CLASSICAL-MECHANICS

From î×ĵ=k̂ to Volumes: The Cross Product in Components

IKey Formulas

Formula	Name	Description
$\hat{i}\times\hat{j}=\hat{k},\ \hat{j}\times\hat{k}=\hat{i},\ \hat{k}\times\hat{i}=\hat{j}$	Cyclic unit products	Right-handed cycle; reverse flips sign; self-product is $0$
$\mathbf{A}\times\mathbf{B}=(A_yB_z-A_zB_y)\hat{i}+(A_zB_x-A_xB_z)\hat{j}+(A_xB_y-A_yB_x)\hat{k}$	Component formula	Each component skips its own axis
$\lvert\mathbf{A}\times\mathbf{B}\rvert=\lvert\mathbf{A}\rvert\lvert\mathbf{B}\rvert\sin\theta$	Magnitude	Area of the spanned parallelogram
$V=\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C})$	Scalar triple product	Signed volume of the parallelepiped

IIInteractive Visualizations

Visualization 1 — Area from the angle between vectors

Why does $\lvert\mathbf{A}\times\mathbf{B}\rvert=\lvert\mathbf{A}\rvert\lvert\mathbf{B}\rvert\sin\theta$ peak at $90^\circ$ and vanish when the two vectors line up?

|A| 3.0 Bx 1.0 By 2.0

💡 Parallel vectors span no area — this is why $\hat{i}\times\hat{i}=\mathbf{0}$ .

Visualization 2 — The cross product is perpendicular to both

The component formula returns a vector at a right angle to both $\mathbf{A}$ and $\mathbf{B}$ — a normal to the plane they span.

Ax 2 Ay 3 Az 0 Bx -1 By 2 Bz 0 view 35°

💡 Normalizing $\mathbf{A}\times\mathbf{B}$ gives a unit normal $\hat{n}=\pm\dfrac{\mathbf{A}\times\mathbf{B}}{\lvert\mathbf{A}\times\mathbf{B}\rvert}$ — the standard way to orient a surface.

Visualization 3 — Volume as base area times height

Shearing the slant edge $\mathbf{A}$ sideways leaves the volume unchanged — only its height along the base normal counts.

Ax 1.0 Ay 0.5 Az (height) 2.5 view 40°

💡 $\mathbf{B}\times\mathbf{C}$ builds the base area; dotting with $\mathbf{A}$ extracts the height.

IIIQuiz Questions

Question 1

Let $\mathbf{A}=2\hat{i}-3\hat{j}+7\hat{k}$ and $\mathbf{B}=5\hat{i}+\hat{j}+2\hat{k}$ .

What is the $\hat{i}$ component of $\mathbf{A}\times\mathbf{B}$ , given by $A_yB_z-A_zB_y$ ?

✅ Correct! $(-3)(2)-(7)(1)=-6-7=-13$ .

❌ Not quite. Use $A_yB_z-A_zB_y$ and watch the sign of $A_y=-3$ .

Show solution

Solution:

The $\hat{i}$ component is $A_yB_z-A_zB_y$ with $A_y=-3$ , $B_z=2$ , $A_z=7$ , $B_y=1$ :

$A_yB_z - A_zB_y = (-3)(2) - (7)(1) = -6 - 7 = -13$

So the $\hat{i}$ component is $-13$ . (The full result is $\mathbf{A}\times\mathbf{B}=-13\hat{i}+31\hat{j}+17\hat{k}$ .)

Question 2

Using the right-handed cycle $\hat{i}\times\hat{j}=\hat{k}$ , $\hat{j}\times\hat{k}=\hat{i}$ , $\hat{k}\times\hat{i}=\hat{j}$ , what is $\hat{j}\times\hat{i}$ ?

✅ Correct! Reversing the order flips the sign: $\hat{j}\times\hat{i}=-\hat{k}$ .

❌ Not quite. The cross product is anti-commutative — reversing order flips the sign.

Show solution

Solution:

The cross product is anti-commutative: $\mathbf{B}\times\mathbf{A}=-(\mathbf{A}\times\mathbf{B})$ . Since $\hat{i}\times\hat{j}=\hat{k}$ , reversing the order flips the sign:

$\hat{j}\times\hat{i} = -(\hat{i}\times\hat{j}) = -\hat{k}$

It is not zero — the two unit vectors are perpendicular, not parallel.

Question 3

A parallelepiped has edge vectors $\mathbf{A}=(1,0,4)$ , $\mathbf{B}=(2,0,0)$ , and $\mathbf{C}=(0,3,0)$ . What is its volume, $V=\lvert\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C})\rvert$ ?

✅ Correct! Base area $6$ times height $4$ gives volume $24$ .

❌ Not quite. Compute $\mathbf{B}\times\mathbf{C}=(0,0,6)$ , then dot with $\mathbf{A}$ .

Show solution

Solution:

First compute the base $\mathbf{B}\times\mathbf{C}$ with $\mathbf{B}=(2,0,0)$ , $\mathbf{C}=(0,3,0)$ :

$\mathbf{B}\times\mathbf{C} = (0\cdot0-0\cdot3,\ 0\cdot0-2\cdot0,\ 2\cdot3-0\cdot0) = (0,0,6)$

Then take the dot product with $\mathbf{A}=(1,0,4)$ :

$\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}) = (1)(0)+(0)(0)+(4)(6) = 24$

The volume is $\lvert 24\rvert = 24$ . (Base area $=6$ , height $=4$ .)

Question 4

True or False: To find a unit vector perpendicular to a plane containing two non-parallel vectors $\mathbf{A}$ and $\mathbf{B}$ , you can compute $\pm\dfrac{\mathbf{A}\times\mathbf{B}}{\lvert\mathbf{A}\times\mathbf{B}\rvert}$ , and the $\pm$ reflects that two opposite directions are both perpendicular.

✅ Correct! The normalized cross product gives a unit normal, with two opposite choices.

❌ Not quite. The cross product is perpendicular to both vectors; normalizing gives a unit normal.

Show solution

Solution:

True. The cross product $\mathbf{A}\times\mathbf{B}$ is perpendicular to both $\mathbf{A}$ and $\mathbf{B}$ , hence normal to the plane they span. Dividing by its magnitude makes it a unit vector:

$\hat{n} = \pm\frac{\mathbf{A}\times\mathbf{B}}{\lvert\mathbf{A}\times\mathbf{B}\rvert}$

The $\pm$ appears because a plane has two opposite unit normals — both are valid perpendicular directions. (This requires $\mathbf{A}$ and $\mathbf{B}$ to be non-parallel, so that $\mathbf{A}\times\mathbf{B}\neq\mathbf{0}$ .)

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