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Numerade Educator



Problem 3 Medium Difficulty

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

$ a = 2j - 4k $ , $ b = -i + 3j + k $


$\mathbf{a} \times \mathbf{b}=14 \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}$

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Video Transcript

Let's try another cross product question where we're taking the cross product of a two J -4 K in B negative I plus three J plus K. In order to solve this, Let's write these down in our matrix two, J -4 K. is the same thing as zero I plus two J minus four K. And B is the same thing as -1 plus three J plus one K. And then we can use the formula laid out in our textbook where we ignore the first column of our matrix And we look at the product two times 1-, maybe four times 3. Let's write that down two kinds one minus negative four times three I minus. And then we'll ignore the second column And look at the product zero times 1 minus negative four times negative one. Let's keep track of those minus signs, zero times one minus negative, four times negative one, jay plus. And the same idea we live nor A. K. The third column. And then look at the product of zero times 3 -22 times negative one giving us zero times three minus two times negative one. Okay, so when we simplify this that gives us two times 1 is two negative four times three is negative 12. So two minus negative 12. That's the same thing as two plus 12. Hi zero times 1 minus negative four times a negative one. That's just positive for jay Plus zero Times 3. That's just zero. And then two times negative one is -2. So zero minus negative too Is the same thing as 0-plus two. Okay, if we simplify this just a little more, That gives us 14 I minus negative four. That's plus four jay plus two. Okay. Which is the same thing as the vector. 14 for two. Thanks for watching.