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(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 (0, -2, 0) $ , $ Q (4, 1, -2) $ , $ R (5, 3, 1) $

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(a) \langle 13,-14,5\rangle(b) $\frac{1}{2} \sqrt{390}$

03:51

Wen Zheng

Calculus 3

Chapter 12

Vectors and the Geometry of Space

Section 4

The Cross Product

Vectors

Lala W.

February 24, 2021

seems like the answer in the description is correct but the video is wrong..

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02:56

In mathematics, a vector (from the Latin word "vehere" meaning "to carry") is a geometric entity that has magnitude (or length) and direction. Vectors can be added to other vectors according to vector algebra. Vectors play an important role in physics, engineering, and mathematics.

11:08

In mathematics, a vector (from the Latin word "vehere" which means "to carry") is a geometric object that has a magnitude (or length) and direction. A vector can be thought of as an arrow in Euclidean space, drawn from the origin of the space to a point, and denoted by a letter. The magnitude of the vector is the distance from the origin to the point, and the direction is the angle between the direction of the vector and the axis, measured counterclockwise.

08:23

(a) Find a nonzero vector …

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02:49

Welcome back to another cross product problem where we're trying to find a new vector Orthogonal to the plane created by these three points. One of the ways we can do that is by finding two vectors in the plane and taking the cross product which is necessarily going to be perpendicular to that plane. So let's look at vector PQ which is just q minus P Or 4 0. One minus negative too, And -2 0. And then Pr is just ar minus P, Giving us 5 0, three minus negative too, And 1 0. Plugging those into our matrix here for three negative two 5 5 and one. We can find the cross product PQ. Cross pr Using the method from our textbook specifically, will ignore the first column And look at three times 1 minus negative, two times five. That's one minus negative, two times five. Hi minus. Then we ignore our second column four times 1 -20 times five, all times J. And lastly we'll ignore the third column And we'll look at four times 5 -5 three times five. Five -3 times five. Okay, if we simplify this down to a single vector, we're looking at three minus negative 10 or 13. Four minus negative 10 is 14. Can't forget to make that negative. and 20-15 fast five. This is perpendicular to the plane defined by those three points. Therefore the negative version of it negative 13 14 -5 or any scalar multiples of this vector will also work. Now if we want to find the area of the triangle defined by three by these three points, we can again use the fact that we know the magnitude of PQ Cross PR has written this is the area of the parallelogram defined by those three points. Sorry, defined by those two vectors. And so we divide that area by two to get us the area of the triangle we want. And so the magnitude of 13 negative 14 5 Is just going to be 13 squared negative 14 squared and five squared all under a square root and divided by two, Giving us the square root of 169 Plus 1 96 plus 25 all over to Which is just the square root of 390 divided by two. Thanks.

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