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Problem 33 Medium Difficulty

Find the volume of the parallelepiped determined by the vectors $ a, b $, and $ c $.

$ a = \langle 1, 2, 3 \rangle $, $ b = \langle -1, 1, 2 \rangle $, $ c = \langle 2, 1, 4 \rangle $


9 units $^{3}$

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Video Transcript

Welcome back to another cross product problem where we'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'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'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't an eye, this is a one. So we're going to multiply by one minus next. We'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'll ignore our third column -1 times 1 -1 times two. All times. Not K, but three. Well notice that since we don't have I. J and K anymore, this is a scalar and not a vector. And so we can add everything up. We're looking at 4 -2 is two. James one minus negative four minus four. That'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