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Numerade Educator

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Problem 28 Hard Difficulty

Find the area of the parallelogram with vertices $ P (1, 0, 2), Q (3, 3, 3), R (7, 5, 8) $, and $ S (5, 2, 7) $.

Answer

$\sqrt{269}$

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Video Transcript

Hello. Welcome to this lesson in this lesson, we'll find, uh, you have the parallelogram. So the area is a p ko across PS, so this requires us to find the victor P ko What? That is three minus one during minus zero. Their minus two. So this gives us 23 and one. The next one that will look for is the victor PS in that days. Mhm. Mhm five minus one to minus zero seven minus two. Yeah. Mhm. And this is for two. Yeah, five. So p kill. Oh, with the peak l across the PS has given us the I, j and K 231 42 and five. Okay, so here we'll have spanned using the cool factor might make it. Where would delete the first column? The second, uh, first column. First rule. They have 32 15 minus G will delete the first rule in a second column. So you have 241 and five. The next one will delete the first rule and the third column. So you have to four, 32 Okay, so here we have 15 minus two minus j. We have 10 minus four than last K. We have four minus 12. So I that becomes a 13 I minus 60 than minus eight k. So this is the P Q. Of course. PS. Now let's take the the magnitude of that because that's a vector. So can take a magnitude of a vector. And that is by squaring the components, all of them. The sum of this class of all of them. So that is one 69 plus 36 plus 64. So that is curled of 26 nine. Okay, so this is the area of the parallelogram. Thanks your time. That's the end of the lesson.