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Problem 6 Medium Difficulty

Find the cross product $ a \times b $ and verify that it is orthogonal to both $ a $ and $ b $.

$ a = ti + \cos tj + \sin tk $ , $ b = i - \sin tj + \cos tk $


$\mathbf{i}+(\sin t-t \cos t) \mathbf{j}-(t \sin t+\cos t) \mathbf{k}$

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Video Transcript

let's do another cross product problem. This time in two parts, we want to take the cross product of the vector T. I plus co sign T. J plus sign T. K. And the vector one I minus sign T. J plus cosign T. Okay. Using the method from our textbook, we're going to ignore the first column and look at the product co sign times co sign minus sine times negative sign. That'll give us Cosine squared t minus sign T times negative sci fi. That will give us negative Science square T. And if we subtract negative science squared T. That's the same thing as plus science square T. All times I minus now. We'll ignore the second column of our matrix. And look at T times co sign t Minour sign Tee Times one going to be T. Co sign t minus one time sci fi. All times J. And lastly, we'll ignore the third column. Well look at T times minus sign T- Cosine T. Times one. That gives us minus T. Sign T. Subtract one. Co sign T. All of that times K. We can simplify this a little bit because cosine squared plus sine squared is just one. Hi. And then we can write minus T. Co sign T plus sign T. All of this time's jay plus negative T. Sign T minus co sign T. All of this times K. And if we want to rewrite this as a vector, we get a cross B is just one than sein T minus T. Times go sign T And then -170 minus coastline T. That's the end of Part one. Now, if we want to verify that this new vector, let's call it C. Is orthogonal to both A and B. We can do that by taking the dot product A dot C and a and B dot C. And if we did this right, our dog products should be zero. So let's take a dot product of A and C will multiply the first term time. The first time is the first term tee times one is just T plus co sign tee times all of this sign t minus T. Uh huh. And co sign T plus sign T and all of this, that'll be negative T. Sign t minus co sign T. Just squeezing that in. And then if we simplify this, this gives us T plus co sign times Sine let's call that sign T. Co sign T minus T times co sign T squared. In the last term We have -1 times sine T squared. That looks promising minus 70 co sign T. So the first thing we notice is we have a plus scientific. 070- Scientific. Oh 70. And then we can simplify this as minus T times Cosine T squared plus sign T squared. And as we already said, Cosine squared plus sine squared is one. So we're really looking at t minus T times one which is indeed zero. So a in C are orthogonal, let's do the same thing for BNC. Look at one times 1 that's just one. And we'll look at negative sign T times 70 minus T co sign T. And then we'll add to that co sign T times all of this. That will be a negative T sign t minus coz I M T. And once again, let's simplify all of this. We're almost done. It's really going at one minus sign is squared T plus T time, cy Inti times co sign t minus T times sci fi times Cosine T. That's great minus Cosine squared T. And once again We have a T 70 co sign T minus T scientific. Oh 70 Leaving us one sine squared T plus co sine squared T. That's 1 -1 is a big old, zero meaning we did our math right and see is perpendicular or orthogonal to be as well, which it should be. Thanks for watching.