TheoremProved

Orthogonality and area property of the cross product

Statement

For any $\vec{u}=(u_1,u_2,u_3)$ and $\vec{v}=(v_1,v_2,v_3)$ , the cross product $\vec{u}\times\vec{v}=(u_2v_3-u_3v_2,\,u_3v_1-u_1v_3,\,u_1v_2-u_2v_1)$ satisfies $(\vec{u}\times\vec{v})\cdot\vec{u}=0$ , $(\vec{u}\times\vec{v})\cdot\vec{v}=0$ , and $|\vec{u}\times\vec{v}|^2=|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2$ .

Why is it true?

The first two equations confirm that $\vec{u}\times\vec{v}$ is perpendicular to both input vectors, so it gives a ready-made normal vector to any plane spanned by $\vec{u}$ and $\vec{v}$ . The third equation rewrites as $|\vec{u}\times\vec{v}|=|\vec{u}||\vec{v}|\sin\theta$ , the base-times-height area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$ .

Proof sketch

Compute the dot product directly from the coordinate definitions: (u x v) . u = (u2 v3 - u3 v2) u1 + (u3 v1 - u1 v3) u2 + (u1 v2 - u2 v1) u3.

Expanding the three products gives u1 u2 v3 - u1 u3 v2 + u2 u3 v1 - u1 u2 v3 + u1 u3 v2 - u2 u3 v1 = 0, since the six terms cancel in pairs. The calculation for (u x v) . v = 0 is identical.

For the magnitude identity, expand |u x v|^2 = (u2 v3 - u3 v2)^2 + (u3 v1 - u1 v3)^2 + (u1 v2 - u2 v1)^2. Meanwhile, expand |u|^2 |v|^2 - (u . v)^2 = (u1^2+u2^2+u3^2)(v1^2+v2^2+v3^2) - (u1 v1 + u2 v2 + u3 v3)^2.

In the second expression the diagonal terms u1^2 v1^2, u2^2 v2^2, u3^2 v3^2 cancel, leaving the exact same six off-diagonal terms u_i^2 v_j^2 - 2 u_i u_j v_i v_j as in the expansion of |u x v|^2. Since u . v = |u||v|cos(theta), this equals |u|^2|v|^2(1 - cos^2(theta)) = (|u||v|sin(theta))^2.

Topics that use this theorem

Coordinates in space

Step-by-step proofs

No step-by-step proof yet for this theorem.