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Find the volume of the parallelepiped determined by the position vectors $\mathbf{u}=\langle 3,1,0\rangle, \mathbf{v}=\langle 2,4,1\rangle,$ and $\mathbf{w}=\langle 1,1,5\rangle$ (seeExercise 63 ).

The volume of the parallelepiped is $48$.

Calculus 3

Chapter 13

Vectors and the Geometry of Space

Section 4

Cross Products

Vectors

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Hello, everybody. In this video, I'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're gonna have to recall a few important results from previous exercises. Let'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'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're looking for. So let'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's attracted from this. We'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'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'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'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's how you solve exercise 64.

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