Question

Find \(u \times v\) and show that it is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). $$ \begin{array}{l} \mathbf{u}=\mathbf{i}+\mathbf{j}+\mathbf{k} \\ \mathbf{v}=2 \mathbf{i}+\mathbf{j}-\mathbf{k} \end{array} $$

Short answer

The cross product \(u \times v\) is \(-2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\) and it is indeed orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\) as their respective dot products with \(u \times v\) is zero.

Step-by-step solution

Calculation of cross product

First, calculate the cross product \(u \times v\) using the determinant of a 3x3 matrix. Arrange the coefficients of \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) within \(\mathbf{u}\) and \(\mathbf{v}\) respectively in the rows of the matrix. Treat \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) as the column headers. Find the determinant of the matrix. \[ \(u \times v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 1 \ 2 & 1 & -1 \end{vmatrix} \) \]\[ = \mathbf{i}(1*(-1) - 1*1) - \mathbf{j}(1*(-1) - 1*2) + \mathbf{k}(1*1 - 1*2) \]\[ = -2\mathbf{i} + 3\mathbf{j} - \mathbf{k} \]

Proving Orthogonality

Then, we have to prove that \(u \times v\) is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). This is done by showing that the dot product of \(u \times v\) with \(\mathbf{u}\) and \(\mathbf{v}\) is zero. This is true for all orthogonal vectors. Calculate the dot product:\[ \mathbf{u} \cdot (u \times v) = (1*-2) + (1*3) + (1*-1) = -2 + 3 - 1 = 0 \]\[ \mathbf{v} \cdot (u \times v) = (2*-2) + (1*3) + (-1*-1) = -4 + 3 + 1 = 0 \] We can see that the value of both the dot products is zero, hence proving that \(u \times v\) is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\).