Question

Find \((a) \mathbf{u} \times \mathbf{v},(b) \mathbf{v} \times \mathbf{u},\) and \((\mathbf{c}) \mathbf{v} \times \mathbf{v}\). $$ \begin{array}{l} \mathbf{u}=3 \mathbf{i}+5 \mathbf{k} \\ \mathbf{v}=2 \mathbf{i}+3 \mathbf{j}-2 \mathbf{k} \end{array} $$

Short answer

\(\mathbf{u} \times \mathbf{v}\) = -15i - 6j + 9k, \(\mathbf{v} \times \mathbf{u}\) = 15i + 6j - 9k, \(\mathbf{v} \times \mathbf{v}\) = \(\mathbf{0}\)

Step-by-step solution

Cross product of \(\mathbf{u}\) and \(\mathbf{v}\)

First, write the cross product as a determinant of a 3x3 matrix, with \(i\), \(j\), and \(k\) as the first row, the coefficients of \(\mathbf{u}\) as the second row, and the coefficients of \(\mathbf{v}\) as the third row. Then, calculate the determinant to get \(\mathbf{u} \times \mathbf{v}\).

Cross product of \(\mathbf{v}\) and \(\mathbf{u}\)

Similarly, form a 3x3 matrix but this time with the coefficients of \(\mathbf{v}\) as the second row and the coefficients of \(\mathbf{u}\) as the third row. Then, calculate the determinant to get \(\mathbf{v} \times \(\mathbf{u}\). Remember, the cross product is not commutative, so \(\mathbf{u} \times \mathbf{v}\) is not always equal to \(\mathbf{v} \times \(\mathbf{u}\).

Cross product of \(\mathbf{v}\) with itself

Here, the second and third rows of our 3x3 matrix will have the same coefficients, since these are for \(\mathbf{v}\). The result will be the zero vector because the cross product of any vector with itself is always the zero vector. Therefore, \(\mathbf{v} \times \mathbf{v}\) = \(\mathbf{0}\).