Finding a Cross Product Using Determinants

If \(\mathbf{v}=2\mathbf{i}-\mathbf{j}+\mathbf{k}\) and \(\mathbf{w}=4\mathbf{i} +2\mathbf{j}-\mathbf{k}\), find

1. \(\mathbf{v}\times \mathbf{w}\)
2. \(\mathbf{w} \times \mathbf{v}\)
3. \(\mathbf{v\times v}\)

Solution (a) \(\mathbf{v}\times \mathbf{w}=\left\vert \begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 4 & 2 & -1 \end{array} \right\vert =(1-2)\mathbf{i}-(-2-4)\mathbf{j}+(4+4)\mathbf{k}=-\mathbf{i}+6 \mathbf{j}+8\mathbf{k}\)

(b) \(\mathbf{w}\times \mathbf{v}=\left\vert \begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 2 & -1 \\ 2 & -1 & 1 \end{array} \right\vert =(2-1) \,\mathbf{i}-( 4+2) \,\mathbf{ j\,+\,}( -4-4) \,\mathbf{k=i}-6\mathbf{j}-8\mathbf{k=-}( \mathbf{v}\times \mathbf{w}) \)

(c) \(\mathbf{v}\times \mathbf{v}=\left\vert \begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 2 & -1 & 1 \end{array} \right\vert =0\,\mathbf{i}-0\,\mathbf{j}+0\mathbf{\,k}=\mathbf{0}\)