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Given that the vector π is equal to negative nine π’ minus π£ plus three π€ and the vector π is equal to three π’ minus two π£ minus seven π€, determine the cross product of π and π, π cross π.
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The cross product of two vectors can be written as the determinant of a matrix, where the entries in 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 this matrix are the coefficients of π’, π£, and π€, respectively, in the first vector, in the cross product π.
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So as π is equal to negative nine π’ minus π£, or one π£, plus three π€, the second row is negative nine, negative one, three.
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And for the third and final row, we just have the coefficients of π’, π£, and π€ in the second vector, in the cross product π.
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Namely, three, negative two, and negative seven.
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And now we have a determinant which we can evaluate, and weβll do this in the traditional way by expanding along the first row.
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The first entry in the first row is the vector π’ and its minor is the determinant that you get by deleting the row and column in which π’ lies.
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So this is the determinant: minus one, three, negative two, negative seven.
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And we multiply these two together.
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And thatβs our first term.
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We do the same thing with the second entry of the first row which is π£, its minor which you get by deleting the second column and the first row in which π£ lies.
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Itβs the determinant negative nine, three, three, negative seven.
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And we remember that we have to subtract this term.
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And finally, we add the term coming from the third entry of the first row which is π€ times its minor, the determinant negative nine, negative one, three, negative two.
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And now we just have to expand these two by two determinants.
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So we have the product of the entries on the main or leading diagonal, negative one times negative seven, minus the terms on the other diagonal, three times negative two.
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And evaluating that, we get 13, and so our term is 13π’.
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And if we also evaluate the two by two determinants in the other two terms, we get 13π’ minus 54π£ plus 21π€ is the vector value of our cross product of π and π.