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Hello everybody.
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In this video, i'm going to be showing you how to solve exercise 68 in chapter 13, section 4 of calculus early transcendental.
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Now, in this problem, they want us to prove that a vector cross -width itself always produces the zero vector, and they want us to prove this in three different ways.
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The first method they want us to use is the definition of the cross -product itself.
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As per this definition, the vector u cross -v between two vectors u and v is a vector that has a magnitude, which is equal to the magnitude of u times the magnitude of v, times the sign of the angle between u and v.
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So this means that u cross with view is a vector that is equal to the magnitude of u times the magnitude of u squared, times the sign of zero, as, of course, the angle of any vector with itself is zero.
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They are not separated by an angle because they're the same vector.
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But the sign of zero is equal to zero, which makes this whole quantity equal to zero.
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And now the only vector that has a magnitude of zero is the zero vector itself.
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So we can conclude that root u cross u equals the zero vector.
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And so that's how you prove this statement using the definition of the cross product.
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But now let's go on to using the determinant method.
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As per theorem 13 .6, we can express u as a vector with components, u1, u2, and u3.
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And this means that u cross u is equal to the determinant of a 3 by 3 matrix with the standard unit vectors in the top row, the components of u in the second row, as well as in the third row.
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And now if we expand this determinant, we have that the first term is equal to i times the determinant.
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Of this 2x 2 by 2 sub matrix, u2, u3, u2, u3, minus a similar quantity for j, j times this the sub -matrix formed by this bottomless sub -colum and bottom right column, which we take the determinant of u1, u3, u3 again...