The Area of a Parallelogram = 4.1 AREAS, VOLUMES, AND CROSS PRODUCTS 239 The parallelogram determined by two nonzero and nonparallel vectors a = [a1, a2] and b vertex at the origin, and we regard the arrows representing a and b as forming [b1, b2] in R2 is shown in Figure 4.1. This parallelogram has a the two sides of the parallelogram having the origin as a common vertex. We can find the area of this parallelogram by multiplying the length ||a|| of its base by the altitude h, obtaining Area = - ||a|| h ||a|| ||b||(sin 0) = ||a|| ||b|| V1 - cos20. Recall from page 24 of Section 1.2 that a b = ||a|| ||b||(cos 0). Squaring our area equation, we have (Area)² = ||a||2||b||2 — ||a||||b||² cos²0 = ||a||||b|| (a b) = (a₁² + a₂²)(b₁² + b₂²) - (a,b, + a2b₂)² = (a,b₂- a₂b₁)². (1) The last equality should be checked using pencil and paper. On taking square roots, we obtain = Areala,ba₂b₁l. The number within the absolute value bars is known as the determinant of the matrix A = [b₁ b₁₂ and is denoted by |A| or det(A), so that x2 h det(A) = = a, a₂ b₁ b₂ ° x1 FIGURE 4.1 The parallelogram determined by a and b.
The Area of a Parallelogram = 4.1 AREAS, VOLUMES, AND CROSS PRODUCTS 239 The parallelogram determined by two nonzero and nonparallel vectors a = [a1, a2] and b vertex at the origin, and we regard the arrows representing a and b as forming [b1, b2] in R2 is shown in Figure 4.1. This parallelogram has a the two sides of the parallelogram having the origin as a common vertex. We can find the area of this parallelogram by multiplying the length ||a|| of its base by the altitude h, obtaining Area = - ||a|| h ||a|| ||b||(sin 0) = ||a|| ||b|| V1 - cos20. Recall from page 24 of Section 1.2 that a b = ||a|| ||b||(cos 0). Squaring our area equation, we have (Area)² = ||a||2||b||2 — ||a||||b||² cos²0 = ||a||||b|| (a b) = (a₁² + a₂²)(b₁² + b₂²) - (a,b, + a2b₂)² = (a,b₂- a₂b₁)². (1) The last equality should be checked using pencil and paper. On taking square roots, we obtain = Areala,ba₂b₁l. The number within the absolute value bars is known as the determinant of the matrix A = [b₁ b₁₂ and is denoted by |A| or det(A), so that x2 h det(A) = = a, a₂ b₁ b₂ ° x1 FIGURE 4.1 The parallelogram determined by a and b.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The proof as shown below needs to be rewritten explaining all the steps and justification.
![The Area of a Parallelogram
4.1 AREAS, VOLUMES, AND CROSS PRODUCTS 239
The parallelogram determined by two nonzero and nonparallel vectors a -
[a₁, a₂] and b = [b₁, b₂] in R² is shown in Figure 4.1. This parallelogram has a
vertex at the origin, and we regard the arrows representing a and b as forming
the two sides of the parallelogram having the origin as a common vertex.
We can find the area of this parallelogram by multiplying the length ||a|| of
its base by the altitude h, obtaining
Area = ||a|| h = ||a|| ||b||(sin 0) = |a|| ||b||√1 – cos²0.
Recall from page 24 of Section 1.2 that a b = ||a|| ||b||(cos 0). Squaring our area
equation, we have
(Area)²= ||a||||b||-||a||||b|| cos20
=
—
||a||2||b||2² — (a • b)²
2
2
-
= (a + a₂²)(b₁² + b²²) − (a,b₁ + a₂b₂)²
=
-
= (a,b₂ — a₂b₁)².
(1)
The last equality should be checked using pencil and paper. On taking square
roots, we obtain
Area =
|aba₂bl.
The number within the absolute value bars is known as the determinant of the
matrix
A
==
a₁ a₂
[b₁ b₂
and is denoted by A or det(A), so that
0
x2
det(A)
=
a₁ a2
b₁ b₂
b
h
0
a
X1
FIGURE 4.1
The parallelogram determined by a and b.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7e1e198c-2ff2-4722-9853-fa8a39957b79%2F1a3b1fca-04b8-4e3f-9239-41985ae8f930%2Fpg6g3j5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The Area of a Parallelogram
4.1 AREAS, VOLUMES, AND CROSS PRODUCTS 239
The parallelogram determined by two nonzero and nonparallel vectors a -
[a₁, a₂] and b = [b₁, b₂] in R² is shown in Figure 4.1. This parallelogram has a
vertex at the origin, and we regard the arrows representing a and b as forming
the two sides of the parallelogram having the origin as a common vertex.
We can find the area of this parallelogram by multiplying the length ||a|| of
its base by the altitude h, obtaining
Area = ||a|| h = ||a|| ||b||(sin 0) = |a|| ||b||√1 – cos²0.
Recall from page 24 of Section 1.2 that a b = ||a|| ||b||(cos 0). Squaring our area
equation, we have
(Area)²= ||a||||b||-||a||||b|| cos20
=
—
||a||2||b||2² — (a • b)²
2
2
-
= (a + a₂²)(b₁² + b²²) − (a,b₁ + a₂b₂)²
=
-
= (a,b₂ — a₂b₁)².
(1)
The last equality should be checked using pencil and paper. On taking square
roots, we obtain
Area =
|aba₂bl.
The number within the absolute value bars is known as the determinant of the
matrix
A
==
a₁ a₂
[b₁ b₂
and is denoted by A or det(A), so that
0
x2
det(A)
=
a₁ a2
b₁ b₂
b
h
0
a
X1
FIGURE 4.1
The parallelogram determined by a and b.
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