Topic 4 / 17intermediate

Cross Product

Compute the cross product via determinant or component formula, use the right-hand rule for direction, and apply it to areas, normal vectors, and parallelism tests.

The cross product takes two vectors in $R^{3}$ and returns a third vector orthogonal to both, with magnitude equal to the area of the parallelogram they span. Unlike the dot product, the result is a **vector** — and only $R^{3}$ admits this construction in this clean form.

Component formula and determinant trick

For

u = ⟨ u_{1}, u_{2}, u_{3} ⟩, v = ⟨ v_{1}, v_{2}, v_{3} ⟩

u \times 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} ⟩ .

The middle term carries a minus sign — easy to forget. The clean way to compute is the symbolic determinant:

u \times v = i u_{1} v_{1} j u_{2} v_{2} k u_{3} v_{3}

Key takeaways

The middle component has a leading minus sign.
Cross product output is a vector — orthogonal to both inputs.

Right-hand rule for direction

Point fingers along

u

, curl them toward

v

; thumb points along

u \times v

. Order matters:

v \times u = - (u \times v)

Magnitude and area

∣ u \times v ∣ = ∣ u ∣∣ v ∣ sin θ

. This is the area of the parallelogram with sides

u, v

. The triangle with these sides has area

\frac{1}{2} ∣ u \times v ∣

Parallelism test

u \times v = 0

iff one vector is a scalar multiple of the other (or one is zero). This is the cleanest test for parallel vectors in

R^{3}

Algebraic properties

v \times u = - (u \times v)

(anticommutative). Distributive:

u \times (v + w) = u \times v + u \times w

u \times u = 0

. **Cross product is NOT associative.**

Strategy tips

If you swap operand order, flip the sign — don't recompute.
$u \times u = 0$ is a fast sanity check.

Worked examples

Example 1

Compute

⟨ 1, 2, 3 ⟩ \times ⟨ 4, 5, 6 ⟩

1
Set up the determinant.
$i 14 j 25 k 36$
2
Expand along the top row.
$i (2 \cdot 6 - 3 \cdot 5) - j (1 \cdot 6 - 3 \cdot 4) + k (1 \cdot 5 - 2 \cdot 4)$
3
Simplify.
$= i (12 - 15) - j (6 - 12) + k (5 - 8) = ⟨ - 3, 6, - 3 ⟩$

Answer

⟨ - 3, 6, - 3 ⟩

Example 2

Find a vector perpendicular to the plane through

A (1, 0, 0), B (2, 3, 0), C (- 1, 1, 4)

, and the area of triangle

A B C

1
Form two edge vectors.
$A B = ⟨ 1, 3, 0 ⟩, A C = ⟨ - 2, 1, 4 ⟩$
2
Take the cross product.
$A B \times A C = ⟨ 3 \cdot 4 - 0 \cdot 1, 0 \cdot (- 2) - 1 \cdot 4, 1 \cdot 1 - 3 \cdot (- 2)⟩ = ⟨ 12, - 4, 7 ⟩$
3
Magnitude gives parallelogram area; triangle is half.
$∣ ⟨ 12, - 4, 7 ⟩ ∣ = 144 + 16 + 49 = 209$
4
Triangle area.
$A_{△} = \frac{1}{2} 209$

Answer

Normal:

⟨ 12, - 4, 7 ⟩

; triangle area:

\frac{1}{2} 209

Example 3

Find a unit vector orthogonal to

a = j + 2 k

and

b = i - 2 j + 3 k

1
Cross product.
$a \times b = i 01 j 1 - 2 k 23 = ⟨ 1 \cdot 3 - 2 \cdot (- 2), 2 \cdot 1 - 0 \cdot 3, 0 \cdot (- 2) - 1 \cdot 1 ⟩ = ⟨ 7, 2, - 1 ⟩$
2
Magnitude.
$∣ ⟨ 7, 2, - 1 ⟩ ∣ = 49 + 4 + 1 = 54$
3
Two unit vectors (one for each direction).
$\pm \frac{1}{54} ⟨ 7, 2, - 1 ⟩$

Answer

\pm \frac{1}{54} ⟨ 7, 2, - 1 ⟩

Interactive visualizations

Loading visualization…

u_x2.00

u_y0.00

u_z0.00

v_x0.00

v_y2.00

v_z0.00

u \times v = ⟨ 0.00, 0.00, 4.00 ⟩

∣ u \times v ∣ = 4.00 = area of parallelogram

Slide the components of

u

and

v

. The red vector is

u \times v

; the blue parallelogram's area equals its magnitude.

Formulas in this topic

Cross product (components)

u \times 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} ⟩

Compute via component formula or determinant.

Cross product (determinant)

u \times v = i u_{1} v_{1} j u_{2} v_{2} k u_{3} v_{3}

Symbolic determinant form.

Cross product magnitude

∣ u \times v ∣ = ∣ u ∣∣ v ∣ sin θ

Area of parallelogram with sides u, v.

Triangle area

A = \frac{1}{2} ∣ u \times v ∣

Half of the parallelogram area.

Component formula and determinant trick

Right-hand rule for direction

Magnitude and area

Parallelism test

Algebraic properties

Worked examples

Interactive visualizations

Formulas in this topic

Related topics