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# Find the cross product $a \times b$ and verify that it is orthogonal to both $a$ and $b$.$a = \langle 4, 3, -2 \rangle$ , $b = \langle 2, -1, 1 \rangle$

## $\mathbf{a} \times \mathbf{b}=\mathbf{i}-8 \mathbf{j}-10 \mathbf{k}$

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

Let's try another cross product question. We're looking at the cross product of A. Which is 43 -2. Let's write that in the second row of our matrix negative two and B. Two negative one negative one. Sorry to negative one positive one. And then let's use the formula provided in the text book. So first we ignore our first column And look at the product three times 1 minus negative two times negative one. That's going to give us three times 1 minus The product of -2 and negative one I minus. Next we'll ignore J. And we'll look at four times 1 -20 Times two. That's four times one minus negative two times two jay. And lastly we'll ignore the third column And we'll look at four times negative one -3 Times two. Is us four times a negative one -3 Times two. Okay if we simplify all of this, that gives us the three times one is three -2 times negative one is 2. It's really an 3 -2. I minus four times 1 is four -2 times two is -4. So we're looking at four minus native for jay plus And we have four times negative one. It's minus four -3 times two minus six. Okay? And so putting this all together we get 3 -2 is one. I we can just straight I four minus negative four is eight, so we're minus eight J And then -4 -6 is -10. Okay? Alternately We can write that as one minus eight -10. All as a vector. Thanks for watching.

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp