Question Video: Finding the Cross Product of Two Vectors in 3D | Nagwa Question Video: Finding the Cross Product of Two Vectors in 3D | Nagwa

# Question Video: Finding the Cross Product of Two Vectors in 3D Mathematics

Let π = π’ and π = 3π’ + 2π£ + 4π€. Calculate π Γ π.

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### Video Transcript

Let π equal π’ and π equal three π’ plus two π£ plus four π€. Calculate π cross π.

We want to calculate the cross product of the vectors π and π. We can write this cross product as the determinant of a three-by-three matrix. The entries in the first row of the matrix are the unit vectors π’, π£, and π€. These are the same π’, π£, and π€ that the vectors π and π are written in terms of. π’, π£, and π€ are perpendicular and point in the π₯-, π¦-, and π§-directions respectively.

We find the entries of the second row of this matrix from the first vector in the cross product, which is π. The entries of the second row are the coefficients of π’, π£, and π€ when π is written in terms of π’, π£, and π€. We already have π written in this way. π is just equal to π’. We can write this in a way that makes the coefficients more obvious. π is equal to one π’ plus zero π£ plus zero π€. And we enter these coefficients β one zero and zero β into our determinant.

The entries of the third and final row of our determinant come from the second vector in the cross product, which is π. Weβre told in the question that π is three π’ plus two π£ plus four π€. And so the entries of the third row are three, two, and four. Now we clear some room to have space to evaluate this determinant. We expand along the first row, getting a term from each entry.

We get a term involving π’, a term involving π£, and a term involving π€. The coefficient of π’ is the determinant of the matrix you get by deleting the row and column which contain π’. Similarly, the coefficient of π£ is the determinant of the matrix you get by deleting the row and column containing π£. We also must subtract this middle term. And finally, the coefficient of π€ is the determinant you get by deleting the row and column containing π€.

We can use a formula to evaluate the two-by-two determinants. Itβs the product of the terms on the leading diagonal minus the product of terms on the other diagonal. And we can simplify to get zero π’ minus four π£ plus two π€. We can write this in component form as zero, negative four, two. The cross products of π and π should be another vector which is perpendicular to both π and π.

While itβs hard to see that the vector we have produced is perpendicular to π. We should be able to see this is perpendicular to π. The π₯-component of our vector is zero, and so itβs perpendicular to π which points in the π₯-direction.