Q5E The tetrahedron defined by three... [FREE SOLUTION]

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Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 6: Q5E (page 306)

The tetrahedron defined by three vectors ${\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3} in R^{3}$ is the set of all vectors of the form $c_{1} {\overset{鈫�}{v}}_{1} + c_{2} {\overset{鈫�}{v}}_{2} + c_{3} {\overset{鈫�}{v}}_{3}$ , where $c_{i} 鈮� 0$ and $c_{1} + c_{2} + c_{3} 鈮� 1$ . Explain why the volume of this tetrahedron is one sixth of the volume of the parallelepiped defined by ${\overset{鈫�}{v}}_{1}, {\overset{鈫�}{v}}_{2}, {\overset{鈫�}{v}}_{3}$ .

Short Answer

Expert verified

Therefore,

${鈭�}_{蠁 (u)} 1 d x d y = \frac{1}{6} d e t [{\overset{鈫�}{v}}_{1} {\overset{鈫�}{v}}_{2} {\overset{鈫�}{v}}_{3}]$

which is one sixth of the volume of the given parallelepiped.

Step by step solution

Known.

From multivariable calculus, we know that, for a differentiable injective function $蠁 : U 鈫� {鈩�}^{n}$ , where $U 鈯� {鈩�}^{n}$ is an open set, and D is the differentiation matrix of $蠁$ , and a continuous function $f : 蠁 (u) 鈫� {鈩�}^{n}$ , applies

${鈭�}_{蠁 (U)} f (x) d x = {鈭�}_{U} f (蠁 (u)) |d e t D (u)| d u$ .

To explain the volume of tetrahedron is one sixth of the volume of parallelepiped.

In this case, we have

$U = \{(c_{1}, c_{2}, c_{3}) 鈭� {[0, 1]}^{3} : c_{1} + c_{2} + c_{3} 鈮� 1\}$ And

$蠁 : U 鈫� {鈩�}^{n}, 蠁 (c_{1}, c_{2}, c_{3}) = c_{1} {\overset{鈫�}{v}}_{1} + c_{2} {\overset{鈫�}{v}}_{2} + c_{3} {\overset{鈫�}{v}}_{3}$

This means that the differentiation matrix of $蠁$ is indeed

$D = [{\overset{鈫�}{v}}_{1} {\overset{鈫�}{v}}_{2} {\overset{鈫�}{v}}_{3}]$

To calculate the volume of the tetrahedron, we must have

$f (x, y, z) = 1, 鈭赌 (x, y, z) 鈭� 蠁 (U)$

So, using the upper theorem, we have

${鈭�}_{蠁 (U)} 1 d x d y = {鈭�}_{u} 1 |d e t D (u)| d u {鈭�}_{蠁 (U)} 1 d x d y = |d e t D| {鈭�}_{0}^{1} {鈭�}_{0}^{1 - c_{1}} {鈭�}_{0}^{c_{1} - c_{2}} 1 d c_{3} d c_{2} d c_{1} {鈭�}_{蠁 (U)} 1 d x d y = |d e t D| . \frac{1}{6} {鈭�}_{蠁 (U)} 1 d x d y = \frac{1}{6} d e t [{\overset{鈫�}{v}}_{1} {\overset{鈫�}{v}}_{2} {\overset{鈫�}{v}}_{3}]$

This is one sixth of the volume of the given parallelepiped.

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Step by step solution

Known.

To explain the volume of tetrahedron is one sixth of the volume of parallelepiped.

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