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Problem 32 Hard Difficulty

(a) Find a nonzero vector orthogonal to the plane through the points $ P, Q, $ and $ R $, and (b) find the area of triangle $ PQR $.

$ P (2, -3, 4) $ , $ Q (-1, -2, 2) $ , $ R (3, 1, -3) $

Answer

a.$\mathbf{i}-23 \mathbf{j}-13 \mathbf{k}$ is perpendicular to the plane containing $P, Q$ and $R$
b.$A r e a=\frac{\sqrt{699}}{2}$

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Video Transcript

Welcome back to another cross product problem where we're trying to find a vector perpendicular to the plane, defined by these three points here. One of the ways we can do that is by finding two vectors in the plane and then taking their cross product, which necessarily produces a new vector perpendicular to the two that we used in the cross product to do that. Let's look at peak you or Q minus P -1 -2, negative two minus negative three And to -4. And then pr that's ar minus P. That gives us 3 -2, one minus negative three And -3 -4. We can take these two vectors and plug them into our matrix here, one for and negative seven so that we can find the cross product PQ cross pr Using the method from our textbook, Remember to do this, we're going to ignore the first column of our matrix and look at one times negative, seven minus negative. Two times four. All multiplied by i minus. Then we'll ignore the second column. We look at negative three times negative seven minus negative. Two times one, negative three times negative. Seven minus -2 times one. Let's keep track of all these minus signs plus. And then we'll add ignore the third column negative three times four minutes. One times one, -3 times four -1 times one. Okay, writing this all as a single vector that's going to give us negative seven minus negative eight. That's just going to be one minus Uh positive 21 minus negative too. So that'll be minus 23 And then negative 12 minus one. The only -13 equivalently we could just use the negative of this negative one, 23 13 or any other vector. That's a scalar times either of these vectors will work. Now if we want to find the area of the triangle peak, you are we can just use the absolute value or the magnitude of the cross product that we just found. PQ Cross PR has written this is going to give us the magnitude of the parallelogram defined by those two vectors. And so we divided by two. And so we're looking at square root of each of the length of this vector squared. What was this factor again? One squared plus negative, 23 squared plus negative 13 squared. All divided by two. That's the same thing as one squared. Sorry, just one plus 529 plus 169. All divided by two. Or equivalently that's the square root of 699 divided by two. Thanks for watching.

Coker College

Topics

Vectors

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