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Use the scalar triple product to determine whether the points $ A (1, 3, 2), B (3, -1, 6), C (5, 2, 0) $, and $ D (3, 6, -4) $ lie in the same plane.

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The given points are coplanar

02:59

Wen Zheng

Calculus 3

Chapter 12

Vectors and the Geometry of Space

Section 4

The Cross Product

Vectors

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Lectures

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.

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Use the scalar triple prod…

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In each part, use a scalar…

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00:41

Find vectors a, b, and c w…

Welcome back to another cross products problem. This time we're trying to determine if the four points given our co planner. Using a triple product method, the way we can do that is first find three vectors in this plane starting at the same origin. So we could look at for example, A B, A C and a D. And then using the triple product method, calculate the volume of the corresponding parallel pipe pad. If that volume is zero that means they're parallel pipe at a squished flat and therefore the points are all co planner. If not, if the parallel pipe it has volume, then they're not co cleaner. So A B is just b minus a three minus one, -1 -3, six minus two. A C is C minus A. That's 5 -1, two minus three. In 0 -2 and 80 Is 3 -1. Six minus three negative four minus two in the triple product method. Instead of doing this formula as a cross product and then a separate thought product, we can just plug in A. B and C directly into this matrix here. So there's a there's be and there. See. And then we can calculate the cross product same as normal except instead of I. J and K will have these three numbers here. Usually what we would do is ignore the first column and look at negative one times negative six minus -2 -3. one times negative six minus -2 times three. And instead of I We're going to multiply by two, Dennis. Then we ignore our second column four times negative six minus negative two times two, four times negative six -20 Times two. And instead of being multiplied by J, We're multiplying by -4 plus. And then we ignore our 3rd column four times three. It is one times 2, all multiplied by four. Since we don't have any I Js or k's. This is not a vector, but just a scalar. So let's add everything up. Six minus negative six. Thank you. 12 times two minus negative, 24 minus negative four. I'll be -20 times minus four and then lastly 12 minus negative two times four. Plugging this all in. We're getting 24 minus 80 plus 56 Which is indeed zero. Therefore, all of these points are actually co planner. Thanks for what?

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