Recipe

Compute a cross product, then find an area or a normal vector

Use when: you have two vectors in space and need a vector perpendicular to both, the area of the parallelogram they span, or a normal vector to the plane they lie in

Goal: the cross product vector itself, plus whichever of area or normal-vector interpretation the problem asks for

Taught in The cross product

1. 1

   Write \mathbf i,\mathbf j,\mathbf k across a top row, \mathbf u's components across the second row, and \mathbf v's components across the third row.

   Example

   For \mathbf u=\langle1,2,2\rangle and \mathbf v=\langle2,-2,1\rangle, the three rows read \mathbf i,\mathbf j,\mathbf k; then 1,2,2; then 2,-2,1.

   Why: the cross product's formula is exactly this determinant's expansion along the top row, and writing it out this way keeps every pairing straight.

2. 2

   Cover row 1 and the \mathbf i column. Cross-multiply the remaining 2\times2 block and subtract: u_2v_3-u_3v_2.

   Example

   (2)(1)-(2)(-2)=2+4=6.

   Why: this is the standard cofactor expansion pattern, applied to the first column.

3. 3

   Cover row 1 and the \mathbf j column, cross-multiply and subtract, then flip the sign: -(u_1v_3-u_3v_1).

   Example

   -\big[(1)(1)-(2)(2)\big]=-[1-4]=3.

   Why: the middle term in a 3\times3 determinant expansion always carries a minus sign, and skipping that flip is the single most common cross-product mistake.

4. 4

   Cover row 1 and the \mathbf k column, cross-multiply and subtract: u_1v_2-u_2v_1.

   Example

   (1)(-2)-(2)(2)=-2-4=-6.

   ⚡ Shortcut

   When both vectors lie flat in a plane parallel to the xy-plane, meaning both have a zero z-component, the \mathbf i and \mathbf j components above vanish automatically and only this \mathbf k component survives. It collapses to the ordinary 2\times2 determinant ad-bc for vectors \langle a,b\rangle and \langle c,d\rangle. For \mathbf a=\langle3,2,0\rangle and \mathbf b=\langle1,4,0\rangle: \mathbf a\times\mathbf b=\langle0,0,(3)(4)-(2)(1)\rangle=\langle0,0,10\rangle, the familiar 2D "cross product" trick revealed as one special case of the full 3D formula.

   Why: same cofactor pattern as the \mathbf i component, applied to the last column.

5. 5

   The components from Steps 2 through 4 already assemble into \mathbf u\times\mathbf v=\langle6,3,-6\rangle; what's left depends on which quantity the problem actually wants.

   Example

   |\langle6,3,-6\rangle|=\sqrt{36+9+36}=\sqrt{81}=9, so the parallelogram's area is 9.

   • asked for the cross product itself, or a normal vector to the plane \mathbf u and \mathbf v span — \mathbf u\times\mathbf v=\langle6,3,-6\rangle is already the answer.
   • asked for the area of the parallelogram \mathbf u and \mathbf v span — take the magnitude of \mathbf u\times\mathbf v instead.

   Why: the cross product's direction is automatically perpendicular to both \mathbf u and \mathbf v, which makes it a normal vector for free, while its length is defined to equal the area of the parallelogram the two vectors span.