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Hello everybody.
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In this video, i'm going to be showing you how to solve exercise 12 in chapter 13, section 4 of calculus early transcendentals.
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Now, this problem gives us two vectors, u with components negative 4 -00, as drawn here, as well as v, with components 0 -02 as drawn here.
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And using this diagram, they want us to identify the cross -product, u cross -v.
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Now for the purposes of this video, i'm using a bit of a color key just to make these directions clear in 3d.
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For vectors that are along the x -axis, i'm going to be drawing them in red.
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Along the y -axis, i'll draw in blue, and along the z -axis, i'll draw in green, as i've already done here for u and v.
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Now to identify u cross v, first note that it is a vector.
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So it has two descriptors, a magnitude and a direction.
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Let's first find its magnitude.
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We can do this using theorem 13 .3, the formula for the magnitude of a cross -product.
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It gives us that u cross -v is equal to the product of the magnitudes of u and v, as well as the sign of their angle of separation...