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# Find the cross product $a \times b$ and verify that it is orthogonal to both $a$ and $b$.$a = \frac{1}{2} i + \frac{1}{3} j + \frac{1}{4} k$ , $b = i + 2j - 3k$

## $\mathbf{a} \times \mathbf{b}=-\frac{3}{2} \mathbf{i}+\frac{7}{4} \mathbf{j}+\frac{2}{3} \mathbf{k}$

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mm. Let's try another cross product problem. This time with fractions. We want to look at the cross product of a time. A cross B. Where a is the vector one half I Plus 1 3rd J Plus 1 4th K. Let's write that in our matrix And B is one I plus two j -3K. Let's write that on our matrix as well. Then the cross product A cross be. We can use the formula from our textbook where we ignore the first column and look at one third times minus three -1 4th Times two. One third times minus three minus 1/4 times two I minus. And then we ignore the second column And look at 1/2 times -3 Times 1 4th -1. Yeah, of one half times minus three minus one. Fourth types one jay. And lastly we add to that, we ignore the third column And look at 1/2 times two -1 3rd Times one. 1/2 times two minus one third times one. Okay simplifying all of us a little bit. One third times minus three is minus 3/3 Or just -1 And 1 4th times two is to over four or one half -1 -1 half. I minus negative three times one half is minus 3/2 In 1 4th times one is just 1/4 so negative 3/2 -1/4 J plus one half times two is 2/2 or just one minus 1/3 times one is just 1/3. Okay. Simplifying this even more We have negative 1 -1 half. That's -3/2. I and then negative 3/2 minus 1/4. That's the same thing as minus 6/4 minus 1/4 Which would give us -7/4 but we're subtracting that so we're gonna say plus 7/4 jay plus one minus one third. That's three thirds minus one third or two thirds. Okay? Or we can write this as a vector -3/2, 7/4 to over three. Thanks for watching.

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