anonymous
  • anonymous
Tough vector question: u and v are vectors such that: (u-v)(u+2v)=-114 (u-v)(u+2v)=-45 uv=4 (In each of these cases the multiplication is the dot product) Therefore |u x v|=? (cross product)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
here's an example u can do it by seeing this In this section we’re going to be taking a look at two special products of vectors, the dot product and the cross product. However, before we look at either one of them we need to get a quick definition out of the way. Suppose that u and v are two vectors in 2-space or 3-space that are placed so that their initial points are the same. Then the angle between u and v is angle that is formed by u and v such that . Below are some examples of angles between vectors. Notice that there are always two angles that are formed by the two vectors and the one that we will always chose is the one that satisfies . We’ll be using this angle with both products. So, let’s get started by taking a look at the dot product. Of the two products we’ll be looking at in this section this is the one we’re going to run across most often in later sections. We’ll start with the definition. Definition 1 If u and v are two vectors in 2-space or 3-space and is the angle between them then the dot product, denoted by is defined as, Note that the dot product is sometimes called the scalar product or the Euclidean inner product. Let’s see a quick example or two of the dot product. Example 1 Compute the dot product for the following pairs of vectors. (a) and which makes the angle between them . (b) and which makes the angle between them . Solution For reference purposes here is a sketch of the two sets of vectors. (a) There really isn’t too much to do here with this problem. (b) Nor is there a lot of work to do here. Now, there should be a question in everyone’s mind at this point. Just how did we arrive at those angles above? They are the correct angles, but just how did we get them? That is the problem with this definition of the dot product. If you don’t have the angles between two vectors you can’t easily compute the dot product and sometimes finding the correct angles is not the easiest thing to do. Fortunately, there is another formula that we can use to compute the formula that relies only on the components of the vectors and not the angle between them. Theorem 1 Suppose that and are two vectors in 3-space then, Likewise, if and are two vectors in 2-space then, Proof : We’ll just prove the 3-space version of this theorem. The 2-space version has a similar proof. Let’s start out with the following figure. So, these three vectors form a triangle and the lengths of each side is , , and . Now, from the Law of Cosines we know that, Now, plug in the definition of the dot product and solve for . (1) Next, we know that and so we can compute . Note as well that because of the square on the norm we won’t have a square root. We’ll also do all of the multiplications. The first three terms of this are nothing more than the formula for and the next three terms are the formula for . So, let’s plug this into (1). And we’re done with the proof. Before we work an example using this new (easier to use) formula let’s notice that if we rewrite the definition of the dot product as follows, we now have a very easy way to determine the angle between any two vectors. In fact this is how we got the angles between the vectors in the first example! Example 2 Determine the angle between the following pairs of vectors. (a) (b) Solution (a) Here are all the important quantities for this problem. The angle is then, (b) The important quantities for this part are, The angle is then, Note that we did need to use a calculator to get this result. Twice now we’ve seen two vectors whose dot product is zero and in both cases we’ve seen that the angle between them was and so the two vectors in question each time where perpendicular. Perpendicular vectors are called orthogonal and as we’ll see on occasion we often want to know if two vectors are orthogonal. The following theorem will give us a nice check for this. Theorem 2 Two non-zero vectors, u and v, are orthogonal if and only if . Proof : First suppose that u and v are orthogonal. This means that the angle between them is and so from the definition of the dot product we have, and so we have . Next suppose that , then from the definition of the dot product we have, and so the two vectors are orthogonal. Note that we used the fact that the two vectors are non-zero, and hence would have non-zero magnitudes, in determining that we must have . If we take the convention that the zero vector is orthogonal to any other vector we can say that for any two vectors u and v they will be orthogonal provided . Using this convention means we don’t need to worry about whether or not we have zero vectors. Here are some nice properties about the dot product. Theorem 3 Suppose that u, v, and w are three vectors that are all in 2-space or all in 3-space and that c is a scalar. Then, (a) (this implies that ) (b) (c) (d) (e) if (f) if and only if v = 0 We’ll prove the first couple and leave the rest to you to prove since the follow pretty much from either the definition of the dot product or the formula from Theorem 2. The proof of the last one is nearly identical to the proof of Theorem 2 in the previous section. Proof : (a) The angle between v and v is 0 since they are the same vector and so by the definition of the dot product we’ve got. To get the second part just take the square root of both sides. (b) This proof is going to seem tricky but it’s really not that bad. Let’s just look at the 3-space case. So, and and the dot product is We can also compute as follows, However, since , etc. (they are just real numbers after all) these are identical and so we’ve got . Example 3 Given , and compute the following. (a) and (b) (c) and Solution (a) Okay, in this one we’ll be verifying part (a) of the previous theorem. Note as well that because the norm is squared we’ll not need to have the square root in the computation. Here are the computations for this part. So, as the theorem suggested we do have . (b) Here’s the dot product for this part. So, it looks like u and w are orthogonal. (c) In this part we’ll be verifying part (d) of the previous theorem. Here are the computations for this part. Again, we got the result that we should expect . We now need to take a look at a very important application of the dot product. Let’s suppose that u and a are two vectors in 2-space or 3-space and let’s suppose that they are positioned so that their initial points are the same. What we want to do is “decompose” the vector u into two components. One, which we’ll denote for now, will be parallel to the vector a and the other, denoted for now, will be orthogonal to a. See the image below to see some examples of kind of decomposition. From these figures we can see how to actually construct the two pieces of our decomposition. Starting at u we drop a line straight down until it intersects a (or the line defined by a as in the second case). The parallel vector is then the vector that starts at the initial point of u and ends where the perpendicular line intersects a. Finding is actually really simple provided we first have . From the image we can see that we have, We now need to get some terminology and notation out of the way. The parallel vector, , is called the orthogonal projection of u on a and is denoted by . Note that sometimes is called the vector component of u along a. The orthogonal vector, , is called the vector component of u orthogonal to a. The following theorem gives us formulas for computing both of these vectors. Theorem 4 Suppose that u and are both vectors in 2-space or 3-space then, and the vector component of u orthogonal to a is given by, Proof : First let then will be the vector component of u orthogonal to a and so all we need to do is show the formula for is what we claimed it to be. To do this let’s first note that since is parallel to a then it must be a scalar multiple of a since we know from last section that parallel vectors are scalar multiples of each other. Therefore there is a scalar c such that . Now, let’s start with the following, Next take the dot product of both sides with a and distribute the dot product through the parenthesis. Now, and because is orthogonal to a. Therefore this reduces to, and so we get, We can also take a quick norm of to get a nice formula for the magnitude of the orthogonal projection of u on a. Let’s work a quick example or two of orthogonal projections. Example 4 Compute the orthogonal projection of u on a and the vector component of u orthogonal to a for each of the following. (a) (b) Solution There really isn’t much to do here other than to plug into the formulas so we’ll leave it to you to verify the details. (a) First, Now the orthogonal projection of u on a is, and the vector component of u orthogonal to a is, (b) First, Now the orthogonal projection of u on a is, and the vector component of u orthogonal to a is, We need to be very careful with the notation . In this notation we are looking for the orthogonal projection of u (the second vector listed) on a (the vector that is subscripted). Let’s do a quick example illustrating this. Example 5 Given and compute, (a) (b) Solution (a) In this case we are looking for the component of u that is parallel to a and so the orthogonal projection is given by, so let’s get all the quantities that we need. The projection is then, (b) Here we are looking for the component of a that is parallel to u and so the orthogonal projection is given by, so let’s get the quantities that we need for this part. The projection is then, As this example has shown we need to pay attention to the placement of the two vectors in the projection notation. Each part above was asking for something different and as shown we did in fact get different answers so be careful. It’s now time to move into the second vector product that we’re going to look at in this section. However before we do that we need to introduce the idea of the standard unit vectors or standard basis vectors for 3-space. These vectors are defined as follows, Each of these have a magnitude of 1 and so are unit vectors. Also note that each one lies along the coordinate axes of 3-space and point in the positive direction as shown below. Notice that any vector in 3-space, say , can be written in terms of these three vectors as follows, So, for example we can do the following, Also note that if we define and these two vectors are the standard basis vectors for 2-space and any vector in 2-space, say can be written as, We’re not going to need the 2-space version of things here, but it was worth pointing out that there was a 2-space version since we’ll need that down the road. Okay we are now ready to look at the cross product. The first thing that we need to point out here is that, unlike the dot product, this is only valid in 3-space. There are three different ways of defining it depending on how you want to do it. The following definition gives all three definitions. Definition 2 If u and v are two vectors in 3-space then the cross product, denoted by and is defined in one of three ways. (a) - Vector Notation. (b) - Using determinants (c) - Using determinants Note that all three of these definitions are equivalent as you can check by computing the determinants in the second and third definition and verifying that you get the same formula as in the first definition. Notice that the cross product of two vectors is a new vector unlike the dot product which gives a scalar. Make sure to keep these two products straight. Let’s take a quick look at an example of a cross product. Example 6 Compute for and . Solution You can use either of the three definitions above to compute this cross product. We’ll use the third one. If you don’t remember how to compute determinants you might want to go back and check out the first section of the Determinants chapter. In that section you’ll find the formulas for computing determinants of both and matrices. When we’re using this definition of the cross product we’ll always get the answer in terms of the standard basis vectors. However, we can always go back to the form we’re used to. Doing this gives, Here is a theorem listing the main properties of the cross product. Theorem 5 Suppose u, v, and w are vectors in 3-space and c is any scalar then (a) (b) (c) (d) (e) (f) The proof of all these properties come directly from the definition of the cross product and so are left to you to verify. There are also quite a few properties that relate the dot product and the cross product. Here is a theorem giving those properties. Theorem 6 Suppose u, v, and w are vectors in 3-space then, (a) (b) (c) - This is called Lagrange’s Identity (d) (e) The proof of all these properties come directly from the definition of the cross product and the dot product and so are left to you to verify. The first two properties deserve some closer inspection. That they are saying is that given two vectors u and v in 3-space then the cross product is orthogonal to both u and v. The image below shows this idea. As this figure shows there are two directions in which the cross product can be orthogonal to u and v and there is a nice way to determine which it will be. Take your right hand and cup your fingers so that they point in the direction of rotation that is shown in the figures (i.e. rotate u until it lies on top of v) and hold your thumb out. Your thumb will point in the direction of the cross product. This is often called the right-hand rule. Notice that part (a) of Theorem 5 above also gives this same result. If we flip the order in which we take the cross product (which is really what we did in the figure above when we interchanged the letters) we get . In other words, in one order we get a cross product that points in one direction and if we flip the order we get a new cross product that points in the opposite direction as the first one. Let’s work a couple more cross products to verify some of the properties listed above and so we can say we’ve got a couple more examples in the notes. Example 7 Given and compute each of the following. (a) and [Solution] (b) [Solution] (c) and [Solution] Solution In the solutions to these problems we will be using the third definition above and we’ll be setting up the determinant. We will not be showing the determinant computation however, if you need a reminder on how to take determinants go back to the first section in the Determinant chapter for a refresher. (a) and Let’s compute first. Remember that we’ll get the answers here in terms of the standard basis vectors and these can always be put back into the standard vector notation that we’ve been using to this point as we did above. Now let’s compute . So, as part (a) of Theorem 5 suggested we got . [Return to Problems] (b) Not much to do here other than do the cross product and note that part (f) of Theorem 5 implies that we should get . So, sure enough we got 0. [Return to Problems] (c) and We’ve already got computed so we just need to do a couple of dot products and according to Theorem 6 both u and v are orthogonal to and so we should get zero out of both of these. And we did get zero as expected. [Return to Problems] We’ll give one theorem on cross products relating the magnitude of the cross product to the magnitudes of the two vectors we’re taking the cross product of. Theorem 7 Suppose that u and v are vectors in 3-space and let be the angle between them then, Let’s take a look at one final example here. Example 8 Given and verify the results of Theorem 7. Solution Let’s get the cross product and the norms taken care of first. Now, in order to verify Theorem 7 we’ll need the angle between the two vectors and we can use the definition of the dot product above to find this. We’ll first need the dot product. All that’s left is to check the formula. So, the theorem is verified.
anonymous
  • anonymous
Is this from Paul's online notes? what page is it?
dumbcow
  • dumbcow
its better to just post the link if you are taking info from another source

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