💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Like

Report

Numerade Educator

Like

Report

Problem 34 Easy Difficulty

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

$ a = i + j $, $ b = j + k $, $ c = i + j + k $

Answer

1 unit $^{3}$

More Answers

Discussion

You must be signed in to discuss.

Video Transcript

Welcome back to another cross product video. We'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'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's get that out of here. Ed, we'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'll use those values instead. When we ignore our first column, We're looking at one times 1 And it'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'll ignore a third column and then we'll look at zero times one -1 times one zero times one, one times 1, all multiplied by zero. Since we don't have any eyes jay's or k's, this is just a scalar and so we can add it all up. We have zero times one minus negative one and it'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.