silemus.

31Cross product of i\mathbf{i} and j\mathbf{j}

Compute (1,0,0)×(0,1,0)\displaystyle (1, 0, 0) \times (0, 1, 0).

Using the determinant rule for the cross product, the i and j components come out 0, and the k component is 1·1 − 0·0 = 1. So (1,0,0) × (0,1,0) = (0,0,1), which checks out as perpendicular to both inputs.

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