Multivariable-Calculus · Unit 2 · Video 5 · Interactive Practice

The Triple Product: How Base Times Height Turns Into a Determinant

IKey Formulas

Formula	Name	What you need
$\text{volume} = \text{area(base)} \times \text{height}$	Volume of a parallelepiped	A base and a height
$\text{area(base)} = \|\mathbf{B} \times \mathbf{C}\|$	Base area from a cross product	$\mathbf{B}$ and $\mathbf{C}$
$\text{height} = \mathbf{A} \cdot \hat{\mathbf{n}}, \qquad \hat{\mathbf{n}} = \dfrac{\mathbf{B} \times \mathbf{C}}{\|\mathbf{B} \times \mathbf{C}\|}$	Signed height along the normal	$\mathbf{A}$ and the base
$\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \det(\mathbf{A}, \mathbf{B}, \mathbf{C}), \qquad \text{volume} = \big\|\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})\big\|$	The triple product	All three vectors

Key Insight: $|\mathbf{B} \times \mathbf{C}|$ appears twice — once as the base area, once in the denominator of $\hat{\mathbf{n}}$ — and the two cancel. What survives is $\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})$ .

IIVisualization 1 — Base Area Times Height

The base is fixed by $\mathbf{B}$ and $\mathbf{C}$ ; only how far $\mathbf{A}$ climbs off that plane changes the volume.

a₂ 1 a₃ 3

IIIVisualization 2 — Cross First, Then Dot

Only two vectors can be crossed, and only a vector can be dotted — so the order is forced.

IVVisualization 3 — One Row, Three Minors, One Minus Sign

The first-row expansion of $\det(\mathbf{A}, \mathbf{B}, \mathbf{C})$ and the components of $\mathbf{B} \times \mathbf{C}$ are the same three minors.

Step 1

Step 1 — The claim

\det(\mathbf{A}, \mathbf{B}, \mathbf{C}) \overset{?}{=} \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})

\mathbf{A} = \langle 2,\, 1,\, 3 \rangle, \qquad \mathbf{B} = \langle 1,\, 0,\, 2 \rangle, \qquad \mathbf{C} = \langle 0,\, 3,\, 1 \rangle

Step 2 — Term 1: delete row 1 and column 1

a_1\begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} = 2\begin{vmatrix} 0 & 2 \\ 3 & 1 \end{vmatrix} = 2\,(0 \cdot 1 - 2 \cdot 3) = 2(-6) = -12

Step 3 — Term 2: delete row 1 and column 2, then subtract

-\,a_2\begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} = -1\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = -1\,(1 \cdot 1 - 2 \cdot 0) = -1

Step 4 — Term 3: delete row 1 and column 3, and add up

a_3\begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} = 3\begin{vmatrix} 1 & 0 \\ 0 & 3 \end{vmatrix} = 3\,(1 \cdot 3 - 0 \cdot 0) = 9

\det(\mathbf{A}, \mathbf{B}, \mathbf{C}) = -12 - 1 + 9 = -4

Step 5 — The same three minors build $\mathbf{B} \times \mathbf{C}$

\mathbf{B} \times \mathbf{C} = \left\langle\, \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix},\;\; -\begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix},\;\; \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} \,\right\rangle = \langle -6,\, -1,\, 3 \rangle

\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 2(-6) + 1(-1) + 3(3) = -12 - 1 + 9 = -4

The minus sign written in front of

a_2

in the expansion is the same minus sign that lives inside the

\hat{\jmath}

component, so the two sides agree term by term: both equal

-4

, and the volume is

|-4| = 4

💡 The geometric route never escapes determinants: the base needs $\mathbf{B} \times \mathbf{C}$ , the unit normal needs it again, and every cross product is three $2 \times 2$ determinants in disguise.

VQuiz Questions

Problem 1 · Volume From a Triple Product

Given: $\mathbf{A} = \langle 1, 1, 3 \rangle$ , $\mathbf{B} = \langle 3, 1, 0 \rangle$ , $\mathbf{C} = \langle 1, 3, 0 \rangle$ — find the volume of the parallelepiped they span.

✅ Correct!

\mathbf{B} \times \mathbf{C} = \langle 0, 0, 8 \rangle

, so

\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 24

— base area

8

times height

3

❌ That is the height alone.

\mathbf{A}

climbs

3

above the base plane, but the base still has an area of its own.

❌ That is the base area alone.

|\mathbf{B} \times \mathbf{C}| = 8

measures the floor; multiply it by the height.

❌ Not quite. Volume is base area times height, not base area plus height.

❌ Not quite. Compute

\mathbf{B} \times \mathbf{C}

first, then dot it with

\mathbf{A}

Show solution

Step 1: the base. $\mathbf{B}$ and $\mathbf{C}$ both lie in the $xy$ -plane, so

\mathbf{B} \times \mathbf{C} = \langle 1 \cdot 0 - 0 \cdot 3,\; 0 \cdot 1 - 3 \cdot 0,\; 3 \cdot 3 - 1 \cdot 1 \rangle = \langle 0,\, 0,\, 8 \rangle

\text{area(base)} = |\mathbf{B} \times \mathbf{C}| = 8

Step 2: the height. $\hat{\mathbf{n}} = \tfrac{1}{8}\langle 0, 0, 8 \rangle = \langle 0, 0, 1 \rangle$ , so

\text{height} = \mathbf{A} \cdot \hat{\mathbf{n}} = \langle 1, 1, 3 \rangle \cdot \langle 0, 0, 1 \rangle = 3

Step 3: multiply — and watch the length cancel.

\text{volume} = |\mathbf{B} \times \mathbf{C}| \left( \mathbf{A} \cdot \frac{\mathbf{B} \times \mathbf{C}}{|\mathbf{B} \times \mathbf{C}|} \right) = \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 8 \times 3 = 24

Directly: $\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 1(0) + 1(0) + 3(8) = 24$ . The volume is 24.

Problem 2 · Signed Triple Product vs. Volume

Given: $\mathbf{A} = \langle 2, 1, 3 \rangle$ , $\mathbf{B} = \langle 1, 0, 2 \rangle$ , $\mathbf{C} = \langle 0, 3, 1 \rangle$ — find the volume of the parallelepiped they span.

✅ Correct!

\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = -4

, and a volume is the absolute value:

|-4| = 4

❌ Close — that is the triple product, not the volume. The height

\mathbf{A} \cdot \hat{\mathbf{n}}

is signed, so

\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \pm\,\text{volume}

. Take the absolute value.

❌ That is only the first term.

a_1

times its minor is

2(-6) = -12

; you still owe

-a_2

times its minor and

a_3

times its minor, which bring the total to

-4

❌ Zero would mean the three vectors are coplanar. Here the triple product is

-4 \ne 0

, so the box is not flat.

❌ Not quite. Build

\mathbf{B} \times \mathbf{C}

from the three

2 \times 2

minors, dot with

\mathbf{A}

, then take the absolute value.

Show solution

Step 1: the cross product.

\mathbf{B} \times \mathbf{C} = \langle 0 \cdot 1 - 2 \cdot 3,\; 2 \cdot 0 - 1 \cdot 1,\; 1 \cdot 3 - 0 \cdot 0 \rangle = \langle -6,\, -1,\, 3 \rangle

Step 2: dot with $\mathbf{A}$ .

\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 2(-6) + 1(-1) + 3(3) = -12 - 1 + 9 = -4

Step 3: read off the volume. The negative sign says $\mathbf{A}$ leans to the far side of the base plane from $\hat{\mathbf{n}}$ , so the signed height is negative:

\text{volume} = \big| \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \big| = |-4| = 4

The volume is 4. (Check by determinant: $2\begin{vmatrix} 0 & 2 \\ 3 & 1\end{vmatrix} - 1\begin{vmatrix} 1 & 2 \\ 0 & 1\end{vmatrix} + 3\begin{vmatrix} 1 & 0 \\ 0 & 3\end{vmatrix} = -12 - 1 + 9 = -4$ .)

Problem 3 · Order and Parentheses

Given: $\mathbf{A} = \langle 1, 1, 3 \rangle$ , $\mathbf{B} = \langle 3, 1, 0 \rangle$ , $\mathbf{C} = \langle 1, 3, 0 \rangle$ , with $\mathbf{A} \times \mathbf{B} = \langle -3, 9, -2 \rangle$ .

Which expression is meaningless?

Evaluate $(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}$ .

✅ Correct!

(\mathbf{A} \cdot \mathbf{B})

is a number, and a number cannot be crossed with anything. Meanwhile

(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C} = 24 = \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})

— the triple product is cyclic.

❌ Check the type of each intermediate result. A cross product needs two vectors in; a dot product needs two vectors in and gives a number out.

❌ Check the arithmetic.

\mathbf{A} \times \mathbf{B} = \langle -3, 9, -2 \rangle

; now dot that with

\mathbf{C} = \langle 1, 3, 0 \rangle

Show solution

Part 1. Read each expression from the inside out and track the type:

$(\mathbf{A} \cdot \mathbf{B}) \times \mathbf{C}$ : $\mathbf{A} \cdot \mathbf{B} = 3 + 1 + 0 = 4$ , a number — and $4 \times \mathbf{C}$ is not a cross product. Meaningless.
$\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})$ : vector dot vector — a number.
$(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}$ : vector dot vector — a number.
$(\mathbf{A} \times \mathbf{B}) \times \mathbf{C}$ : vector cross vector — a vector. Well formed, though it is not a volume.

Part 2. With $\mathbf{A} \times \mathbf{B} = \langle -3, 9, -2 \rangle$ :

(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C} = (-3)(1) + (9)(3) + (-2)(0) = -3 + 27 + 0 = 24

And from Problem 1, $\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 24$ as well: both compute $\det(\mathbf{A}, \mathbf{B}, \mathbf{C})$ , so both give the same signed volume.

Problem 4 · Making the Box Collapse

Given: $\mathbf{A} = \langle 1, t, 2 \rangle$ , $\mathbf{B} = \langle 2, 1, 0 \rangle$ , $\mathbf{C} = \langle 0, 1, 1 \rangle$ , where $t$ is a parameter.

What is the base area $|\mathbf{B} \times \mathbf{C}|$ ?

For which $t$ do $\mathbf{A}$ , $\mathbf{B}$ , $\mathbf{C}$ lie in one plane?

✅ Correct!

\mathbf{B} \times \mathbf{C} = \langle 1, -2, 2 \rangle

has length

3

, and

\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 5 - 2t

, which vanishes exactly at

t = \tfrac{5}{2}

— the box flattens to zero height.

❌ Check the length.

\mathbf{B} \times \mathbf{C} = \langle 1, -2, 2 \rangle

, so

|\mathbf{B} \times \mathbf{C}| = \sqrt{1 + 4 + 4}

— and that square root comes out whole.

❌ Set the triple product to zero. Coplanar means zero volume, so solve

\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 0

for

t

Show solution

Step 1: the base.

\mathbf{B} \times \mathbf{C} = \langle 1 \cdot 1 - 0 \cdot 1,\; 0 \cdot 0 - 2 \cdot 1,\; 2 \cdot 1 - 1 \cdot 0 \rangle = \langle 1,\, -2,\, 2 \rangle

|\mathbf{B} \times \mathbf{C}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3

Step 2: the signed volume as a function of $t$ .

\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = (1)(1) + (t)(-2) + (2)(2) = 1 - 2t + 4 = 5 - 2t

Step 3: collapse the box. Three vectors lie in one plane exactly when the parallelepiped they span has zero volume:

5 - 2t = 0 \quad \Longrightarrow \quad t = \frac{5}{2}

Equivalently, the height $\mathbf{A} \cdot \hat{\mathbf{n}} = \tfrac{5 - 2t}{3}$ is zero there, so $\mathbf{A}$ lies in the plane of $\mathbf{B}$ and $\mathbf{C}$ . Note $t = 2$ gives $5 - 4 = 1 \ne 0$ : the box is thin, but not flat.

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