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# Find the cross product $a \times b$ and verify that it is orthogonal to both $a$ and $b$.$a = \langle 2, 3, 0 \rangle$ , $b = \langle 1, 0, 5 \rangle$

## $15 i-10 j-3 k,$

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Let's try a cross product question where we take the cross product of two vectors A equals 230. Let's write that in the second layer of our matrix And B equals 105. Let's write that in the third layer of our matrix. Now, if we use the strategy laid out in the text book in order to calculate the cross product first, we ignore the first column and we'll take the cross product three times five minus zero times zero, That's 5 0 times zero I minus. Next we'll ignore the second column. And we'll look at happens when we look at two times five zero Times 1 That's two times five zero times 1 jay. And lastly will ignore the third column And multiply two times 0 -3 times one giving us two times zero -3 times one. Okay, and now we just simplify three times 5 is 15 minus zero, I minus can minus zero jay plus zero minus three. Okay, And when we put that all together, that leaves us with 15 I minus 10 jay plus, nope screwed that up minus three jay, which is also just the vector 15 minus 10 minus three. Thanks for watching.

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