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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
$$ \langle 1, 2, 1 \rangle \times v = \langle 3, 1, 5 \rangle $$


a. $\mathbf{v}=<x, 2 x-5, x-1>$
b. no possible solutions

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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.