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Numerade Educator

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Problem 4 Easy Difficulty

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

$ a = 3i + 3j - 3k $ , $ b = 3i - 3j + 3k $

Answer

$a \times b=0 i-18 j-18 k$

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Video Transcript

Let's try another cross product question. We're taking a cross product of two vectors A&B. And let's write them in our Matrix A. Is three, I plus three J minus three K. And b. is three I minus three. J plus three K. And we can use the formula in the textbook. We ignore our first column and we look at three times 3 minus negative three times negative three. Let's keep track of all these minus signs. three times 3 -3 negative three times minus three I minus. And then we ignore the second column and look at the product of three times three minus negative three times three. So that's three times three minus -3 times three jay plus. And then we'll look at the third column or will ignore the third column And look at three times -3 Times three times 3, three times -3 -3 times three. There we go, That's three times -3 minus three times three. Okay, let's start to simplify this, three times three is nine And -3 times -3 is also nine. So 9 -9 I -3 times three is 9 -3 times three is -9. So we have nine minus minus nine. There's nine plus nine. Okay. Plus three times -3 is minus nine minus three times 3 is nine. Okay Simplifying this a little more, we know that 9 -9 is zero. I so we don't even need to write that if we don't want minus nine plus nine is 18 J And then -9 -9. That's another -18. Okay Or we can write that as the vector zero minus 18 minus 18. Thanks for watching.