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Question

Find the volume of the parallelepiped determined by the vectors a, $\mathbf{b},$ and $\mathbf{c} .$

$$\mathbf{a}=\langle 1,2,3\rangle, \quad \mathbf{b}=\langle- 1,1,2\rangle, \quad \mathbf{c}=\langle 2,1,4\rangle$$

Vikash Ranjan · Accepted Answer

we have to find the border Deputy will do more. Cannot be dropped. C This includes 2123 minus 112 to 4. This is given as one into four minus two. My student minus two or minus four. Last three to minus one minus Students is given to black 16 minus nine. This is people equal to nine. Unit. You This is dancer.

Dylan Bates · Answer

Welcome back to another cross product problem where we&#x27;re going to calculate the volume of the parallel pipe ad defined by these three vectors. Now the textbook gives us a nice formula that we can use to calculate volume in that it&#x27;s the absolute value of a dotted with the vector B crust see But instead of calculating across product and then adopt product. What we can do is called the triple product. Where instead of having a metrics with I. J and K at the top directly plug in A. B. And see and then calculate the cross product. Keeping in mind our unusual top row in this case, that&#x27;s going to normally we ignore the first row and then calculate the product One times 4 -2 times one. And normally we would write an eye here because I is the top row of the column that we ignored. This isn&#x27;t an eye, this is a one. So we&#x27;re going to multiply by one minus next. We&#x27;ll ignore that second row. Keep that too in mind Giving us -1 times four and a few times too times. And instead of J. Multiplying by Q plus. And then lastly we&#x27;ll ignore our third column -1 times 1 -1 times two. All times. Not K, but three. Well notice that since we don&#x27;t have I. J and K anymore, this is a scalar and not a vector. And so we can add everything up. We&#x27;re looking at 4 -2 is two. James one minus negative four minus four. That&#x27;s going to be negative eight times two plus -1 -2, three times three. So that will be too plus 16 minus nine, Which is just nine, nine units cubed, is the volume of our parallel pipe bed using the triple product method. Thanks for

Dylan Bates · Answer

Welcome back to another cross product video. We&#x27;re going to try to calculate the volume of a parallel pipe bed defined by these three vectors here. The textbook gives us a nice formula that we can use which says that it&#x27;s the absolute value of a dotted with a vector secrecy. But instead of calculating across product and then adopt product, we can do this using a triple product where instead of using I J and K in our matrix. Let&#x27;s get that out of here. Ed, we&#x27;re going to directly plug in are vectors A. B and see So doing that. We get 1 I1J zero. Okay. zero I 1 J one K and one I one J one. Kay. Then we can calculate the cross products same as normal but anywhere we have an I. J and K. We&#x27;ll use those values instead. When we ignore our first column, We&#x27;re looking at one times 1 And it&#x27;s one times 1. One minus one times not I but rather no. Yes. Now we ignore our second column And we look at zero times 1 -1 times one zero times one -1 times one. All multiplied by one us. We&#x27;ll ignore a third column and then we&#x27;ll look at zero times one -1 times one zero times one, one times 1, all multiplied by zero. Since we don&#x27;t have any eyes jay&#x27;s or k&#x27;s, this is just a scalar and so we can add it all up. We have zero times one minus negative one and it&#x27;s one plus negative one time zero at zero plus one plus zero, Giving us a volume of one unit cubed. Using the triple product method. Thanks for watching.

Vikash Ranjan · Answer

fun before. So it is getting asked one. Why do you? You don&#x27;t want one? 11 is equal still you know my ass wanting to minus one this deep roots to want unit you says dancer.

Ryan Mcalister · 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.

Jose Hannan · Answer

Hello, everybody. In this video, I&#x27;m gonna be showed you had to solve Exercise 64 chapter 13 Section four of calculus early, Transcendental. And this problem were given a parallel a pipe head that is formed by the vectors U V and W where you has components 31 and zero. He has components to 41 and W has components 11 and five. And with this information that wants to find the volume of the parallel a pipe head. To do this, we&#x27;re gonna have to recall a few important results from previous exercises. Let&#x27;s start with the result of Exercise 63 which states that the volume of such a parallel a pipe head is equal to you dotted with across W. And so this is a quantity that we want to calculate. To help calculate this. We&#x27;re also going to use a result from exercise 62 which states that the quantity in the absolute value signs you v cross W is equal to the determinant of a three by three matrix. His top row has the components of you three 10 Who second row has components of E to for one and his last year, has the components of W 11 and five. So what we want to do is calculate this determinant and then take its absolute value in order to get the volume that we&#x27;re looking for. So let&#x27;s break down this determinant. This is gonna be equal to three times the determinant of the two by two sub matrix on the bottom. Right here for one one. It&#x27;s attracted from this. We&#x27;re gonna have one times the determinant of the sub matrix form by this bottom left sub column. In this bottom writes a column to one one in five and in the last term, zero times another constant. So we can neglect this term as it is multiplication by zero. So now let&#x27;s expend out these two determinants in the first term. We have the three in front times four times five, which is 20 minus one times one, which is one it&#x27;s attracted from this we have two times five, which is 10 minus one times one, which is one. And so now carrying out these operations, this first term becomes three times 20 minus one, which is 19 and subjected from that, we have 10 minus one, which is nine and now three times 1957. It&#x27;s attracting nine from 57. We get 48 and so to conclude we have that. The volume of the parallel a pipe head is equal to the absolute value of 48 which is just 48 as it is already positive. And so here we have the volume of the parallel a pipette, and that&#x27;s how you solve exercise 64.

Problem

Find the volume of the parallelepiped determined …

Problem 33 Easy Difficulty

Find the volume of the parallelepiped determined by the vectors a, $\mathbf{b},$ and $\mathbf{c} .$

$$\mathbf{a}=\langle 1,2,3\rangle, \quad \mathbf{b}=\langle- 1,1,2\rangle, \quad \mathbf{c}=\langle 2,1,4\rangle$$

Answer

9 units $^{3}$

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9 units $^{3}$

Essential Calculus Early Transcendentals

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Anna Marie V.

Kayleah T.

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Joseph L.

James Stewart

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