three-vectors-mathbfa-mathbfb-and-mathbfc-are-given-a-find-their-scalar-triple-product-mathbfa-cdotm

Question

Three vectors $\mathbf{a}, \mathbf{b},$ and $\mathbf{c}$ are given. (a) Find their scalar triple product $\mathbf{a} \cdot(\mathbf{b} 	imes \mathbf{c}) .$ (b) Are the vectors coplanar? If not, find the volume of the parallelepiped that they determine.
$$
\mathbf{a}=\langle 1,2,3angle, \quad \mathbf{b}=\langle- 3,2,1angle, \quad \mathbf{c}=\langle 0,8,10angle
$$

Ryan Mcalister · Accepted Answer

this question starts by asking us to find a, uh, scaler triple product of three vectors. The vectors that it gives are A which is equal to 123 be, which is equal to negative 32 one and see which is equal to 08 and 10. Now the scale a triple product is going to be a dot be cross sea and order of operations will still apply here. We&#x27;re going to do the parentheses first, be cross, see and then take the dot product with a So let&#x27;s jump into it the same way we always do with cross products when you use a three by three matrix, the first row of which will be our unit vectors I J. K. And the second and third row will be the vectors himself. So negative. 321 and 08 and 10. When we take the determine of this three by three matrix, we&#x27;re going to dio co factor decomposition along the first row. That means we&#x27;re going to take ah minor 11 which is just the matrix without Row one and column one that is 281 and 10. The bottom right corner multiply that by I. Then we&#x27;ll subtract. Remember, we&#x27;re subtracting this next term because of where it is in the Matrix. Since J is Element 12 and one plus two is three, which is an odd number. We subtract this term this co factor and that co factors gonna be found by taking the determinant of minor one to which again is the Matrix without Row one and column to so negative 301 and 10 times element one to which, of course, is J. And then we&#x27;ll finally add the co factor. 13 which is the determinant of minor 13 negative 30 to 8 times the unit vector K. So if we do all of this really quickly, what we&#x27;ll find is we have 20 minus eight, which is 12 I minus negative 30 which is positive 30 j and minus 24 plus zero. So native 24 minus 24 k So this is our cross product. We have 12 i 30 j 24 k We can turn that into a vector. Just 12. 30 negative 24. And now what we want to do is take the doubt product with a So the dot product a dot be cross C is going to be equal to 123 dot 12. 30 is the negative negative 24 in the dot product. Just to review is we multiply each component and then add them together. So what we&#x27;re gonna have is it&#x27;s not a vector, it&#x27;s a scaler quantity. We&#x27;re gonna have one times 12 plus two times 30 plus three times negative 24 So will be left with 12 plus 60 60 60 minus 72 which is equal to zero. So our scaler triple product is equal to zero. As we found part A and part B asked us, Are these all co plainer? Well, when we have a scale trip of product equal to zero, we know that yes, they are co planner. Co planner means that all three of these vectors can be contained within the same plane. And I&#x27;ll, uh, give a little bit of explanation Is too. How we know this. Yes. So, yes, they are all cope later. So how do we know that skeletal product of zero means they are complaining? Well, let&#x27;s take a look when we found be cross see first we said, for example, here&#x27;s B, here&#x27;s C and we found Be crust. See here and be crust see with equal to or perpendicular to B and C Let&#x27;s be cross e. So no matter of what, When we have two vectors, we can always put a plane around them and I&#x27;m gonna draw that. Now let&#x27;s say we have a plane that contains B and C, and it looks something like this. Um, this is you have to believe me. Drawing in three dimensions is not easy. Um, so let&#x27;s just say that&#x27;s the plane that contains B and C. And now we know that be cross, see is perpendicular is orthogonal to that plane, so it&#x27;s sticking straight out. What we did is we found that a dot beak Rossi is equal to zero. Well, we know that any any two vectors let&#x27;s call them you and v u dot v equals zero means that you is perpendicular to V. So wouldn&#x27;t we found that? Be cross si dot a is equal to zero. We found that a is perpendicular to be across C. That means that a is perpendicular to this vector that we drew here. And the only way to be perpendicular to it is to fall into this plane that we just drew around B and C. So since a is perpendicular to be cross, see a is complainer with B and C, and that is how we know that all three of these factors are indeed cope later. And that&#x27;s your final answer.

Ryan Mcalister · Answer

this question asks us to find the scaler triple product of three factors. The vectors that were given our vector A is equal to 23 negative too. Vector B is able to negative one for zero in Vector C is equal to three negative. 13 Now we know from our book that the scaler triple product is gonna be vector A dotted with the cross product of vectors B and C So because we&#x27;re going to take the preserve the order of operations, even for our back your functions, we use the print that will do the parentheses functions first. Well, first to be cross. See, we&#x27;re gonna do that in the normal method. Um, where we use a three by three matrix whose first row is I. J. K. Our ah unit vectors in the second row will, of course, be negative 140 and the third row will be C three negative 13 So when we do our co affect, your decomposition will find that this determined is equal to the determinant of matrix. For negative ones, there are three times I minus the determinant of negative 1303 times J plus the determinant of negative +134 negative one times K. And when we find our determinants of these two by two, major cities will get a value of 12 I plus number. Plus, we had to flip the sign because the determinant is negative, but we&#x27;re subtracting. It was plus three j minus 11 K In other words, we have our cross product. 12 3 negative 11. So that is our cross product of BNC. Now to find the scale a triple product, we just need to doubt that with a So a dot be cross see is equal to a, which is 23 negative, too dot be cross C 12 3 negative 11. Remember, the dot product is the product of the air is the sum of the products of corresponding components. So what we&#x27;ll do is we&#x27;ll most play two times 12 which is 24 will add three times three, which is just nine. And then we&#x27;ll add negative two times negative 11 which is positive. 22. When we add all those up, we get a dot product of 55. So that is our scaler triple value. It&#x27;s 55 now. What does this. Tell us, Since the scaler triple product is not equal to zero, we know that these are not co plainer, that is, there&#x27;s no plane that holds all three of these vectors. The second part of this question asks us if the, uh, vectors aren&#x27;t complainer. What is the value? Or does the volume of the parallel a pipe head that these three vectors describe? Well, let me try to draw a picture here. Of course, this is in three dimensions, and I&#x27;m only drawing in too. So won&#x27;t be great, but I&#x27;ll try a parallel pipe head. We said, Let&#x27;s say this is B and this is C. If we have vector a here, which is not co plainer, so it&#x27;s not in the same plane. This is Victor A. Then we&#x27;re gonna have a parallel a pipe it. I&#x27;ll try to draw this as best I can with a base that looks like this. And, uh, height, something like this. It&#x27;s not perfect, but what it is is it&#x27;s a parallel your, um imagine we had a parallelogram prism and we shifted it. So all the sides are gonna be parallel a grams will we know from our book that the volume of this parallel a pipe head is actually just the scale. A triple product. So the volume of this we already know is 55. But how? How do we know that? Why is that true? Well, when refining the volume of anything, we start with base times height. In this case, base is going to be the, um be and see parallelogram that we have here. Let me draw it now. This is, um, flat. Let&#x27;s pretend this is the two dimensional base here. Got B and C, and it goes up like this to form a parallelogram. We re friend when we find be cross See, what we&#x27;re doing is the magnitude that is the magnitude of B times, the magnitude of C times, the sine of the angle in between. So if this is state of here, be cross. See is the length of b times See? Signed. Faded. Now what is C sine theta? Well, if we were to make this into a right triangle, this is a right angle. See, science data is just this component. While Seiko sine theta would be this. So we have B times c sine theta, and that is exactly what we need to find the area of this parallelogram. So the magnitude of be cross sea is also the area of the base of this parallel a pipette. The special thing about the cross product is that it also gives us the direction that is perpendicular to this parallelogram. So in our first diagram will show where are cross product was It was sticking straight out like this. The remaining thing we need for the volume of our parallel a pipe ed is the height. We already have the base. We know it&#x27;s the magnitude of the cross product. We just need the height. Well, if we have an angle between a and this be cross see right here and we can make this into a right triangle like this. So if this is Fada in this angle here and this is the right angle, the height perpendicular to B B and C is gonna be this or a to sign Fada. Well, when we found the magnet er, because he&#x27;s not even the magnitude when we found the dot product, we know that a dot be cross see is equal to the magnitude of a times the magnitude of be cross see times the co sign of theater. So we already know that. Be cross. See, the magnitude is airbase and a co Cynthia is our height. So what we&#x27;ve essentially found is just the base times the height or the volume of this parallel of pipe head. And that is what is described with our scaler triple product. So this is the volume of what we&#x27;re looking for and that your final answer

Ryan Mcalister · Answer

Yeah, this question asks us to find this gala triple value product of three vectors. The vectors that are given are as follows. Vector A is equal to three zero. Thank you for Victor. B is equal to 111 and Vector C is equal to 740 Our scale, a triple product, is found by taking the dot product of a with the cross product of be cross seat. So order of operations still holds in our vector functions. So we&#x27;ll do the, uh, parentheses first with full find. Be cross. See where we&#x27;re going to do that is our normal method of using a three by three matrix and finding the determinant be cross C is going to be equal to the determinant of the matrix. With the first row being I j k are unit vectors. The second row being be 111 and the third rule being see 740 will do co factor decomposition. To get that, this is equal to determine of one for 10 I, minus 1710 J plus 1714 Okay, now, when we solve for the determinants of thes two by two matrices. We&#x27;ll find that we&#x27;re have a resulting vector of negative four I plus seven j minus three K. In other words, are vector are cross product is negative? 47 negative three. So to find the scale a triple the product, we just need to take a and the dot product of this factor that we just found. So a debt be crusty is going to be equal to 30 negative four dot negative for seven Negative three. And when we do this, what we&#x27;ll find is stop product is equal to this. Some of the products of the corresponding, uh, components of three times negative for is negative. 12 plus zero time +70 plus negative four times negative three, Which is positive. 12. When we add all those up, we get zero. So we found that our scaler triple product is equal to zero. What does this mean? What would we have? A scale? A triple product equal to zero. We know that all three of our vectors are gonna be co planner co planner, just meaning that they&#x27;re in the same plane and I&#x27;ll try to draw a little picture to explain how we know. First we found the cross product of B and C. So let&#x27;s draw our vector, be here and say See is right there. Um, imagine this is a plain and C is a little bit behind B. So if we were to draw like a plane with containing both of these vectors, it would look something like this. When we found the cross product, we found a vector that was perpendicular to both them that is sticking straight up out of this plane that we just true. So it was at a right angle with this one and a right angle with this one. So that would be cross, see? And then we took the dot product with a when he took the doubt product with a we got a value of zero. Something we know from earlier chapters is that if you dot V is equal to zero, then you is perpendicular to V. So a is going to be perpendicular to this green vector that we just found as be cross see, which means it has to fall somewhere in this plane to because it will be at a right angle and the only place where it could be at a right angle is in this plane. So since the scaler triple product was zero, we know that all three of the victors are in the same plane, and that is your final answer.

Ryan Mcalister · Answer

this question asked us to find the scaler triple product of three vectors. Vectors were given our A which is one that Nick of 10 Vector B which is negative on 01 and Vector C, which is zero negative 11 So we have those three factors and we want to find the scale A triple product. Of course, the scale a triple product from our book we know is a dot Be cross See So we&#x27;re going to start with what&#x27;s in the parentheses. We&#x27;re gonna start by finding Be cross, See? So be cross See, we&#x27;re gonna find using our standard method with the three by three determinant whose first rose i. J k our unit vectors and second row will be be so negative 101 The third row will be C zero negative one What? We want to find the determinant of this so we&#x27;ll use co factor decomposition. The way that&#x27;s gonna look is we will take the determinant of the matrix zero negative 111 times I subtract from that the main determinant of matrix negative 1011 times j. Then we add the determinant of matrix negative 1001 negative one times k and that is splitting it up into our co factors. When we find the determinants of each of these two by two, Major sees what we&#x27;ll get is a positive one. Eyes just I we&#x27;ll have negative one j but we&#x27;re subtracting it. So we end up adding JSA Plus one J and then again, plus K. So all this is is I j. K. Another way of writing that if we were to treat each of these as a unit vector, it&#x27;s 100 plus 010 plus 001 So be crusty. We know it was just 11 one. So this is our cross product now what we need to do is dot it with Ace, where we have a dot be cross see which is equal to one negative 10 dot 111 And when we do that, we&#x27;ll find that the dot product is equal to one times one which is one plus one times negative one which is negative one plus one time zero, which is zero. So one plus negative one plus zero is equal to zero. So our scaler triple product is equal to zero now what does that mean? We know from our book that if we have a scale a triple product of zero, the vectors are co plainer. But how do we really describe that? Complainer just means that they&#x27;re in. All of three vectors are in the same plane, so let&#x27;s draw B and C first. We know that from any two vectors, we can always make a plane that fits both of them, no matter what. If you have two vectors, there&#x27;s a plane that can contain both. So all draw the that plane right now something like this, and it&#x27;s hard to do on a two dimensional surface. Draw on three D. But this is as close as we get. So B and C are in this plane and we found Be cross, see, be crossed. See, we know is a vector that is perpendicular to both B and C, which means it sticks straight up out of the plane. It&#x27;s orthogonal to that plane, so there&#x27;s a right angle with B and C. We also found that be cross sea dotted with a is equal to zero. What does that mean? Well, from her book In an earlier chapter, we learned that you dot v equals zero is the same thing is saying you is perpendicular to V. Well, if a is perpendicular to be across C than a has to be somewhere in this plane, because if it&#x27;s at a right angle with be cross see, it has to be somewhere 90 degrees from it, and the only things that are 90 degrees from it are in this plane. So we have found that a is contained in the same plane that contains B and C, and that is your final answer.

Ryan Mcalister · Answer

this question wants us to find the scaler triple product of a of three vectors vectors that it gives our vector A which is equal to to I minus two J minus three k. I&#x27;m gonna rewrite this as just one vector. Right now, it&#x27;s a what&#x27;s called a linear combination of three different unit vectors, all scaled and added together. I&#x27;m gonna rewrite This is just one vector, and that&#x27;s going to be just a simple s too negative to negative three. Because as we know, I is 100 j 010 and k is 001 when we scale them and added together, this is what we get. I&#x27;ll do the same thing for the other vectors. B, which is, um three. I minus j minus K is just equal to three negative one negative one and see, which is just, Ah six. I is just going to be 600 So these are our vectors and were asked to find a scaler triple product. The scale a triple product is, of course, from your book a dotted with Be cross. See? No, let&#x27;s start with finding be cross e b crusty. We&#x27;re gonna find using our regular method of the determinant of a three by three Matrix will have the first row. B i j K are unit vectors. Our second and third row will be B and C. So three negative one negative one and of course, zero or excuse me six 00 When we find the determinant of this will do a co factor, decomposition along the first row will get the determinant of negative one negative 100 times I minus. Remember, we subtract when we&#x27;re doing the J term here. 36 negative 10 times j And then we add the determinant of 36 negative 10 times k. Now we know how to find determines of to a two. Major sees it. This is just going to be zero I, uh, minus six J because we have a positive six as their determinant and we are subtracting it and then we&#x27;re adding six case. Let me rewrite this and a little bit prettier of fashion is just gonna be negative. Six j plus six k. And in other words, that is zero negative. Six positive six. So we have our cross product now all that&#x27;s left to do is dot it with our a vector, remember, Are a vector was too negative to negative three So too negative to negative. Three dotted with zero negative. Six negative positive. Six. And what we do here is we take our, uh we add up the products of the corresponding, uh, components. So are I. Components multiplied together is to time zero, which is zero r j components multiple together. Negative two times. Think of six is positive 12 and R K components negative. Three times six isn&#x27;t negative. 18. So what that is 12 miners 18 is, of course, negative six. And that is our scaler triple product a dotted with the across C. Now the next part of this question asks us if this is co plainer. And if not, it wants us to find the volume of a shape that these three, uh, factors make what we know right away that since the scale a triple product is negative six. It&#x27;s not complainer. Any skeletal product that&#x27;s not zero is from three factors that are not co plainer. So a and B and C or not complaining, which means they are not all in the same plane, so they formed a parallel, a pipette apparent little pipe ed. If I can try to draw it on a two dimensional screen, will look something like this. Let&#x27;s say this is B and C is somewhere behind it. So let&#x27;s imagine this is ah, a little bit behind B. And there&#x27;s an angle here and this is in one plane. Of course, any two vectors, um, can always be put in tow. One plane and now a we know is not in this plan, so it&#x27;ll be sticking out a little bit. Let&#x27;s say it&#x27;s up here. The parallel of hype Ed, this defines is something like this. It has a base of B and C that will look like this, let&#x27;s say and then height defined by a like this, and we can connect this all here, and it&#x27;s not very well drawn, but you can imagine what it really would look like. So it&#x27;s essentially a parallelogram prism, but then you slanted a little bit with a This is Specter A So how are we gonna find the volume of this? Well, there&#x27;s a trick, but I want to show you the long way first, and we&#x27;ll get, um we&#x27;ll meet up with the trick at the very end, and it&#x27;s pretty exciting. So, uh, buckle in first, we want to find for any prism. We&#x27;re always gonna find the volume as base times height. So let&#x27;s find the volume of the base. First, we&#x27;re gonna treat B and C as the base. So here is vector B, and this is now the two dimensional, like looking straight at the base from above. And here, see? And we&#x27;ve got, like, something like this and something like this. So when we&#x27;re looking at B and C, let&#x27;s say this is data DiVall In the area of a parallelogram like this is the base times the height perpendicular to that base of this right here. Well, we know what see is and we know if it is. So this height right here is simply see sign Seda, Remember, since we&#x27;re finding the opposite side of this right triangle here that&#x27;s gonna be see Scient. Ada. So the height or excuse me? The area of this parallelogram is the magnitude of B times, the magnitude of C times sign of data. And now this is where it gets exciting because you probably already recognize this as the magnitude of be cross. See? So the magnitude to be cross see is what we&#x27;re looking for as the base. Well, we had we be cross. He was part of this scale, a triple product. So let&#x27;s see how this will play into it. We have the base and we want the height. While the height again is not gonna be a it&#x27;s going to be the, uh, the portion of a the component of a that is perpendicular to our base or orthogonal from our base. What we know that be cross see, is orthogonal to this Plan B is B cross. He is perpendicular to be and to see. So if there&#x27;s an angle fader between be cross C and A, we want to find the component of A that is alongside this, uh, this value here alongside Be cross, See? Well, that&#x27;s gonna be a times this co sign of data. The blue part is eight times the co sign of Fada. So we want to multiply the base times this height. The base we know is the magnitude of be cross. See, I&#x27;m gonna write this in. Ah, let me to read here. The base is a magnitude of be crossed, see? And the height is the manager of a times the co sign of Fada. And now this is where it gets really, really exciting. Because this is exactly the formula for a dotted with be cross. See? Well, hold on. We just figured out what a daughter be cross. He was. We said it was, um we said it was negative. Six. So we&#x27;re finding the magnitude of negative six. Well, magnitude just means absolute value. The absolute value of six is, of course, positive. Six. So we found that the volume of this scale a triple proud of this peril pipette is equal to the absolute value of the scale. A triple product, which is six. And that&#x27;s gonna be true for any three non co plainer, uh, vectors. The area of the volume of this parallel pipe. It will be the absolute value of the scale. A triple product. So this is the really exciting part. Um, I hope you enjoyed. That&#x27;s your final answer.

Megan Mcfarland · Answer

in this problem, you&#x27;re given a three vectors and asked to find that you don&#x27;t that in parentheses v across the W. And so to do that first we&#x27;re going to determine what be cross W. Is. And to do that, we&#x27;re going to create a matrix with its first row as I, J and K. It&#x27;s second row with the components of the because we&#x27;re doing it be cross W. And so that&#x27;s going to be negative. Three, two and one. And the third row is going to have the components of W zero eight and 10. No, you&#x27;re the cross product. And to get eyes component first, we&#x27;re going to a cross out the first column and take the determinants of this two by two matrix just going to be two times 10 minus one times eight. That is 20 minus eight that is 12. Then we&#x27;re going to subtract OJ&#x27;s component, as we always do. Then we went across at the middle column and take the determines of this two by two Matrix is going to be noted three times 10 minus one times zero. That&#x27;s negative. 30 minus zero. That&#x27;s negative. Three negatives. Cancel out to create a positive and then to get K&#x27;s component. We&#x27;re going to cross out the third comma and we want to take the determinants of this two by two matrix and we want to get negative three times eight minus two times zero. That&#x27;s negative. 24 minus zero. That&#x27;s just negative. 24. Okay, And now we have found our doctor for V Cross W, which is 12 30 Negative 24. And we&#x27;re going to dock product that with you, which is 123 Okay. And when we do that, we&#x27;re going to multiply one times 12 plus two times 30 Those three times negative 24 one times 12 is 12 plus two times 30 12 plus 60 72 24 times three is also 72. So we get 70 to minus 72 that is equal to zero. So our answer for part A is going to be zero now. Part B then asks about whether the vectors are co plainer and for the doctors to be co plainer. You dot the cross w has to be equal to zero. Sure enough, that&#x27;s what we found. And so we found the directors are indeed co plainer. And because they&#x27;re co planner, we don&#x27;t need to find the volume of the parallel pipe. And so that is our answer to party.

Problem

Three vectors $\mathbf{a}, \mathbf{b},$ and $\mat…

Problem 29 Hard Difficulty

Answer

$\vec{a} \cdot(\vec{b} \times \vec{c})=0$
$\vec{a}, \vec{b}, \vec{c}$ are coplaner.

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$\vec{a} \cdot(\vec{b} \times \vec{c})=0$$\vec{a}, \vec{b}, \vec{c}$ are coplaner.

Algebra and Trigonometry

Catherine R.

Heather Z.

Caleb E.

Samuel H.

Vectors Intro

Vector Basics - Example 1

James Stewart, Lothar Redlin, Saleem WatsonAlgebra and Trigonometry

Catherine R.

Heather Z.

Caleb E.

Samuel H.

$\vec{a} \cdot(\vec{b} \times \vec{c})=0$
$\vec{a}, \vec{b}, \vec{c}$ are coplaner.

James Stewart, Lothar Redlin, Saleem Watson

Algebra and Trigonometry