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Use Theorem 3 to prove the Cauchy-Schwarz Inequal…

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Problem 60 Hard Difficulty

Suppose that all sides of a quadrilateral are equal in length and opposite sides are parallel. Use vector methods to show that the diagonals are perpendicular.

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Video Transcript

So if we want to show diagonals of a foresighted quadrilateral herb one that has all the same size I should say are equal. We can go ahead and start by labeling a quadrilateral with the same size. So the only wants the adverse where and rhombus, uh, someone to label it a b like this, a b center, Not like that e and a And then over here, I'll say a e e a ghost Recall that for a quadrilateral besides, cross from it should be parallel, So these should be the same factor. Um, so they all have the same magnitude. So maybe I should say, up here, Magnitude of a is equal to the magnitude of B. We just have where they're oriented differently. So I'm just going to do everything with the square up here. But you can also do with the rhombus, and it will give you the same thing. So our first diagonal is going to go straight across like this. So I want to look at this factor here, so I'll call this deep one. Well, d one is just going to be a plus B. So we have a plus B now for our second diagonal. Um, we can go ahead and have where it goes like this. So knows we're starting at a so d two is going to be a, um and then we go ahead and we're actually sorry. It wouldn't be starting from me. Um, we're going over to be. So that's where we're landing on. So it would be and then minus where we're starting, which would be a I don't know why I had such a hard time seeing that, because this is the end. And then over here, this is the start. So d two is just going to be b minus a, uh, and you can also do that same one over here as well, Because our second diagonal you would see, actually, if I'm going to do it that way, it would start from a go to be so you. So you would just be be might say again. So that's d two. And maybe I'll fill this in over here. Just kind of for completion. Six. So you can see how doesn't matter which one we use. Okay. Now, if we want to show these are perpendicular, we can take the dot product of them and show it is zero. So let's look at that. So d one dotted with D two is going to be equal to a plus B started with B minus A, and we can distribute the dot product just like we could multiplication and do the following. So this is going to be a dotted with B. Well, actually, me spend the screen all so a daughter would be plus B started with, um, Earth. Let me do it this way. So a daughter with the and then plus actually minus a dotted with a and then loss be dotted with the and then minus B dotted with a. But remember, since the dot product is communicative, it doesn't matter if we do a deal would be or be dotted with a So those cancel out? Yeah, and we're going to get that this is equal to just be daughter would be minus a daughter with a But this dot product here is actually just the magnitude of the squared. If you got it with itself and then we have the same thing over here. It's the dot product with itself and if you recall, will be set up here A is equal to be so we can just go ahead and replace that. So be a squared minus he squared. And that is just going to be zero. So d one dotted with d two music zero, which implies, say to is going to be, uh, 90 degrees because remember, this would be to the magnitude of the one times the magnitude of the to co sign of data and co sign is equal to 90 degrees, or the only time coastlines equals zero is at 90 degrees, at least in the first quadrant, and this here proves that they are going to be for particular. So you put your little proof box and it's my face because you're glad you're able to prove it.

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This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Kayleah Tsai

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Joseph Lentino

James StewartCalculus: Early Transcendentals

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