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Dale;Given that &_ and are ornot_Justify your answee nch -zero vectors; with determine whether written explanation; each expression makes gen88 Maked sketches...

Question

Dale;Given that &_ and are ornot_Justify your answee nch -zero vectors; with determine whether written explanation; each expression makes gen88 Maked sketches_examples etc: Expression Senae? YNN JustificationIal--3YINb) a+b-7YIN[6/+/e/-7YINa+b+c=0YINla+b-/a-6|YIN

Dale; Given that &_ and are ornot_Justify your answee nch -zero vectors; with determine whether written explanation; each expression makes gen88 Maked sketches_examples etc: Expression Senae? YNN Justification Ial--3 YIN b) a+b-7 YIN [6/+/e/-7 YIN a+b+c=0 YIN la+b-/a-6| YIN



Answers

Determine which of the following are defined for nonzero vectors $\mathbf{u}, \mathbf{v},$ and $\mathbf{w} .$ Explain your reasoning. (a) $\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})$ (b) $(\mathbf{u} \cdot \mathbf{v}) \mathbf{w}$ (c) $\mathbf{u} \cdot \mathbf{v}+\mathbf{w}$ (d) $\|\mathbf{u}\| \cdot(\mathbf{v}+\mathbf{w})$

All right. Given the vectors X and redd and Y and blue on the left. And we are told given these pictures alone to draw X plus Y, X minus Y, y minus X and negative Y plus negative X. This question is testing our ability to perform vector addition specifically because we aren't giving components for X and Y. We're just giving the vectors graphically or visually. It's testing our ability to perform graphical vector addition. So remember that vector addition graphically is done for two vectors A plus B. Where you place vector A on your graph. And then to add vector B. You put the starting point of be on the tip of A. And the addition is from the starting point A. To the endpoint of be. So for instance, for X plus Y and x minus Y, you first put X and then we put Y on top of X of the tip. Which gives us this edition Director here. Similarly for x minus Y, we take the negative of Y, which is just the same length, but off of the direction between X minus right here. Next for y minus X and negative Y plus negative X. We follow the same steps where for negative X, we invert the direction of X, so y minus X, gives the image of the left a negative one plus negative X. We add xy together, but taking the inverse for both obtaining the black vector on the bottom line.

So for this problem we're asked to take to ah, multiplication zen dot products. The 1st 1 will be a dot be time See And the second is a dotty Times B and were given our vectors on top here in the form of I plus J I minus K and I'm on this two K. So one thing that will just help us out here is to write this as ah there I j k forms in purely numeric form. So that will be just Ah, I hear Jay's minus one K's and when I zero jet ah, Jays and minus two K's. And so now that we have that ah, we can figure out what a dot B is. So we'll do that by pairing up terms here. So one times one is one plus zero plus zero is going to be one times our vector C, which is 10 minus two or 10 minus two for the total. And then here we're going to take the dot product of A and C, so one times one is 11 times 000 times minus 20 So it's once again that's still going to be one. And this time we're going to multiply that by B, which is 10 minus one. And so I total will be 10 My this one.

Hello there. So in this exercise we have some relations um related with the norm of the some of these factors A and E. And we need to find some examples for these two vectors but they should be unitary or in case it is not possible explain why it's not possible. So in this case we have we need to give them an example of two vectors and unitary vectors that the norm of the some of them is going to be equal to zero. So let's remember that the only vector that is equal to the norms of the victory is close to zero is the zero vector and the zero vector is just zero. Is your Okay? So how can we define these two vectors? Well, We need to define two vectors a plus B. Such that the sum is the cost of the zero vector. So that means that the vector B is equal to mine is a victory. So this makes things easier. And then we just need to pick only one vector when unitary vector and then the only the other one will be just the same minus that factory. So let's choose A equals to the vector 10 This unitary. So then the vector B will be minus 10 And you can observe that the norm of these two vectors 10 sorry. In the other invitation. All right. And the victory will be -1. So I minus I. The norm. It's going to be equal to the norm of zero vector and that is equal to. Then we need to find two vectors A and B that satisfied the same condition. Both should be unitary And the sun should be equal to one. The norm of the Sun should be close to. So in this case let's choose for example the vector A Equals to the square root of three over to I plus yes plus one half jay. And the veteran B Will be equals 2 -1 jay. Clearly these two vectors are unitary, the length of both A is equal to the length of B that is equals to one. So both are unitary vectors. But some in particular in this case is equal to the victor A plus the vector B is equal to the square root of three over to the I direction -1/2 in the J Direction. And if this is unitary then this victory will be also unitary. You can observe because they are the same. There is only a change of sign here. So the length or the No I'm sorry of the some of these two pictures will correspond to taking the norm of the vector squared of 3/2, I -1/2. And the norm of this vector is equal. Two. The next relation is given bye. The some of the norm of the some of these two factors should be equal to do So here I should point something. So let's remember that the these norms satisfy the traveler inequality. That means that A plus B is less or equal than the sum of the norms. Okay. And in this case they are unitary vectors. So that means that these both norms are equal to one. So in other words, this relation here say that some of the two vectors A and B should be less or equal than to. So the value of two is the maximum possible value that can take this. These these normal of the some of the selectors that our unity and that the quality in this case in the triangular inequality is satisfied if these two vectors are part. Okay, so the vector A and B point in the same direction. So in that case we just need to take any victory. Yeah. Eight. And the vector B will be the same because they should be part. Okay? So let's choose something Is the easiest one. So the vector i. So the factory B will be also I. And then you can observe that norm of A plus B will be the norm of the vector to I. And this clearly is a constitution And the last one say that we need to the that too B say that we should find the two vectors to unitary vectors that satisfied that the some of the norm The norm of this one is equal to three, but this clearly is not possible because as I mentioned before, the the sum of these two vectors, because they should be unit serving will be less or equal than two, so three is greater than two, so that will violate the triangular inequality in that is a contradiction, and that is not possible. So in this case is not possible.

Hello there. So for this exercise, we need to consider two points on the plane. So technically, we got a vector. Are that there is a generic point X y, uh, on the plane, that means are to on another point or zero. That is just like initial points also are part off are zero our escort are to Okay, So what we want to know is what is the meaning off the following expression? So the first expression say that what is going to be our minor are zero equals one. So remember that this part this vector here are like a generic point on our to. So what we're saying is that the distance between any point X and y the plane Let's see here let's suppose that here is zero y zero. So we way need to find the points x and y such that they have a length one in distance with with 00 So here we're dealing with the with including Norm. So basically this when we solve this part here, you think the square root Oh x minus x zero Where plus why minus y zero square is equals to one and then we can take the square on both sides and we obtain the following equation X minus zero square Waas. Why mine otherwise zero square is equals to one on what we can observe here is that this correspondent to the equation off circle with center on X zero y zero net zero. So it means that from zero we obtain here a zero CO on it makes sense because every point that these on the circle has the same distance with respect to the 0.0 y zero. So that is exactly what this statement sit here. So this at the end, corresponds to the circle around a 0.0 y zero or center on our zero. That is easy to say then. So if then we got the following statement that say that the norm off the difference between these vectors is less or equal than one. So if we have already know what what is the meaning off this? So R minus are zero is equals to one Corresponds to Here is our zero on. We got a circle right off radius one. Okay, so if we got this, uh, what is the meaning? That off this less or equal. Well, it means that we're also going to consider the values that the radios off the circle that is less than one. So that means that we're going to feel these, uh, the circle on. We're going to have a ball. Okay, so at the end, this expression here means that we got a bowl center, uh, are zero. And with maximum reduce one. Finally, the last statement that we need Thio illustrate it. Say what? What is going to be our minus r zero greater than the one. So, technically, this is like the compliment off what we've seen before. So we got here. Some ball center are zero on. We're going to consider now the point that are outside off this boat. So that means that we're going to consider the whole space except course for thes bo here. So all the values inside are not going to be considered. Uh, yeah, that's it. So it's just like the compliment off the ball that we see on the previews part


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