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# If $a \cdot b = \sqrt{3}$ and $a \times b = \langle 1, 2, 2 \rangle$, find the angle between $a$ and $b$.

## $60^{\circ}$

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Welcome back to another cross product video. This time we're trying to find the angle between two vectors. But all we know is the value of their dot product dot be is the square root of three. And their cross product a Cross B is the vector 1 to 2. We don't know A and B. In fact, we're not going to figure out A and B. We can jump straight to what is the angle between them? How do we do that? Well, let's start with some things that we know about. Dot products and cross products. We know that the doc product between two vectors is the magnitude of a times the magnitude of B. Times cosine of the angle between them. That's going to help us out. We also know that the magnitude of the cross product magnitude of a cross B will end up being the magnitude of a magnitude be times sine theta. And we can calculate the magnitude of this vector here as the square root of one squared plus two squared plus two squared that's the square root of one plus four plus four. Or the squared of nine is 3. So what do we know? Well, we can substitute a dot be for square 23 equals magnitude of a magnitude is B. Cosign theta and for the magnitude of a Crosby, we know that three is the magnitude of a magnitude of b sine theta and we'll notice that both of these terms have a magnitude of a at times of magnitude is being them, then we don't really need. Therefore, if we solve for that unknown value on one side, we get that. This is route three over Go Science data and on the other This is three over sine theta. And so now we have two quantities that are equal to each other. If we rearrange this a little bit, we can write for example three over Route three, which is the same thing as three. Route three over Route three. Route three or just square at three. This is equal to signed data over co signed data just by rearranging, multiplying both sides by Signed data, dividing both sides by Square 2, 3. And that is tangent. Theta square root three equals tangent data. That tells us that fada has to be arc tangent of squared of three, which is going to be pi over three or 60 degrees. How do we know that? This is in the first quadrant? Well, since a dot be is positive, we know that we have to be in the first quadrant. And so we're taking the positive version of this. Thanks for watching.

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