Recipe

Compute a cross product in R3

Use when: you have two vectors in \mathbb R^3 and need a third vector perpendicular to both of them, or that vector's length for an area

Goal: the vector \mathbf u\times\mathbf v, verified against both perpendicularity conditions, ready to use directly or to take the length of for an area

Taught in The cross product in R3

1. 1

   Cross-multiply (u_2,u_3) with (v_2,v_3) diagonally and subtract: u_2v_3-u_3v_2.

   Example

   For \mathbf u=(2,-1,3), \mathbf v=(4,0,-2): (-1)(-2)-(3)(0) = 2-0 = 2.

   Why: the first output entry, by the formula's own construction, never involves subscript 1 — it's built entirely from the two vectors' other two entries.

2. 2

   Cross-multiply (u_3,u_1) with (v_3,v_1) diagonally and subtract: u_3v_1-u_1v_3.

   Example

   (3)(4)-(2)(-2) = 12-(-4) = 16.

   Why: writing this entry with the subscripts in the same order as the first and third entries — u_1v_3-u_3v_1 instead of u_3v_1-u_1v_3 — is the single most common sign slip in the whole formula, since the middle entry's order is reversed relative to the other two.

3. 3

   Cross-multiply (u_1,u_2) with (v_1,v_2) diagonally and subtract: u_1v_2-u_2v_1.

   Example

   (2)(0)-(-1)(4) = 0-(-4) = 4.

   Why: the same diagonal pattern as the first entry, now applied to the two vectors' first two entries.

4. 4

   Collect the three entries into one vector, then confirm \mathbf u\cdot(\mathbf u\times\mathbf v)=0 and \mathbf v\cdot(\mathbf u\times\mathbf v)=0.

   Example

   \mathbf u\times\mathbf v=(2,16,4). Check: \mathbf u\cdot(2,16,4)=4-16+12=0 and \mathbf v\cdot(2,16,4)=8+0-8=0.

   Why: perpendicularity to both original vectors is guaranteed for a correctly computed cross product — if either dot product comes out nonzero, one of the three entries above has an arithmetic or sign error.

5. 5

   Check what the question is actually asking for before reporting a final answer.

   Example

   For \mathbf a=(1,2,2), \mathbf b=(3,0,-4): \mathbf a\times\mathbf b=(-8,10,-6), and \lVert\mathbf a\times\mathbf b\rVert=\sqrt{64+100+36}=\sqrt{200}=10\sqrt2\approx14.14, the parallelogram's area.

   • the question asks for a perpendicular direction (a normal vector, a torque, a plane's orientation) — Step 4's result is already the final answer.
   • the question asks for an area or a magnitude — take \lVert\mathbf u\times\mathbf v\rVert, the ordinary \mathbb R^3 length, applied to Step 4's result.

   Why: the cross product's length is exactly the area of the parallelogram the two vectors span — a fact proven once and reused here, not re-derived every time.