The Area of a Parallelogram

Section 1.4 The Area of a Parallelogram

The purpose of this section is to derive a formula for the area of a parallelogram spanned by two vectors in the plane, in space or more generally in

R^{N} .

The result will be essential for the understanding of the transformation formula for double integrals and the definition of surface integrals!

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The area of a parallelogram is given by the product of its base and its height. If we look at the parallelogram spanned by two vectors

x

and

y

then the base is

‖ x ‖,

and the height is

‖ n ‖,

where

n

is the projection of the vector

y

into the direction orthogonal to

x

as shown in Figure 1.15.

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Figure 1.15. The Parallelogram spanned by $x$ and $y .$
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To compute

‖ n ‖

note that

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\begin{aligned} ‖ n ‖^{2} & = ‖ y - p ‖^{2} \\ = y \cdot y - 2 \frac{x \cdot y}{‖ x ‖^{2}} x \cdot y + \frac{(x \cdot y)^{2}}{‖ x ‖^{4}} x \cdot x \\ = ‖ y ‖^{2} - 2 \frac{(x \cdot y)^{2}}{‖ x ‖^{2}} + \frac{(x \cdot y)^{2}}{‖ x ‖^{4}} ‖ x ‖^{2} \\ = ‖ y ‖^{2} - \frac{(x \cdot y)^{2}}{‖ x ‖^{2}} . \end{aligned}

(This is the same computation as done in the proof of the Cauchy-Schwarz inequality).

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Hence the area,

A,

of the parallelogram under consideration is

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\begin{aligned} A & = ‖ x ‖ ‖ n ‖ \\ = ‖ x ‖ \sqrt{‖ y ‖^{2} - \frac{(x \cdot y)^{2}}{‖ x ‖^{2}}} \\ (1.6) & = \sqrt{‖ x ‖^{2} ‖ y ‖^{2} - (x \cdot y)^{2}} . \end{aligned}

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By definition of the determinant of a

2 \times 2

-matrix and the norm of a vector we see that

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\begin{aligned} A^{2} & = det [\begin{array}{c} ‖ x ‖^{2} & x \cdot y \\ x \cdot y & ‖ y ‖^{2} \end{array}] \\ = det [\begin{array}{c} x \cdot x & x \cdot y \\ x \cdot y & y \cdot y \end{array}] . \end{aligned}

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Now let

J := [\begin{matrix} x & y \end{matrix}]

denote the matrix with columns

x

and

y,

and let

J^{T}

be its transposed matrix. Taking into account Note 1.6 we find that

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[\begin{matrix} x^{T} x & x^{T} y \\ x^{T} y & y^{T} y \end{matrix}] = J^{T} J .

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Hence we have proved the following theorem.

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Theorem 1.16. Area of parallelogram.

Suppose that

x, y \in R^{N} .

Then the area of the parallelogram spanned by

x

and

y

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\begin{matrix} (1.7) & A = \sqrt{det (J^{T} J)}, \end{matrix}

where

J := [\begin{matrix} x & y \end{matrix}]

is the

N \times 2

-matrix with columns

x

and

y .

N = 2,

that is, for vectors in the plane we have

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A = | det (J) | .

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Proof.

See the above discussion. The last assertion follows as

det (J^{T} J) = det J^{T} det J = (det J)^{2} .

Remark 1.17.

Note that in the formula (1.7) it is essential to compute

J^{T} J

and not

J J^{T} .

The matrix

J^{T} J

is always symmetric, so only three entries need to be computed.

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In a first example we compute the area of a parallelogram spanned by two vectors in the plane.

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Example 1.18.

Compute the area of the parallelogram spanned by the vectors

(1, 3)

and

(2, 4) .

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Answer.

2

square units.

Solution.

According to Theorem 1.16 the area is given by

| det [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}] | = | 4 - 6 | = 2 square units.

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In a second example we compute the area of a parallelogram spanned by two vectors in three dimensional space.

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Example 1.19.

Compute the area of the parallelogram spanned by

(1, 2, 4)

and

(1, 1, 3) .

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Answer.

\sqrt{6}

square units.

Solution.

We form the matrix

J

with the given vectors as its columns and its transposed matrix:

J = [\begin{matrix} 1 & 1 \\ 2 & 1 \\ 4 & 3 \end{matrix}] J^{T} = [\begin{matrix} 1 & 2 & 4 \\ 1 & 1 & 3 \end{matrix}] .

Then we form their product:

J^{T} J = [\begin{matrix} 1 & 2 & 4 \\ 1 & 1 & 3 \end{matrix}] [\begin{matrix} 1 & 1 \\ 2 & 1 \\ 4 & 3 \end{matrix}] = [\begin{matrix} 21 & 15 \\ 15 & 11 \end{matrix}] .

Finally we take the determinant of the last matrix:

det [\begin{matrix} 21 & 15 \\ 15 & 11 \end{matrix}] = 231 - 225 = 6 .

According to Theorem 1.16 the area of the parallelogram is

\sqrt{det (J^{T} J)} = \sqrt{6} square units.

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