The Main Idea 

The cross product multiplies two vectors to produce a third vector that’s perpendicular to both inputs. While the dot product gives a scalar, the cross product gives a vector.

Cross Product Formula

For [latex]\mathbf{u} = \langle u_1, u_2, u_3 \rangle[/latex] and [latex]\mathbf{v} = \langle v_1, v_2, v_3 \rangle[/latex]:

[latex]\mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle[/latex]

The result is orthogonal to both [latex]\mathbf{u}[/latex] and [latex]\mathbf{v}[/latex].

Direction: The Right-Hand Rule

1. Point your right hand fingers along [latex]\mathbf{u}[/latex]
2. Curl them toward [latex]\mathbf{v}[/latex]
3. Your thumb points in the direction of [latex]\mathbf{u} \times \mathbf{v}[/latex]

Key Properties

• Anticommutative: [latex]\mathbf{u} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{u})[/latex] (order matters!)
• Self cross product: [latex]\mathbf{v} \times \mathbf{v} = \mathbf{0}[/latex]
• Zero vector: [latex]\mathbf{u} \times \mathbf{0} = \mathbf{0}[/latex]

Unit Vector Cross Products

Follow the cyclic pattern [latex]\mathbf{i} \to \mathbf{j} \to \mathbf{k} \to \mathbf{i}[/latex]:

• Forward: [latex]\mathbf{i} \times \mathbf{j} = \mathbf{k}[/latex], [latex]\mathbf{j} \times \mathbf{k} = \mathbf{i}[/latex], [latex]\mathbf{k} \times \mathbf{i} = \mathbf{j}[/latex]
• Backward (negative): [latex]\mathbf{j} \times \mathbf{i} = -\mathbf{k}[/latex], etc.

Magnitude

[latex]||\mathbf{u} \times \mathbf{v}|| = ||\mathbf{u}||  ||\mathbf{v}||  \sin \theta[/latex]

Compare to the dot product (uses cosine). The cross product uses sine.

Cross products require three dimensions—there’s no direction simultaneously perpendicular to two 2D vectors without leaving the plane.

The Main Idea 

The cross product formula [latex]\mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle[/latex] is hard to remember. Determinant notation provides an organized, memorable alternative.

Quick Determinant Review

2×2 determinant: [latex]\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} = a_1b_2 - b_1a_2[/latex]

3×3 determinant (expand along first row with alternating signs): [latex]\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix} = a_1\begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3\begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix}[/latex]

Cross Product Using Determinants

For [latex]\mathbf{u} = \langle u_1, u_2, u_3 \rangle[/latex] and [latex]\mathbf{v} = \langle v_1, v_2, v_3 \rangle[/latex]:

[latex]\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}[/latex]

Process:

1. First row: [latex]\mathbf{i}[/latex], [latex]\mathbf{j}[/latex], [latex]\mathbf{k}[/latex]
2. Second row: [latex]\mathbf{u}[/latex] components
3. Third row: [latex]\mathbf{v}[/latex] components
4. Expand along first row with alternating signs ([latex]+[/latex], [latex]-[/latex], [latex]+[/latex])

[latex]\mathbf{u} \times \mathbf{v} = \mathbf{i}\begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} - \mathbf{j}\begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix} + \mathbf{k}\begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix}[/latex]

Expanding the determinant gives exactly the same components as the original formula, but in an organized, memorable format.

Note: This is technically a mnemonic device (memory aid), not a true determinant, since determinants contain only numbers, not vectors.

The Main Idea 

The cross product solves key geometric problems in 3D: finding perpendicular vectors, calculating areas of parallelograms and triangles, and more.

Finding Orthogonal Vectors

Since [latex]\mathbf{u} \times \mathbf{v}[/latex] is always perpendicular to both [latex]\mathbf{u}[/latex] and [latex]\mathbf{v}[/latex], it’s the go-to method for finding orthogonal vectors.

Process: Calculate [latex]\mathbf{u} \times \mathbf{v}[/latex], then normalize if you need a unit vector: [latex]\frac{\mathbf{u} \times \mathbf{v}}{||\mathbf{u} \times \mathbf{v}||}[/latex]

Both [latex]\mathbf{u} \times \mathbf{v}[/latex] and [latex]\mathbf{v} \times \mathbf{u}[/latex] are orthogonal, but point in opposite directions.

Calculating Areas

The magnitude of the cross product gives areas directly:

• Parallelogram: If [latex]\mathbf{u}[/latex] and [latex]\mathbf{v}[/latex] form adjacent sides: [latex]\text{Area} = ||\mathbf{u} \times \mathbf{v}||[/latex]
• Triangle: Half the parallelogram area: [latex]\text{Area} = \frac{1}{2}||\mathbf{u} \times \mathbf{v}||[/latex]

Example: To find a triangle’s area from three vertices [latex]P[/latex], [latex]Q[/latex], [latex]R[/latex]:

1. Form vectors [latex]\overrightarrow{PQ}[/latex] and [latex]\overrightarrow{PR}[/latex]
2. Calculate [latex]\overrightarrow{PQ} \times \overrightarrow{PR}[/latex]
3. Area = [latex]\frac{1}{2}||\overrightarrow{PQ} \times \overrightarrow{PR}||[/latex]

The Main Idea 

The triple scalar product combines a dot product with a cross product: [latex]\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})[/latex]. The result is a scalar with important geometric meaning.

Determinant Formula

Instead of computing [latex]\mathbf{v} \times \mathbf{w}[/latex] first, then taking the dot product, use a determinant:

[latex]\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix}u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}[/latex]

This is much faster—just set up the determinant with vector components as rows and evaluate.

Order Matters

Swapping vectors changes the determinant’s sign (like swapping rows):

• [latex]\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = -\mathbf{u} \cdot (\mathbf{w} \times \mathbf{v})[/latex] (switching cross product order)
• [latex]\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) = \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v})[/latex] (cyclic permutations give same value)

Volume of a Parallelepiped

A parallelepiped is a 3D prism where all faces are parallelograms. When vectors [latex]\mathbf{u}[/latex], [latex]\mathbf{v}[/latex], and [latex]\mathbf{w}[/latex] form its edges:

[latex]V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})|[/latex]

Use absolute value because volume is always positive.

Quick Process:

1. Set up determinant with vector components
2. Evaluate the determinant
3. Take absolute value for volume

The Main Idea 

Cross products solve practical problems in physics and engineering, from testing if vectors lie in the same plane to calculating rotational forces.

Testing for Coplanar Vectors

Three vectors are coplanar if they all lie in the same plane. The test: calculate the triple scalar product.

[latex]\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = 0[/latex] if and only if the vectors are coplanar.

If the parallelepiped volume is zero, one dimension must be zero—the vectors lie flat in a plane.

Finding Vectors Perpendicular to Planes

To find a vector perpendicular to a plane, take the cross product of any two non-parallel vectors in that plane.

Given three points [latex]P[/latex], [latex]Q[/latex], [latex]R[/latex] in a plane, form vectors [latex]\overrightarrow{PQ}[/latex] and [latex]\overrightarrow{QR}[/latex], then [latex]\overrightarrow{PQ} \times \overrightarrow{QR}[/latex] is perpendicular to the plane.

Torque: Measuring Rotational Force

Torque ([latex]\tau[/latex]) measures how effectively a force causes rotation. When you turn a wrench or tighten a screw, you’re applying torque:

[latex]\tau = \mathbf{r} \times \mathbf{F}[/latex]

where [latex]\mathbf{r}[/latex] is the position vector from the rotation axis to where force [latex]\mathbf{F}[/latex] is applied.

Key Facts:

• Magnitude: [latex]|\tau| = |\mathbf{r}| \cdot |\mathbf{F}| \sin \theta[/latex]
• Direction: Points along the rotation axis (right-hand rule)
• Maximum torque: When force is perpendicular to [latex]\mathbf{r}[/latex] ([latex]\theta = 90°[/latex])
• Zero torque: When force points toward/away from axis ([latex]\theta = 0°[/latex] or [latex]180°[/latex])

Units: Newton-meters (N⋅m) or foot-pounds (ft⋅lb)

Example: Longer wrench = greater torque. That’s why you can tighten a bolt more easily with a longer wrench—you’re increasing [latex]|\mathbf{r}|[/latex].