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Problem 47 Medium Difficulty

Show that $ {\mid a \times b \mid}^2 = {\mid a \mid}^2 {\mid b \mid}^2 - (a \cdot b)^2 $.

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Video Transcript

Welcome back to another cross product problem. This time we're trying to prove an identity about the magnitude of a cross B squared. So the first thing we can note is that this is just the magnitude of a cross B multiplied by the magnitude of a cross B. And then we start, I need to start using things that we know about cross products. Here's some things that we know, we know that the magnitude of a cross B is magnitude magnitude of be signed Ceta, meaning if this is squared then we can just square each of the terms in this identity. So let's make that a little more explicit. It is going to be the magnitude of a squared, the magnitude of B squared times sine squared sarah. Now one of the things we know about science squared zeta is that we can right signs squared as one minus cosine squared theta. And then we can expand these parentheses to get a squared B squared times one minus a squared b squared, I'm cosine squared. So let's write that out magnitude of a squared magnitude of B squared times one minus magnitude of a squared magnitude of b squared, go sine squared tha tha. And hold up. We know that magnitude of a magnitude of be cosign theta is just a dot be, we have all of this squared, Therefore that's the same thing as squaring the other side. And we're left with the magnitude of a squared magnitude of B squared minus a dot be squared, which is what the identity was we were trying to prove. And so the left side magnitude of across B squared is equal to this whole thing. Thanks for watching.