Multivariable-Calculus · Unit 2 · Video 4 · Interactive Practice

Area With a Direction: The Cross Product and the Right-Hand Rule

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	The components of both vectors in rows 2 and 3
$\vec{A} \times \vec{B} = \langle\, a_2b_3 - a_3b_2,\ \ -(a_1b_3 - a_3b_1),\ \ a_1b_2 - a_2b_1 \,\rangle$	Expansion along row 1	Three $2 \times 2$ minors, the middle one negated
$\|\vec{A} \times \vec{B}\| = \|\vec{A}\|\,\|\vec{B}\|\sin\theta = \text{area of the parallelogram}$	The length payoff	Both lengths and the angle between them
$\vec{A} \times \vec{B} \perp \vec{A}$ and $\vec{A} \times \vec{B} \perp \vec{B}$	The direction payoff	The right-hand rule to pick which of the two normals

Key Insight: The minus sign on the $\hat{j}$ term belongs to the formula, not to the numbers. And notice what is missing: no $\pm$ out front. A length is never negative, so unlike the $2 \times 2$ area there is nothing to repair afterward.

IIVisualization 1 — Expanding the Symbolic Determinant

Row one holds $\hat{i}, \hat{j}, \hat{k}$ ; expanding along it turns a symbolic array into an honest vector.

Vectors Step 1

1 · the memory device

\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}

2 · the $\hat{i}$ term (sign $+$ )

+\,\hat{i}\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\,\hat{i}

3 · the $\hat{j}$ term (sign $-$ )

-\,\hat{j}\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} = -(a_1b_3 - a_3b_1)\,\hat{j}

4 · the $\hat{k}$ term (sign $+$ )

+\,\hat{k}\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} = (a_1b_2 - a_2b_1)\,\hat{k}

5 · an honest vector

\vec{A} \times \vec{B} = \langle\, a_2b_3 - a_3b_2,\ -(a_1b_3 - a_3b_1),\ a_1b_2 - a_2b_1 \,\rangle

IIIVisualization 2 — The Length Is the Area

The parallelogram's area is one number; the cross product carries it, and points which way the plane faces.

A: a₁ 4 A: a₂ 1 B: b₁ 2 B: b₂ 3

💡 This is the parallelogram. The triangle with the same two sides has exactly half this area, as it did in the plane.

IVVisualization 3 — The Right-Hand Rule on the Unit Vectors

Geometry and the determinant must agree on the simplest case: one unit vector crossed with another.

First vector A Second vector B

💡 Only in three dimensions do the directions perpendicular to a plane form a single line — just two opposite arrows, and the right hand names one of them; that is why the cross product lives nowhere else.

VQuiz Questions

Problem 1 · Expanding the Determinant

Given: $\vec{A} = \langle 2, 0, 1 \rangle$ and $\vec{B} = \langle 1, 3, 0 \rangle$ — find $\vec{A} \times \vec{B}$ .

✅ Correct! Check it the way you always can:

\vec{A}\cdot(\vec{A}\times\vec{B}) = -6 + 0 + 6 = 0

and

\vec{B}\cdot(\vec{A}\times\vec{B}) = -3 + 3 + 0 = 0

❌ The middle term lost its minus. The

\hat{j}

minor is

a_1b_3 - a_3b_1 = (2)(0) - (1)(1) = -1

, and the formula negates it:

-(-1) = +1

❌ That is $\vec{B} \times \vec{A}$ . Every component has flipped sign, which is what swapping the last two rows of a determinant does. Put

\vec{A}

in row 2.

❌ Those are componentwise products.

\langle a_1b_1, a_2b_2, a_3b_3\rangle

is not the cross product — each component mixes the other two coordinates.

❌ Not quite. Expand along the first row:

\hat{i}

minor, then the negated

\hat{j}

minor, then the

\hat{k}

minor.

Show solution

Write the symbolic determinant and expand along row 1:

\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 3 & 0 \end{vmatrix}

Term by term:

\hat{i}:\ (0)(0) - (1)(3) = -3 \qquad \hat{j}:\ -\big[(2)(0) - (1)(1)\big] = +1 \qquad \hat{k}:\ (2)(3) - (0)(1) = 6

So $\vec{A} \times \vec{B} = \langle -3, 1, 6 \rangle$ . The free check costs two dot products, and both must vanish: $\vec{A}\cdot(\vec{A}\times\vec{B}) = -6 + 0 + 6 = 0$ , $\vec{B}\cdot(\vec{A}\times\vec{B}) = -3 + 3 + 0 = 0$ .

Problem 2 · The Middle Term

Given: $\vec{A} = \langle 4, -1, 2 \rangle$ and $\vec{B} = \langle 3, 5, -2 \rangle$ — find the $\hat{j}$ -component of $\vec{A} \times \vec{B}$ .

✅ Correct! The minor is

-14

, and the formula's own minus sign turns it into

+14

— two negatives that are easy to collapse into one.

❌ That is the minor, not the component.

(4)(-2) - (2)(3) = -14

is only half the story: the

\hat{j}

term carries a minus sign, so the component is

-(-14) = 14

❌ Wrong minor.

(-1)(-2) - (2)(5) = -8

deletes column 1, so that is the

\hat{i}

-component. For

\hat{j}

, delete column 2: the rows left are

4, 2

and

3, -2

❌ That is the $\hat{k}$ -component.

(4)(5) - (-1)(3) = 23

comes from deleting column 3. The

\hat{j}

minor uses the first and third columns.

❌ Not quite. Delete row 1 and column 2, take the

2 \times 2

determinant, then negate it.

Show solution

Deleting row 1 and column 2 leaves the first and third columns:

\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} = \begin{vmatrix} 4 & 2 \\ 3 & -2 \end{vmatrix} = (4)(-2) - (2)(3) = -8 - 6 = -14

The $\hat{j}$ term of the expansion carries a minus sign, so the component is

-(a_1b_3 - a_3b_1) = -(-14) = 14

For the record, the whole product is $\vec{A} \times \vec{B} = \langle -8, 14, 23 \rangle$ . The classic mistake is reporting the minor itself and getting $-14$ ; the sign is part of the formula, not part of the arithmetic inside the bars.

Problem 3 · Area of a Triangle in Space

Given: the triangle with vertices $P = (1, 1, 1)$ , $Q = (2, 3, 1)$ and $R = (1, 3, 2)$ — find its area.

✅ Correct!

\vec{PQ} \times \vec{PR} = \langle 2, -1, 2 \rangle

has length

3

— the parallelogram — and the triangle is half of it.

❌ That is the parallelogram.

|\vec{PQ} \times \vec{PR}| = 3

is the area spanned by the two edges; the triangle with those same two sides is half of it.

❌ The square root was skipped.

2^2 + (-1)^2 + 2^2 = 9

is the length squared. Take the root first:

|\langle 2, -1, 2\rangle| = 3

, then halve.

❌ The $\sin\theta$ is missing.

|\vec{PQ}|\,|\vec{PR}| = \sqrt{5}\cdot\sqrt{5} = 5

is only the base times the slant side; the area needs the height, which is where

\sin\theta

enters.

❌ Not quite. Build two edge vectors from the same vertex, cross them, take the length, then halve.

Show solution

Both edges must leave the same corner. From $P$ :

\vec{PQ} = \langle 1, 2, 0 \rangle, \qquad \vec{PR} = \langle 0, 2, 1 \rangle

Expand the symbolic determinant along the first row:

\hat{i}:\ (2)(1) - (0)(2) = 2 \qquad \hat{j}:\ -\big[(1)(1) - (0)(0)\big] = -1 \qquad \hat{k}:\ (1)(2) - (2)(0) = 2

\vec{PQ} \times \vec{PR} = \langle 2, -1, 2 \rangle, \qquad |\vec{PQ} \times \vec{PR}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

That $3$ is the area of the parallelogram on $\vec{PQ}$ and $\vec{PR}$ . The triangle is half of it:

\text{Area} = \tfrac{1}{2}\,|\vec{PQ} \times \vec{PR}| = \tfrac{3}{2}

Problem 4 · Order, Sign and Area

Given: $\vec{A} = \langle 2, 0, 0 \rangle$ and $\vec{B} = \langle 0, 5, 0 \rangle$ — find $\vec{B} \times \vec{A}$ and the area of the parallelogram the two vectors span.

✅ Correct! Order flips the vector, never the area: the direction reverses to

-\hat{k}

while the length stays

10

❌ That is $\vec{A} \times \vec{B}$ . Right hand along

\vec{B}

(the

y

-axis), fingers curling toward

\vec{A}

(the

x

-axis): the thumb points down, so the answer is

\langle 0, 0, -10 \rangle

❌ An area is never negative. The

\pm

that haunted the

2 \times 2

determinant is gone here: the area is

|\langle 0, 0, -10\rangle| = 10

, and the sign has moved into the direction of the vector.

❌ Zero is the parallel case. These two are perpendicular, which makes

\sin\theta = 1

— the largest the cross product can be, not the smallest.

❌ Not quite. Expand with

\vec{B}

in row 2, then take the length of what comes out.

Show solution

Put $\vec{B}$ in row 2 and $\vec{A}$ in row 3:

\vec{B} \times \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 5 & 0 \\ 2 & 0 & 0 \end{vmatrix} = \hat{i}\,(0 - 0) - \hat{j}\,(0 - 0) + \hat{k}\,(0 - 10) = \langle 0, 0, -10 \rangle

The other order gives $\vec{A} \times \vec{B} = \langle 0, 0, 10 \rangle$ : swapping the two rows of a determinant flips its sign, so $\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B})$ .

The area is the length of either one:

|\vec{B} \times \vec{A}| = \sqrt{0 + 0 + 100} = 10 = |\vec{A}|\,|\vec{B}|\sin 90\degree = (2)(5)(1)

Both orders report the same area $10$ — the $2 \times 5$ rectangle they span — and disagree only about which way it faces.

Solved: 0 / 4