The Cross Product and the Right-Hand Rule

The cross product of two 3D vectors $\vec A$ and $\vec B$ produces a third vector that is perpendicular to both. Its magnitude is:

$|\vec A \times \vec B| = |\vec A|\,|\vec B|\sin\theta$

— where $\theta$ is the angle between $\vec A$ and $\vec B$. Geometrically, this is the area of the parallelogram spanned by the two vectors. The direction is given by the famous right-hand rule: point your fingers along $\vec A$, curl them toward $\vec B$, and your thumb points in the direction of $\vec A \times \vec B$.

The Component Formula

For $\vec A = (A_x, A_y, A_z)$ and $\vec B = (B_x, B_y, B_z)$:

$\vec A \times \vec B = (A_y B_z - A_z B_y,\; A_z B_x - A_x B_z,\; A_x B_y - A_y B_x)$

For 2D vectors lying in the $xy$-plane (so $A_z = B_z = 0$), the cross product is purely in the $z$-direction:

$(\vec A \times \vec B)_z = A_x B_y - A_y B_x$

This single number is positive if the rotation from $\vec A$ to $\vec B$ is counterclockwise, negative if clockwise.

Step-by-Step Solution

Given: $\vec A = (3, 1, 0)$ and $\vec B = (1, 2, 0)$ (both in the $xy$-plane).

Find: $\vec A \times \vec B$, its magnitude, and direction.

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Step 1 — Compute the cross product component-by-component.

Using the formula above:

$\vec A \times \vec B = (A_y B_z - A_z B_y,\; A_z B_x - A_x B_z,\; A_x B_y - A_y B_x)$

$= ((1)(0) - (0)(2),\; (0)(1) - (3)(0),\; (3)(2) - (1)(1))$

$= (0,\; 0,\; 5)$

The result is purely in the $+z$ direction (out of the screen).

Step 2 — Compute the magnitude.

$|\vec A \times \vec B| = \sqrt{0^{2} + 0^{2} + 5^{2}} = 5$

Step 3 — Verify with the geometric formula.

First find the angle between $\vec A$ and $\vec B$:

$\vec A \cdot \vec B = (3)(1) + (1)(2) + 0 = 5$

$|\vec A| = \sqrt{9 + 1} = \sqrt{10},\quad |\vec B| = \sqrt{1 + 4} = \sqrt{5}$

$\cos\theta = \dfrac{5}{\sqrt{10}\sqrt{5}} = \dfrac{5}{\sqrt{50}} = \dfrac{5}{5\sqrt{2}} = \dfrac{1}{\sqrt{2}}$

$\theta = 45°$

Now verify:

$|\vec A|\,|\vec B|\sin\theta = \sqrt{10}\sqrt{5} \cdot \dfrac{1}{\sqrt{2}} = \sqrt{50}/\sqrt{2} = \sqrt{25} = 5 \;\;\checkmark$

Both methods give the same magnitude.

Step 4 — Right-hand rule for direction.

Point your right hand's fingers along $\vec A = (3, 1, 0)$. Curl them toward $\vec B = (1, 2, 0)$ — the curl is counterclockwise as seen from above. Your thumb points up out of the page, in the $+z$ direction. ✓

The cross product is anti-commutative: $\vec B \times \vec A = -\vec A \times \vec B = (0, 0, -5)$, pointing into the page.

Step 5 — Geometric interpretation.

The magnitude 5 is also the area of the parallelogram spanned by $\vec A$ and $\vec B$. (Think of it as base $|\vec A|$ times "height" $|\vec B|\sin\theta$ — the perpendicular distance from $\vec B$'s tip to the line through $\vec A$.)

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Answer:

$\boxed{\;\vec A \times \vec B = (0, 0, 5),\quad |\vec A \times \vec B| = 5\;}$

The cross product is perpendicular to both $\vec A$ and $\vec B$, points in the $+z$ direction (out of the screen by the right-hand rule), and has magnitude equal to the area of the parallelogram they span.

Try It

• Adjust the components of A and B with the sliders.
• The visualization shows the parallelogram spanned by the two vectors with its area equal to $|\vec A \times \vec B|$.
• The HUD shows whether the cross product points OUT (+z) or INTO (−z) the screen, and computes the magnitude.
• When you make $\vec A$ and $\vec B$ parallel, the cross product collapses to zero.