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Problem

(a) Let $ P $ be a point not on the line $ L $ th…

05:42

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Problem 44 Hard Difficulty

(a) Find all vectors $ v $ such that
$$ \langle 1, 2, 1 \rangle \times v = \langle 3, 1, -5 \rangle $$
(b) Explain why there is no vector $ v $ such that
$$ \langle 1, 2, 1 \rangle \times v = \langle 3, 1, 5 \rangle $$


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Wen Zheng

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Calculus 3

Calculus: Early Transcendentals

Chapter 12

Vectors and the Geometry of Space

Section 4

The Cross Product

Related Topics

Vectors

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Vector Basics Overview

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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Problem 54

Video Transcript

Welcome back to another cross products problem. This time we're trying to find all vectors V. That's right, that is X. Y. Z. Such that 1- one. Cross v equals 3. -5. Let's go ahead and write these in our matrix here. 1 to 1. And we said V was X Y Z. So we can calculate the cross product using the method in our textbook. Remember we eliminate the first column of our matrix and look at two Z minus one Y to z minus. Why? All multiplied by high minus. Then we ignore the second column. Look at one Z -1 X. The minus X. Time's jay. And lastly we'll ignore the third column And look at one way and a few X. I -2 x. Okay. Simplifying this a little bit gives us the vector to Z minus. Why negative Z minus X. That's x minus C. And why -2 X. Which should give us the same thing as the answer that we want. 31 negative five. This means that we have three linear equations here to the -Y equals three X -Z equals one. And Why -2? X equals negative five. So we have three equations with three unknowns. There's a couple different ways we can proceed here. But since we want something in the form X, Y. Z. Let's go ahead and write Y and Z. In terms of X. It's using these two equations. We know that Z is equal to x minus one, adding a Zeon, subtracting one from both sides. And we know that why is equal to two x minus five. One of the things we can do is check what's going on up here, You know that two times E minus? Why should equal three. And sure enough, two x -2 minus two, x minus five. These cancel -2 -5, no negative two plus five. There we go, negative two plus five Equals three. So these are consistent and we are looking at vectors in the form x x is just equal to x. Why is equal to two x -5 and Z is equal to x minus what? So for any value of x We can write v as X 2, -5, X -1. And it will satisfy this equation up here. Now what if we change things around a little bit? What if we want 1-1? Cross v To equal 3. 1 positive five. How would that work? The same idea? We would still want uh to z minus Y equals three. We would still want x minus Z equals one. And this time we would want y minus two X To equal positive five. OK, easy enough. We still get Z Equals X -1 and this time we get y equals two X plus five. But what happens when we check this equation for consistency? Where we get to z That's two times X minus one minus Y. That's two X plus five. We want this to equal three. But what do we get? 2? X -2 minus two. X minus five. Well that's the same thing as two X -2 x. We're left with -7 equals three. And that doesn't work. That doesn't make any sense. Negative seven does not equal three. Therefore this equation has no solutions. There are no vectors V. Such that 1- one. Cross v equals 315. Thanks for watching.

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