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

$ a = \langle t, 1, \frac{1}{t} \rangle $ , $ b = \langle t^2, t^2, 1 \rangle $

$(1-t) \hat{i}-0 \hat{j}+\left(t^{3}-t^{2}\right) \hat{k}$

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Let's try another cross product problem where we take the cross product of two vectors. And then after this is a two part problem, will try and verify that our solution is indeed perpendicular To both of those two original vectors. So we're looking to take the cross product of vector A. That's T. One one over T. As we're writing in this matrix here in B. That's T squared t squared one. Let's use the same technique that are textbook specifies where we ignore the first column And then look at one times 1 -1 of her T and his T squared. That gives us one times 1 minus one over T. Times T squared all times I minus. Then we'll ignore the second column And look at T. Times one minus one of her tee times T squared T groups tee times one and it's one of her T times T squared. All of that is J. Or is times J. Plus. Then it will ignore the third column, like at T times t squared -1 times t squared. And that gives us tee times t squared -1 times t squared all of that times K. If we simplify this we're left with one times one is just one minus T squared over. Tea is just one of her tee times T squared is T squared over tea. Mhm one minus T. I minus two times one minus one of her tee times T squared. That's again T squared over. Tea is just tea jay plus tee times T squared is T cubed -1 times t squared is t squared. Okay? And if we write this as a vector, That's 1- T. Remember, T minus T. Is just zero and then T cubed minus T squared. If we want to verify that this is orthogonal to both A and B. We can do that by looking at the dot product any times see our new vector and and be dot C again our new vector. And we should get zero for each of these if they are indeed orthogonal. So let's look at the dot product. A dot C will multiply t Times 1 -1 Plus one times 0 Plus one over T times T cubed minus t squared. And if we simplify this this gives us t minus t squared plus zero plus t cubed over T. That's T squared minus t squared over tea. That's just T. So now we have t minus t t squared minus t squared plus zero equals zero. Just as we expect. Now, if you look at be dot C will do the same idea. Really. Can t squared Times 1 -1 b squared Times 1 -1 plus t squared time zero plus one times t cubed minus t squared, expanding everything out. That gives us T squared times one minus T squared times T. That's T cute plus zero Plus one T cubed minus T squared, and once again t squared minus t squared t cubed minus T cubed, plus zero equals zero, indicating that our cross product is indeed perpendicular To our original two vectors.