3.5 Cross Product
A Vector Operation in 3-Space
The cross product is a vector operation unique to 3-dimensional space. Unlike the dot product, which results in a scalar, the cross product produces a vector. It’s crucial for understanding concepts like torque, angular momentum, and magnetic forces in physics and engineering.
If \(\mathbf{u} = (u_{1}, u_{2}, u_{3})\) and \(\mathbf{v} = (v_{1}, v_{2}, v_{3})\) are vectors in 3-space, then the cross product \(\mathbf{u} \times \mathbf{v}\) is defined as:
\[ \mathbf{u} \times \mathbf{v} = (u_{2}v_{3} - u_{3}v_{2}, u_{3}v_{1} - u_{1}v_{3}, u_{1}v_{2} - u_{2}v_{1}) \]
This can be remembered using a determinant format:
\[ \mathbf{u} \times \mathbf{v} = \biggl( \left| \begin{array}{cc} u_{2} & u_{3} \\ v_{2} & v_{3} \end{array} \right|, - \left| \begin{array}{cc} u_{1} & u_{3} \\ v_{1} & v_{3} \end{array} \right|, \left| \begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array} \right| \biggr) \quad \text{(1)} \]
Tip
Mnemonic for Calculation: Form a \(2 \times 3\) matrix: \(\left[ \begin{array}{lll}u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{array} \right]\)
Find \(\mathbf{u} \times \mathbf{v}\), where \(\mathbf{u} = (1, 2, -2)\) and \(\mathbf{v} = (3, 0, 1)\).
Solution:
Using the determinant notation or mnemonic:
\[ {\begin{array}{r l}&{\mathbf{u}\times\mathbf{v}={\bigg(}{\bigg|}{\begin{array}{l l}{2}&{-2}\\ {0}&{1}\end{array}}{\bigg|},-{\bigg|}{\begin{array}{l l}{1}&{-2}\\ {3}&{1}\end{array}}{\bigg|},{\bigg|}{\begin{array}{l l}{1}&{2}\\ {3}&{0}\end{array}}{\bigg|}{\bigg)}}\\ &{\qquad=(2,-7,-6)}\end{array}} \]
Enter the components of \(\mathbf{u}\) and \(\mathbf{v}\) to calculate their cross product.
The cross product has fundamental relationships with the dot product and important geometric implications.
If \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in 3-space:
Important
Properties (a) and (b) are key: The cross product of two vectors is always orthogonal (perpendicular) to both original vectors. This is its defining geometric characteristic.
Let’s verify properties (a) and (b) using the vectors from Example 1.
\(\mathbf{u} = (1, 2, -2)\), \(\mathbf{v} = (3, 0, 1)\), and \(\mathbf{u} \times \mathbf{v} = (2, -7, -6)\).
The cross product also has several algebraic properties.
If \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are any vectors in 3-space and \(k\) is any scalar:
The standard unit vectors are \(\mathbf{i} = (1,0,0)\), \(\mathbf{j} = (0,1,0)\), \(\mathbf{k} = (0,0,1)\).
Calculations show:
And due to anti-commutativity:
Also, for self-cross products: - \(\mathbf{i}\times \mathbf{i} = \mathbf{0}\) - \(\mathbf{j}\times \mathbf{j} = \mathbf{0}\) - \(\mathbf{k}\times \mathbf{k} = \mathbf{0}\)

Figure 3.5.2: Mnemonic for unit vector cross products.
The cross product can also be expressed symbolically using a \(3 \times 3\) determinant.
\[ \mathbf{u}\times \mathbf{v} = \left| \begin{array}{lll}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3} \end{array} \right| = \left| \begin{array}{lll}u_{2} & u_{3}\\ v_{2} & v_{3} \end{array} \right|\mathbf{i} - \left| \begin{array}{lll}u_{1} & u_{3}\\ v_{1} & v_{3} \end{array} \right|\mathbf{j} + \left| \begin{array}{lll}u_{1} & u_{2}\\ v_{1} & v_{2} \end{array} \right|\mathbf{k} \quad \text{(4)} \]
Example: \(\mathbf{u} = (1,2,-2)\), \(\mathbf{v} = (3,0,1)\)
\[ \mathbf{u}\times \mathbf{v} = \left| \begin{array}{lll}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 2 & -2\\ 3 & 0 & 1 \end{array} \right| = (2)(1) - (-2)(0)\mathbf{i} - ((1)(1) - (-2)(3))\mathbf{j} + ((1)(0) - (2)(3))\mathbf{k} = 2\mathbf{i} - 7\mathbf{j} - 6\mathbf{k} \]
Warning
The cross product is not associative! \(\mathbf{u}\times (\mathbf{v}\times \mathbf{w}) \neq (\mathbf{u}\times \mathbf{v})\times \mathbf{w}\) For example, \(\mathbf{i}\times (\mathbf{j}\times \mathbf{j}) = \mathbf{0}\), but \((\mathbf{i}\times \mathbf{j})\times \mathbf{j} = -\mathbf{i}\).
The direction of \(\mathbf{u} \times \mathbf{v}\) for non-zero vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by the right-hand rule.
This rule is a convention that establishes the orientation of the resulting vector in 3-space.
Figure 3.5.3: The Right-Hand Rule.
The magnitude of the cross product has a direct geometric meaning.
From Lagrange’s identity: \(\| \mathbf{u}\times \mathbf{v}\|^{2} = \| \mathbf{u}\|^{2}\| \mathbf{v}\|^{2} - (\mathbf{u}\cdot \mathbf{v})^{2}\). Using \(\mathbf{u}\cdot \mathbf{v} = \| \mathbf{u}\| \| \mathbf{v}\| \cos \theta\): \[ \| \mathbf{u}\times \mathbf{v}\| = \| \mathbf{u}\| \| \mathbf{v}\| \sin \theta \quad \text{(6)} \] This formula is precisely the area of the parallelogram determined by vectors \(\mathbf{u}\) and \(\mathbf{v}\).
Theorem 3.5.3: If \(\mathbf{u}\) and \(\mathbf{v}\) are vectors in 3-space, then \(\| \mathbf{u}\times \mathbf{v}\|\) is equal to the area of the parallelogram determined by \(\mathbf{u}\) and \(\mathbf{v}\).
Figure 3.5.4: Area of parallelogram.
Find the area of the triangle determined by points \(P_1(2,2,0)\), \(P_2(-1,0,2)\), and \(P_3(0,4,3)\).
Form vectors: \(\overrightarrow{P_1P_2} = (-1-2, 0-2, 2-0) = (-3, -2, 2)\) \(\overrightarrow{P_1P_3} = (0-2, 4-2, 3-0) = (-2, 2, 3)\)
Calculate cross product: \(\overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3} = ((-2)(3) - (2)(2), (2)(-2) - (-3)(3), (-3)(2) - (-2)(-2))\) \(= (-6-4, -4+9, -6-4) = (-10, 5, -10)\)
Calculate magnitude: \(\| \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3}\| = \sqrt{(-10)^2 + 5^2 + (-10)^2} = \sqrt{100 + 25 + 100} = \sqrt{225} = 15\)
Area of triangle: \(A = \frac{1}{2} \| \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3}\| = \frac{1}{2} (15) = \frac{15}{2}\)
Figure 3.5.5: Triangle formed by three points.
Enter the coordinates of three points to calculate the area of the triangle they form.
The scalar triple product involves three vectors and results in a scalar.
Definition 2: If \(\mathbf{u}, \mathbf{v}, \mathbf{w}\) are vectors in 3-space, then \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\) is called the scalar triple product.
It can be calculated using a \(3 \times 3\) determinant: \[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \left| \begin{array}{ccc}u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \\ w_{1} & w_{2} & w_{3} \end{array} \right| \quad \text{(7)} \]
For \(\mathbf{u} = (3, -2, -5)\), \(\mathbf{v} = (1, 4, -4)\), \(\mathbf{w} = (0, 3, 2)\):
\[ \begin{array}{r l} & {\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w}) = \left| \begin{array}{l l l}{3} & {-2} & {-5}\\ {1} & {4} & {-4}\\ {0} & {3} & {2} \end{array} \right|}\\ & {\qquad = 3(4 \cdot 2 - (-4) \cdot 3) - (-2)(1 \cdot 2 - (-4) \cdot 0) + (-5)(1 \cdot 3 - 4 \cdot 0)}\\ & {\qquad = 3(8 + 12) + 2(2) - 5(3)}\\ & {\qquad = 3(20) + 4 - 15 = 60 + 4 - 15 = 49} \end{array} \]
The absolute value of determinants has profound geometric meaning.
Theorem 3.5.4 (a): The absolute value of \(\operatorname{det}\left[ \begin{smallmatrix}u_{1} & u_{2} \\ v_{1} & v_{2} \end{smallmatrix} \right]\) is the area of the parallelogram in 2-space determined by \(\mathbf{u}=(u_1, u_2)\) and \(\mathbf{v}=(v_1, v_2)\).
Theorem 3.5.4 (b): The absolute value of \(\operatorname{det}\left[ \begin{smallmatrix}u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3}\\ w_{1} & w_{2} & w_{3} \end{smallmatrix} \right]\) is the volume of the parallelepiped in 3-space determined by \(\mathbf{u}, \mathbf{v}, \mathbf{w}\). \[ V = |\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})| \quad \text{(9)} \]
Figure 3.5.7: (a) Area of 2D parallelogram, (b) Volume of 3D parallelepiped.
Enter the components of three vectors \(\mathbf{u}, \mathbf{v}, \mathbf{w}\) to calculate the volume of the parallelepiped they determine.
The scalar triple product provides a simple test for whether three vectors lie in the same plane.
If the vectors \(\mathbf{u}, \mathbf{v}, \mathbf{w}\) have the same initial point, then they lie in the same plane if and only if:
\[ \mathbf{u}\cdot (\mathbf{v}\times \mathbf{w}) = \left| \begin{array}{lll}u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3}\\ w_{1} & w_{2} & w_{3} \end{array} \right| = 0 \]
Note
Coplanar points: A set of points is coplanar if you can draw a single flat surface (a plane) that contains all of them.
Coplanar vectors: Vectors are coplanar if their linear combinations lie within the same plane.
Geometrically, if three vectors are coplanar, they cannot form a parallelepiped with a positive volume. Hence, their scalar triple product (which is the signed volume) must be zero.
Important
The cross product and scalar triple product are fundamental tools in ECE for understanding 3D geometry, forces, fields, and spatial relationships.