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Chapter 8: Problem 11

Find the volume \(V(S)\) of the parallelepiped \(S\) in \(\mathbf{R}^{3}\) determined by the vectors: (a) \(u_{1}=(1,1,1), u_{2}=(1,3,-4), u_{3}=(1,2,-5)\) (b) \(u_{1}=(1,2,4), u_{2}=(2,1,-3), u_{3}=(5,7,9)\)

Short Answer

Expert verified
The volumes of the parallelepipeds are: (a) \(V(S) = 1\) (b) \(V(S) = 132\)

Step by step solution

01

Compute the Cross Product

Compute the cross product \(\vec{n} = \vec{u_{2}}\times\vec{u_{3}}\): \[ \vec{n} = \begin{vmatrix} \mathbf{i} &\mathbf{j} &\mathbf{k}\\ 1 &3 &-4\\ 1 &2 &-5 \end{vmatrix} \] To calculate this determinant, we will use cofactor expansion: \(\vec{n} = \Big| (3)(-5) - (-4)(2)\Big|\mathbf{i} - \Big| (1)(-5) - (-4)(1)\Big| \mathbf{j} + \Big| (1)(2)-(3)(1) \Big|\mathbf{k}\) \(\vec{n} = (-15+8)\mathbf{i} +(5+4) \mathbf{j} +(2-3)\mathbf{k} = (-7, 9, -1)\)
02

Compute the Scalar Triple Product

Compute the scalar triple product: \(V(S) =|\vec{u_{1}}\cdot(\vec{u_{2}} \times \vec{u_{3}})|=|(1,1,1)\cdot(-7, 9, -1)|\) So, \(V(S) =|(1(-7) + 1(9) + 1(-1))|=|(-7+9-1)|=|1|\) The volume of parallelepiped \(S\) in (a) is 1. (b) We have \(\vec{u_{1}}=(1,2,4), \vec{u_{2}}=(2,1,-3)\) and \(\vec{u_{3}}=(5,7,9)\).
03

Compute the Cross Product

Compute the cross product \(\vec{n} = \vec{u_{2}}\times\vec{u_{3}}\): \[ \vec{n} = \begin{vmatrix} \mathbf{i} &\mathbf{j} &\mathbf{k}\\ 2 &1 &-3\\ 5 &7 &9 \end{vmatrix} \] To calculate this determinant, we will use cofactor expansion: \(\vec{n} = \Big| (1)(9) - (-3)(7)\Big|\mathbf{i} - \Big| (2)(9) - (-3)(5)\Big| \mathbf{j} + \Big| (2)(7)-(1)(5) \Big|\mathbf{k}\) \(\vec{n} = (9+21)\mathbf{i} +(18+15) \mathbf{j} +(14-5)\mathbf{k} = (30, 33, 9)\)
04

Compute the Scalar Triple Product

Compute the scalar triple product: \(V(S) =|\vec{u_{1}}\cdot(\vec{u_{2}} \times \vec{u_{3}})|=|(1,2,4)\cdot(30, 33, 9)|\) So, \(V(S) =|(1(30) + 2(33) + 4(9))|=|(30+66+36)|=|132|\) The volume of parallelepiped \(S\) in (b) is 132.

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Most popular questions from this chapter

Evaluate: (a) \(\left|\begin{array}{rrrr}2 & -1 & 3 & -4 \\ 2 & 1 & -2 & 1 \\ 3 & 3 & -5 & 4 \\ 5 & 2 & -1 & 4\end{array}\right|\) (b) \(\left|\begin{array}{rrrr}2 & -1 & 4 & -3 \\ -1 & 1 & 0 & 2 \\ 3 & 2 & 3 & -1 \\ 1 & -2 & 2 & -3\end{array}\right|\) (c) \(\left|\begin{array}{rrrr}1 & -2 & 3 & -1 \\ 1 & 1 & -2 & 0 \\ 2 & 0 & 4 & -5 \\ 1 & 4 & 4 & -6\end{array}\right|\)

Let \(\sigma=24513\) and \(\tau=41352\) be permutations in \(S_{5} .\) Find \((\mathrm{a}) \quad \tau \circ \sigma,(\mathrm{b}) \quad \sigma^{-1}\)

Let \(V\) be the space of \(m\) -square matrices (as above), and suppose \(D: V \rightarrow K\). Show that the following weaker statement is equivalent to \(D\) being alternating: $$D\left(A_{1}, A_{2}, \ldots, A_{n}\right)=0 \quad \text { whenever } \quad A_{i}=A_{i+1} \text { for some } i$$

Find the determinant of \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) defined by $$ F(x, y, z)=(x+3 y-4 z, 2 y+7 z, x+5 y-3 z) $$The determinant of a linear operator \(F\) is equal to the determinant of any matrix that represents \(F\). Thus first find the matrix \(A\) representing \(F\) in the usual basis (whose rows, respectively, consist of the coefficients of $x, y, z$ ). Then $$ A=\left[\begin{array}{rrr} 1 & 3 & -4 \\ 0 & 2 & 7 \\ 1 & 5 & -3 \end{array}\right], \quad \text { and so } \quad \operatorname{det}(F)=|A|=-6+21+0+8-35-0=-8 $$

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