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Numerade Educator

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Problem 17 Easy Difficulty

If $ a = \langle 2, -1, 3 \rangle $ and $ b = \langle 4, 2, 1 \rangle $, find $ a \times b $ and $ b \times a $.

Answer

$a \times b=-7 i+10 j+8 k$
$b \times a=7 i-10 j-8 k$

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Video Transcript

let's do another cross product problem where we're looking at the cross product of vector A. That's too negative one, three and vector B is for two. What? We'll write those in our matrix just like this so that we can use the method described in our textbook where we ignore the first column of our matrix, then look at -1 times one -3 times two -1. Types one -3 times two I minus. Then we'll ignore the second column and look at two times one -3 times four and is one -3 times four jay lastly will ignore the third column And look at two times 2 minus negative one times four. Giving us it comes to minus -1 Times four. Okay, if we simplify all of this, that gives us negative one minus six. Hi minus two minus 12 jay, plus for minus negative four. That's for plus four. Not quite enough from there. Okay, so we can simplify this and say that a cross B Is in Vector Form -1 -6. That's minus seven, 2 -12 is -10. So negative negative 10 it's 10 and four plus 4 is it? Now if we wanted to calculate be cross A. We have properties of cross products that tell us that be cross A. It's the same thing as negative A. Cross B. So we don't need to go through all this work again. Rather we can just right That this is seven negative 10 negative eight. And if you want you can verify that the slow way. Thanks for watching.