components focusing. So let's start by calculating the products. So you be one is equal 21 plus three. Physicals to four. The inner part of you would be too is equal to and then the norm the corresponding norms the square of the time. So the square of the norms why is equivalent to say is equals to say the inequality one he did so And this is equal to one was war one which at the end become sex and the same for you too. It's just four was one is equal to five. So here we have all the components required for targeting the projection of you and w. So we obtain the projection of the victory. You innocent space of you complain is equal 246 is actually two thirds times one minus 21 plus the fifth times two. Why? So basically what I did here was taking B2 square is equal to five. So that's why we have these five here during the you would be two is equal to do. So we have here two fields and same here the the square The norm is equal to six. The Interpol to the cost of four. So it has become for sex and days after simplifying is the 2/3 that we have here And here. We have B one and B two. Just applying, replacing the said that. We have calculated here on the phone of the protection. And as a final result we obtained this is equal 2 1/15 Times are back for 22 minus work team. So this response is a projection of you on to this point of view. And now we need to calculate the alt component of U. With respect to this subspace up basically, that means it's obstructing the protection you mind as a projection of you in this space or playing this. After placing data is equal to one 03 minus the projection that we have already calculated here. 1/15, two -14 and 10. And the result of this is equal two one over 15 times minus seven, 14 35.

Hello there. So for this exercise we have respect for you to find out 102. And the soup space David, that actually corresponds to plane defined by the vectors we want and beachy. So we need to calculate the projection of this vector here onto this point of view. The orthogonal projection and that is equal to take in to project you each the components that expand this this based on. So here we have the formal So let's calculate the inner roads first. So you'll be one is equal to 3-plus 4. That is equal to seven. Then the book of you would see becomes -1. Let's do that at the end is just one and then the corresponding it's part of the knobs. So the square of the number the one is equals to taking dinner. Part of the one with itself that ever it's too mhm. and the square of the 42nd fracture becomes three. Okay, so we just need to replace the data and that gave us objection of the victory on this space is equal to seven over for correspondent entity. These provisions Times The one veteran of these two. Two plus one. Number three. And the second vector that is my. So what happened here is that here? We can simplify this with this that this The end is one how and after solving decimation of structures You obtain this is equal to one Over six times 7. Right? So this responsible projection of these victor you on this point. And then we can take the what's your name component of the vector view. Basic structure protection. This We have all the data already so you remember is a better 10 two. And then we subscribe what we have a range calculated that is the projection That is mine and 1/6 seven 58 And the final result is 16 Mine is one minus five.

Okay, we have vectors V n W. They're both to non zero ones. And we're going to show that vector v minus eight times W um is orthogonal to W if a equals the dot product divided by the magnitude of w squared. Okay, so, let's look at our formula are dot product is the magnitude of the times the magnitude of W times the co sign of the angle. Now, also, we want to look at our actual uhm projection of the on W. That would be the magnitude of the times the cosine of the angle. Or also that's the same thing as the dot product divided by the magnitude of W. So, once you square, that you um would be able to multiply it by the vector W to come up with that, that V one component. So, V two ends up being the difference between our vector V. And are you know what we just came up with with the projection. And really what we just came up with the projection is the exact thing that they state up here for a it's that dot product divided by um the magnitude of w squared, so that just would then get multiplied by the vector W. And it's really the definition. So when we went through this process, we're just explaining, you know, kind of the definition of how to find the component that actually is orthogonal.


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