WEBVTT
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Given that the vector π has components negative five, negative nine, negative one and the vector π has components two, negative one, negative seven, find the cross product of π and π.
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The cross product of two vectors can be written as the determinant of a three-by-three matrix where the entries of the first row of the matrix are the unit vectors π’, π£, and π€ which point in the π₯-, π¦-, and π§-directions, respectively.
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The entries of the second row of the matrix are the components of the first vector in our cross product, so the vector π with components negative five, negative nine, and negative one.
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And the third and final row of our matrix contains the components of the second vector in our cross product, the vector π with components two, negative one, and negative seven.
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Now we have our three-by-three determinant.
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We can evaluate it by expanding along the first row.
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The first entry of the first row is the vector π’ and the minor of π’, that is the determinant of the elements you get if you delete the row and column containing π’ from the matrix.
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Itβs the determinant negative nine, negative one, negative one, negative seven.
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And so the first term of our expansion is this: determinant negative nine, negative one, negative one, negative seven multiplied by the vector π’.
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The next term comes from the next entry in the first row of the matrix which is π£ multiplied by its minor, the determinant negative five, negative one, two, negative seven.
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And we remember that we have to subtract this term.
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And the final term comes from the final entry in the first row of our matrix, π€, multiplied by its minor, the determinant negative five, negative nine, two, negative one.
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And this term we add.
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We have expanded the three-by-three determinant along its first row, and we can now expand each term in this expansion.
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We evaluate the first determinant first and we multiply the two entries on the leading diagonal, thatβs negative nine and negative seven together, like so.
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And from this we subtract the product of the other two entries, negative one and negative one, like so.
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So thatβs the determinant expanded.
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Remember that the first term is this determinant times π’.
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And we can evaluate negative nine times negative seven minus negative one times negative one; this is just 62.
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So the first term is just 62π’.
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And so we get our final answer: 62π’ minus 37π£ plus 23π€.
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In component form, this has components 62, negative 37, and 23.