The dot product turns two vectors into a number. In space there is a second product that turns two vectors into a vector: the cross product. It answers a question you will ask constantly in this course: which direction is perpendicular to both of these? Its formula is a 3 × 3 determinant.
Idea clave
$$\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = \langle u_2v_3 - u_3v_2,\ u_3v_1 - u_1v_3,\ u_1v_2 - u_2v_1 \rangle.$$
It is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$, points by the right-hand rule, and its length $|\mathbf{u}||\mathbf{v}|\sin\theta$ is the area of the parallelogram they span.
Por qué funciona
Perpendicular: dot the result with $\mathbf{u}$:
$$u_1(u_2v_3 - u_3v_2) + u_2(u_3v_1 - u_1v_3) + u_3(u_1v_2 - u_2v_1) = 0,$$
because the six products cancel in pairs (it is a determinant with two equal rows). The same happens with $\mathbf{v}$.
Length: multiplying out gives Lagrange's identity $$|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2|\mathbf{v}|^2 - (\mathbf{u} \cdot \mathbf{v})^2 = |\mathbf{u}|^2|\mathbf{v}|^2(1 - \cos^2\theta),$$ so $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\sin\theta$: base $|\mathbf{u}|$ times height $|\mathbf{v}|\sin\theta$.
Length: multiplying out gives Lagrange's identity $$|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2|\mathbf{v}|^2 - (\mathbf{u} \cdot \mathbf{v})^2 = |\mathbf{u}|^2|\mathbf{v}|^2(1 - \cos^2\theta),$$ so $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\sin\theta$: base $|\mathbf{u}|$ times height $|\mathbf{v}|\sin\theta$.
- $\mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v})$: order matters (anticommutative).
- $\mathbf{u} \times \mathbf{u} = \mathbf{0}$, and parallel vectors have cross product $\mathbf{0}$ ($\sin 0 = 0$).
- $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, $\mathbf{k} \times \mathbf{i} = \mathbf{j}$ (go around the circle $\mathbf{i} \to \mathbf{j} \to \mathbf{k}$).
- The volume of the box (parallelepiped) on $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ is $|\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})|$.
Worked examples
Ejemplo resuelto: Compute and check
Find $\langle 1, 2, 3 \rangle \times \langle 4, 5, 6 \rangle$.
- First: $u_2v_3 - u_3v_2 = 2 \cdot 6 - 3 \cdot 5 = -3$.
- Second (careful): $u_3v_1 - u_1v_3 = 3 \cdot 4 - 1 \cdot 6 = 6$.
- Third: $u_1v_2 - u_2v_1 = 1 \cdot 5 - 2 \cdot 4 = -3$.
- Check with a dot product: $$\langle -3, 6, -3 \rangle \cdot \langle 1, 2, 3 \rangle = -3 + 12 - 9 = 0.$$
Respuesta: $\langle -3, 6, -3 \rangle$
Ejemplo resuelto: Area of a triangle
Find the area of the triangle with vertices $P(1, 0, 0)$, $Q(0, 2, 0)$, $R(0, 0, 3)$.
- Two sides from $P$: $\overrightarrow{PQ} = \langle -1, 2, 0 \rangle$ and $\overrightarrow{PR} = \langle -1, 0, 3 \rangle$.
- $$\overrightarrow{PQ} \times \overrightarrow{PR} = \langle 2 \cdot 3 - 0 \cdot 0,\ 0 \cdot (-1) - (-1) \cdot 3,\ (-1) \cdot 0 - 2 \cdot (-1) \rangle = \langle 6, 3, 2 \rangle.$$
- Its length is $\sqrt{36 + 9 + 4} = 7$: the parallelogram's area. The triangle is half: $\frac{7}{2}$.
Respuesta: $\frac{7}{2}$
Ejemplo resuelto: Volume of a box
Find the volume of the parallelepiped with edges $\mathbf{u} = \langle 1, 1, 0 \rangle$, $\mathbf{v} = \langle 0, 1, 1 \rangle$, $\mathbf{w} = \langle 1, 0, 1 \rangle$.
- $$\mathbf{v} \times \mathbf{w} = \langle 1 \cdot 1 - 1 \cdot 0,\ 1 \cdot 1 - 0 \cdot 1,\ 0 \cdot 0 - 1 \cdot 1 \rangle = \langle 1, 1, -1 \rangle\text{:}$$ the base area times the unit normal.
- $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = 1 + 1 + 0 = 2$: base area times height.
- Volume $= |2| = 2$.
Respuesta: ${}2$
Error común
Assuming $\mathbf{u} \times \mathbf{v} = \mathbf{v} \times \mathbf{u}$. Swapping the order turns the vector around: the right-hand rule now curls the other way. Losing the minus sign in the middle. Expanding the determinant along the top row gives $+\mathbf{i}(\ldots) - \mathbf{j}(\ldots) + \mathbf{k}(\ldots)$. If you write the middle component as $u_3v_1 - u_1v_3$ you have already included that sign. Always finish with the check: your answer dotted with each factor must give ${}0$.