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

Turn One Vector Ninety Degrees: How the 2×2 Determinant Measures Area

IKey Formulas

Formula	Name	What you need
$\text{triangle} = \tfrac{1}{2}\|A\|\|B\|\sin\theta$	Area from two sides	Two sides and the angle between them
$A' = \langle -a_2,\ a_1 \rangle$	$A$ turned $90^\circ$ counterclockwise	The components of $A$
$\det(A, B) = \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1$	The $2 \times 2$ determinant	Main diagonal minus antidiagonal
$\text{parallelogram} = \bigl\|\det(A, B)\bigr\|$	Area from the determinant	The determinant, then its absolute value

Key Insight: $|A||B|\sin\theta = |A'||B|\cos\theta' = A' \cdot B = a_1b_2 - a_2b_1$ . A quarter turn converts the sine into a cosine, and a cosine is exactly what the dot product already computes. Read $\theta$ as the angle swept counterclockwise from $A$ to $B$ : when $B$ lies clockwise from $A$ that sweep is negative and so is the determinant, which is why an area needs $\bigl|\det(A,B)\bigr|$ .

IIVisualization 1 — A Quarter Turn in Components

Turning $A$ by $90^\circ$ swaps its two components and flips exactly one sign — the box decides which.

Turn

IIIVisualization 2 — The Sign the Bars Carry

The determinant is the parallelogram's area with a sign: positive exactly when $B$ sits counterclockwise from $A$ .

IVVisualization 3 — Base Times Height Becomes a Determinant

Five moves carry base times height for $A = \langle 3, 1 \rangle$ , $B = \langle 1, 2 \rangle$ down to $3 \cdot 2 - 1 \cdot 1$ .

Step 1

\text{area} = \text{base} \times \text{height}

\text{base} = |A| = \sqrt{10}

h = |B|\sin\theta = \tfrac{1}{2}\sqrt{10}

|A|\,h = \sqrt{10}\cdot\tfrac{1}{2}\sqrt{10} = 5

A' = \langle -1,\ 3 \rangle,\quad |A'| = |A|

|A||B|\sin\theta = |A'||B|\cos\theta'

|A'||B|\cos\theta' = A' \cdot B

A' \cdot B = (-1)(1) + (3)(2) = 5

A' \cdot B = a_1b_2 - a_2b_1

\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = 3 \cdot 2 - 1 \cdot 1 = 5

💡 Any polygon splits into triangles by drawing diagonals from one vertex, so this single determinant is enough to get the area of a polygon with any number of sides.

VQuiz Questions

Problem 1 · Evaluate a Determinant

Given: $A = \langle 4, 1 \rangle$ and $B = \langle 2, 3 \rangle$ — find $\det(A, B)$ .

✅ Correct! Main diagonal

4 \cdot 3 = 12

, antidiagonal

1 \cdot 2 = 2

, and

12 - 2 = 10

❌ Not quite. That is

12 + 2

. The determinant subtracts the antidiagonal from the main diagonal.

❌ Close — check the order.

-10

\det(B, A)

, the antidiagonal minus the main diagonal. With

A

in the top row the products go

a_1b_2 - a_2b_1

❌ Those are the wrong pairs.

a_1b_1 - a_2b_2 = 8 - 3

multiplies matching components — that is the dot-product pattern. The determinant multiplies crosswise.

❌ Not quite. Main diagonal minus antidiagonal:

4 \cdot 3 - 1 \cdot 2

Show solution

Stack $A$ on top of $B$ and take main diagonal minus antidiagonal:

\det(A, B) = \begin{vmatrix} 4 & 1 \\ 2 & 3 \end{vmatrix} = 4 \cdot 3 - 1 \cdot 2 = 12 - 2 = 10

The value is positive, so $B$ lies counterclockwise from $A$ , and the parallelogram spanned by $A$ and $B$ has area $10$ .

Problem 2 · Area, Not Determinant

Given: $A = \langle 2, 5 \rangle$ and $B = \langle 6, 1 \rangle$ — find the area of the parallelogram they span.

✅ Correct!

\det(A, B) = 2 - 30 = -28

, and the area is

|-28| = 28

; the minus sign only records that

B

lies clockwise from

A

❌ That is the determinant, not the area. The vertical bars around a determinant are not absolute-value bars — the determinant is

\pm

the area, so finish with

|-28| = 28

❌ Not quite.

2 \cdot 1 + 5 \cdot 6 = 32

adds the two diagonal products; the determinant subtracts them.

❌ That is the triangle. The determinant already gives the whole parallelogram; halve it only when you want the triangle.

❌ Not quite. Compute

2 \cdot 1 - 5 \cdot 6

first, then take the absolute value.

Show solution

Step 1 — the determinant:

\det(A, B) = \begin{vmatrix} 2 & 5 \\ 6 & 1 \end{vmatrix} = 2 \cdot 1 - 5 \cdot 6 = 2 - 30 = -28

Step 2 — the area: an area is never negative, so

\text{parallelogram} = \bigl|\det(A, B)\bigr| = 28

The negative determinant says $B$ sits clockwise from $A$ . Feeding the vectors in the other order gives $\det(B, A) = 6 \cdot 5 - 1 \cdot 2 = 28$ — the same area with the opposite sign.

Problem 3 · Rotate, Then Dot

Given: $A = \langle 3, 4 \rangle$ and $B = \langle -1, 2 \rangle$ — turn $A$ by $90^\circ$ counterclockwise, then take $A' \cdot B$ .

What is $A'$ ?

What is $A' \cdot B$ ?

✅ Correct!

A' = \langle -4, 3 \rangle

and

A' \cdot B = 4 + 6 = 10

— the same value as

\det(A, B) = 3 \cdot 2 - 4 \cdot (-1)

❌ That is the clockwise turn. Clockwise gives

\langle a_2, -a_1 \rangle

; counterclockwise swaps and negates the first component:

\langle -a_2, a_1 \rangle

❌ Check the rotation. The box picture sends

\langle a_1, a_2 \rangle

\langle -a_2, a_1 \rangle

: swap the components, then flip one sign.

❌ Watch $b_1$ .

b_1 = -1

, so the first product is

(-4)(-1) = +4

, not

-4

❌ That is $\det(B, A)$ . Reversing the rows flips the sign; here

A

is on top.

❌ Not quite. Multiply matching components of

A' = \langle -4, 3 \rangle

and

B = \langle -1, 2 \rangle

, then add.

Show solution

Step 1 — rotate. A quarter turn counterclockwise sends $\langle a_1, a_2 \rangle$ to $\langle -a_2, a_1 \rangle$ :

A' = \langle -4,\ 3 \rangle

Step 2 — dot with $B$ :

A' \cdot B = (-4)(-1) + (3)(2) = 4 + 6 = 10

Step 3 — check against the determinant:

\det(A, B) = \begin{vmatrix} 3 & 4 \\ -1 & 2 \end{vmatrix} = 3 \cdot 2 - 4 \cdot (-1) = 6 + 4 = 10

The two routes agree because $A' \cdot B$ is $a_1b_2 - a_2b_1$ written out. The parallelogram on $A$ and $B$ has area $10$ .

Problem 4 · A Triangle From Three Points

Given: the triangle with vertices $P(1, 1)$ , $Q(4, 2)$ and $R(2, 5)$ — find its area.

✅ Correct!

\overrightarrow{PQ} = \langle 3, 1 \rangle

\overrightarrow{PR} = \langle 1, 4 \rangle

\det = 12 - 1 = 11

, and the triangle is half of that.

❌ That is the parallelogram.

|\det| = 11

covers both halves; the triangle is

\tfrac{1}{2}|\det|

❌ Those are the wrong vectors.

\det(\langle 4, 2 \rangle, \langle 2, 5 \rangle) = 16

uses position vectors from the origin. The two sides must be measured from the same vertex of the triangle.

❌ Not quite.

12 + 1 = 13

adds the diagonal products; the determinant subtracts:

12 - 1 = 11

❌ Not quite. Build two edge vectors from

P

, take their determinant, then halve its absolute value.

Show solution

Step 1 — two sides from one vertex. Working from $P$ :

\overrightarrow{PQ} = \langle 4 - 1,\ 2 - 1 \rangle = \langle 3, 1 \rangle, \qquad \overrightarrow{PR} = \langle 2 - 1,\ 5 - 1 \rangle = \langle 1, 4 \rangle

Step 2 — the determinant:

\begin{vmatrix} 3 & 1 \\ 1 & 4 \end{vmatrix} = 3 \cdot 4 - 1 \cdot 1 = 12 - 1 = 11

Step 3 — halve it. The two edge vectors span a parallelogram of area $11$ , and the triangle is exactly half:

\text{area} = \tfrac{1}{2}\bigl|\det\bigr| = \tfrac{11}{2}

Starting from $Q$ or $R$ instead changes both vectors but not the answer: from $Q$ , $\overrightarrow{QP} = \langle -3, -1 \rangle$ and $\overrightarrow{QR} = \langle -2, 3 \rangle$ give $-9 - 2 = -11$ , and $\tfrac{1}{2}|-11| = \tfrac{11}{2}$ again.

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