Compute u×v\mathbf{u} \times \mathbf{v}u×v where u=(1,0,2)\mathbf{u} = (1, 0, 2)u=(1,0,2) and v=(3,1,0)\mathbf{v} = (3, 1, 0)v=(3,1,0), and verify orthogonality.
u×v=(−2,6,1)\mathbf{u} \times \mathbf{v} = (-2, 6, 1)u×v=(−2,6,1) and both (u×v)⋅u=0(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{u} = 0(u×v)⋅u=0 and (u×v)⋅v=0(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{v} = 0(u×v)⋅v=0
u×v=(2,−6,−1)\mathbf{u} \times \mathbf{v} = (2, -6, -1)u×v=(2,−6,−1) and (u×v)⋅u≠0(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{u} \neq 0(u×v)⋅u=0
u×v=(−2,6,1)\mathbf{u} \times \mathbf{v} = (-2, 6, 1)u×v=(−2,6,1) and ∣u×v∣=0|\mathbf{u} \times \mathbf{v}| = 0∣u×v∣=0
u×v=(1,6,2)\mathbf{u} \times \mathbf{v} = (1, 6, 2)u×v=(1,6,2) and (u×v)∥u(\mathbf{u} \times \mathbf{v}) \parallel \mathbf{u}(u×v)∥u