Cross Product (Vector Product)

Follow the cyclic scheme: the first coordinate of a⃗×b⃗ is a₂b₃ − a₃b₂, the second a₃b₁ − a₁b₃, the third a₁b₂ − a₂b₁. For each result coordinate you cover the corresponding row and cross-multiply the other two, forwards minus backwards. Example: a⃗ = (1|2|0), b⃗ = (3|0|1): first coordinate $2\cdot{}1 - 0\cdot{}0 = 2$, second $0\cdot{}3 - 1\cdot{}1 = -1$, third $1\cdot{}0 - 2\cdot{}3 = -6$, so a⃗×b⃗ = (2|−1|−6). Always check via dot product: $(2|-1|-6)\cdot{}(1|2|0) = 2 - 2 + 0 = 0$ ✓ and likewise 0 with b⃗ ✓. This ten-second check catches almost all arithmetic and sign errors.

Its main job is the normal vector: a⃗×b⃗ is perpendicular to both factors. If you have a plane in parametric form with spanning vectors u⃗ and v⃗, n⃗ = u⃗×v⃗ delivers the normal vector for the coordinate or normal form in one step; without the cross product you would have to solve a system of two dot-product conditions. Second, the magnitude |a⃗×b⃗| measures the area of the spanned parallelogram, half of it being the triangle area; this is needed for area tasks in space. Third, a⃗×b⃗ = 0⃗ tests for parallelism. And combined with the dot product you get the scalar triple product (a⃗×b⃗)·c⃗, whose absolute value gives the volume of the parallelepiped, one sixth of it for the pyramid on three edge vectors.

|a⃗×b⃗$| = |a$⃗$|\cdot{}|b$⃗$|\cdot{}\sin{} \varphi{}$ is the area of the parallelogram spanned by the two vectors: base |a⃗| times height |b⃗$|\cdot{}\sin{} \varphi{}$. Example: a⃗ = (1|2|0) and b⃗ = (3|0|1) have a⃗×b⃗ = (2|−1|−6) with magnitude $\sqrt{(4 + 1 + 36)} = \sqrt{41} \approx{} 6.4$ area units. For triangles, the most common exam case, you halve: $A = \frac{1}{2}|a$⃗×b⃗|, where a⃗ and b⃗ are two side vectors from the same vertex. Two limiting cases make the formula plausible: for parallel vectors $\sin{} \varphi{} = 0$, the parallelogram degenerates to a segment, area 0. For perpendicular vectors $\sin{} \varphi{} = 1$ and the area is simply the product of the lengths, as for a rectangle.

Only in three-dimensional space is there exactly one direction perpendicular to two (non-parallel) directions; in $R^{2}$ there is none at all, in $R^{4}$ infinitely many perpendicular directions, so such a product cannot be defined uniquely there. In the plane, the determinant a₁b₂ − a₂b₁ takes over the area role; it is precisely the third coordinate of the cross product if you embed the vectors in space. The right-hand rule fixes which of the two possible perpendicular directions the result takes: if the thumb of the right hand points along a⃗ and the index finger along b⃗, the middle finger points along a⃗×b⃗. Anticommutativity follows directly: b⃗×a⃗ = −(a⃗×b⃗), swapping reverses the orientation.

With two dot products: the result n⃗ = a⃗×b⃗ must be perpendicular to both input vectors, so n⃗·a⃗ and n⃗·b⃗ must both be exactly zero. Example: for a⃗ = (1|0|2), b⃗ = (0|1|1) one computes n⃗ = (0·1−2·1 | 2·0−1·1 | 1·1−0·0) = (−2|−1|1); check: n⃗·a⃗ = −2 + 0 + 2 = 0 ✓ and n⃗·b⃗ = 0 − 1 + 1 = 0 ✓. If it is not zero, the error is almost always in the middle coordinate, whose order is reversed (a₃b₁ − a₁b₃). Extra checks: for obviously parallel vectors the zero vector must result, and a plausibility glance at the size of the magnitude (area!) never hurts. These checks cost seconds but regularly save entire geometry tasks, because everything that follows depends on the normal vector.