Question Video: Determining the Cross Product of Vectors | Nagwa Question Video: Determining the Cross Product of Vectors | Nagwa

# Question Video: Determining the Cross Product of Vectors Mathematics • Third Year of Secondary School

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Given that π = β9π’ β π£ + 3π€ and π = 3π’ β 2π£ β 7π€, determine π Γ π.

02:48

### Video Transcript

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.

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. 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. 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. 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. Namely, three, negative two, and negative seven.

And now we have a determinant which we can evaluate, and weβll do this in the traditional way by expanding along the first row. 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. So this is the determinant: minus one, three, negative two, negative seven. And we multiply these two together. And thatβs our first term.

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. Itβs the determinant negative nine, three, three, negative seven. And we remember that we have to subtract this term. 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.

And now we just have to expand these two by two determinants. 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.

And evaluating that, we get 13, and so our term is 13π’. 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.

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