Multivariable-Calculus · Unit 3 · Video 1 · Interactive Practice

Swap the Order, Flip the Sign: The Cross Product in Review

IKey Formulas

Formula	Name	What you need
$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$	The symbolic determinant	$\vec{A}$ in row 2, $\vec{B}$ in row 3
$\vec{A} \times \vec{B} = (a_2b_3 - a_3b_2)\,\hat{i} \;-\; (a_1b_3 - a_3b_1)\,\hat{j} \;+\; (a_1b_2 - a_2b_1)\,\hat{k}$	Expansion along row 1	Three $2 \times 2$ minors, the middle one negated
$\|\vec{A} \times \vec{B}\| = \text{area of the parallelogram}$ , and $\vec{A} \times \vec{B} \perp \vec{A},\ \vec{B}$	The geometric definition	The right-hand rule to choose between the two normals
$\vec{A} \times \vec{B} = -\,\vec{B} \times \vec{A}, \qquad \vec{A} \times \vec{A} = \vec{0}$	Anti-commutativity	Swap the order, put in a minus sign

Key Insight: Swapping $\vec{A}$ and $\vec{B}$ exchanges rows 2 and 3 of the determinant, and exchanging the two rows of a $2 \times 2$ minor reverses its subtraction. All three components change sign together, which is why the whole vector does.

IIVisualization 1 — Exchanging the Two Rows

Swapping the vectors exchanges rows 2 and 3, and every $2 \times 2$ minor reverses its subtraction.

Vectors

IIIVisualization 2 — Same Parallelogram, Opposite Thumb

The swap leaves the parallelogram untouched; only the side the thumb points to changes.

B: b₁ −1 B: b₂ 2

IVVisualization 3 — When B Lands On A

As $\vec{B}$ slides onto $\vec{A}$ the parallelogram flattens, and the cross product flattens with it.

B → A 0.00

B has landed on A — both rows are now the components of A

\vec{A} \times \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ a_1 & a_2 & a_3 \end{vmatrix} = (a_2a_3 - a_3a_2)\,\hat{i} \;-\; (a_1a_3 - a_3a_1)\,\hat{j} \;+\; (a_1a_2 - a_2a_1)\,\hat{k} = \vec{0}

💡 That perpendicular direction is what the next video needs: a normal vector is exactly what turns three points into the equation of a plane.

VQuiz Questions

Problem 1 · The Cost of a Swap

Given: $\vec{A} \times \vec{B} = \langle 5, -5, 5 \rangle$ — find $\vec{B} \times \vec{A}$ .

✅ Correct! Same length, same plane, opposite side:

\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B})

❌ Not quite. Ordinary multiplication does not care about order, but the cross product does — swapping costs a minus sign.

❌ Close, but… Negating a vector reverses the sign of each component; it does not make every component negative. The middle entry was already

-5

❌ Not quite. That is

\vec{A} \times \vec{A}

. Here

\vec{A}

and

\vec{B}

span a genuine parallelogram, so neither order gives the zero vector.

Show solution

Each component of a cross product is a $2 \times 2$ minor of the symbolic determinant. Swapping $\vec{A}$ and $\vec{B}$ exchanges rows 2 and 3, so each minor has its two rows exchanged — and exchanging the rows of a $2 \times 2$ determinant reverses its sign. Every component flips:

\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B}) = -\langle 5, -5, 5 \rangle = \langle -5, 5, -5 \rangle

The picture says the same thing. $\vec{B}$ and $\vec{A}$ span the same parallelogram — same area, same plane — so the length and the line of the answer cannot change. Only the right hand changes: sweeping from $\vec{B}$ to $\vec{A}$ turns the hand over, and the thumb points the other way.

Problem 2 · The Middle Minus

Given: $\vec{A} = \langle 1, -2, 3 \rangle$ and $\vec{B} = \langle 2, 1, -1 \rangle$ — find $\vec{A} \times \vec{B}$ .

✅ Correct! The standing check works:

\vec{A} \cdot (\vec{A} \times \vec{B}) = -1 - 14 + 15 = 0

and

\vec{B} \cdot (\vec{A} \times \vec{B}) = -2 + 7 - 5 = 0

❌ Check the $\hat{j}$ term. Its minor evaluates to

(1)(-1) - (3)(2) = -7

, and the expansion puts a minus in front of it:

-(-7) = +7

❌ Close, but… That is

\vec{B} \times \vec{A}

— every component has the opposite sign. You were asked for

\vec{A} \times \vec{B}

, so

\vec{A}

belongs in row 2.

❌ Not quite. The cross product is not computed componentwise; each entry mixes components of both vectors through a

2 \times 2

minor.

Show solution

Put $\vec{A}$ in row 2 and $\vec{B}$ in row 3, then expand along the top row:

\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix}

$\hat{i}$ : $\ (-2)(-1) - (3)(1) = 2 - 3 = -1$

$\hat{j}$ : $\ -\big[(1)(-1) - (3)(2)\big] = -[-1 - 6] = 7$

$\hat{k}$ : $\ (1)(1) - (-2)(2) = 1 + 4 = 5$

\vec{A} \times \vec{B} = \langle -1, 7, 5 \rangle

The middle minus is the easiest sign in the subject to drop. It belongs to the expansion, not to the numbers, and it is there whatever the entries are.

Problem 3 · The Length Does Not Swap

Given: $\vec{A} = \langle 2, 2, 1 \rangle$ and $\vec{B} = \langle 1, -2, 2 \rangle$ — find $|\vec{B} \times \vec{A}|$ .

✅ Correct! Here

\vec{A} \cdot \vec{B} = 2 - 4 + 2 = 0

, so the parallelogram is a rectangle and its area is just

|\vec{A}|\,|\vec{B}| = 3 \cdot 3 = 9

❌ Not quite. A length is never negative — that is the whole reason the cross product needs no

\pm

repair. Swapping the order turns the arrow around; it cannot shorten it.

❌ Close, but…

36 + 9 + 36 = 81

is the sum of the squares. The length is its square root.

❌ Not quite. That is the area of the triangle with sides

\vec{A}

and

\vec{B}

. The cross product measures the full parallelogram, twice as large.

Show solution

Expand with $\vec{A}$ in row 2:

$\hat{i}$ : $\ (2)(2) - (1)(-2) = 4 + 2 = 6$ ; $\hat{j}$ : $\ -\big[(2)(2) - (1)(1)\big] = -3$ ; $\hat{k}$ : $\ (2)(-2) - (2)(1) = -6$

\vec{A} \times \vec{B} = \langle 6, -3, -6 \rangle \quad\Longrightarrow\quad \vec{B} \times \vec{A} = \langle -6, 3, 6 \rangle

|\vec{B} \times \vec{A}| = \sqrt{(-6)^2 + 3^2 + 6^2} = \sqrt{81} = 9

You never needed the swap: the two vectors span the same parallelogram in either order, so $|\vec{B} \times \vec{A}| = |\vec{A} \times \vec{B}|$ always. Negating every component leaves the length alone.

Problem 4 · When the Answer Is the Zero Vector

Given: $\vec{A}$ and $\vec{B}$ are both nonzero, and $\vec{A} \times \vec{B} = \vec{0}$ — what must be true?

✅ Correct! Both arrows lie along one line, the parallelogram is flat, and the only vector of length

0

\vec{0}

❌ That is the opposite extreme. Perpendicular vectors give the largest cross product,

|\vec{A}|\,|\vec{B}|

. It is the dot product that vanishes when they are perpendicular.

❌ Close, but…

\vec{A} \times \vec{A} = \vec{0}

is true, but it is only one case.

\langle 1, 2, 3 \rangle \times \langle 2, 4, 6 \rangle = \vec{0}

too, with

\vec{A} \neq \vec{B}

❌ Not quite. Equal lengths say nothing about direction:

\langle 3, 0, 0 \rangle

and

\langle 0, 3, 0 \rangle

have the same length and their cross product is

\langle 0, 0, 9 \rangle

Show solution

Read the geometric definition backwards. The length of the cross product is the area of the parallelogram,

|\vec{A} \times \vec{B}| = |\vec{A}|\,|\vec{B}|\sin\theta = 0

With $|\vec{A}| \neq 0$ and $|\vec{B}| \neq 0$ this forces $\sin\theta = 0$ , so $\theta = 0$ or $\theta = 180\degree$ : the two arrows lie along a single line and $\vec{B}$ is a scalar multiple of $\vec{A}$ . The "parallelogram" they span is completely flat, with zero area.

$\vec{A} \times \vec{A} = \vec{0}$ is the case $c = 1$ . In the determinant it shows up as two identical rows, and every $2 \times 2$ minor of the form $a_ia_j - a_ja_i$ is zero.

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