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The cross product in${\mathit{R}}^{\mathbf{n}}$. Consider the vectors ${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}$, ${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{3}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{n}}$ in role="math" localid="1660390620385" ${\mathit{R}}^{\mathbf{n}}$ . The transformation $\mathit{T}\mathbf{\left(}\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathit{d}\mathit{e}\mathit{t}\left[\begin{array}{ccccc}1& |& |& & |\\ \stackrel{鈫�}{x}& {\stackrel{鈫�}{v}}_2& {\stackrel{鈫�}{v}}_3& ...& {\stackrel{鈫�}{v}}_n\\ |& |& |& & |\end{array}\right]$ is linear. Therefore, there exists a unique vector role="math" localid="1660390632609" $\stackrel{\mathbf{鈫�}}{\mathbf{u}}$ in role="math" localid="1660390628687" ${\mathit{R}}^{\mathbf{n}}$such that$\mathit{T}\mathbf{\left(}\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{\right)}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{.}\stackrel{\mathbf{鈫�}}{\mathbf{u}}$for all $\stackrel{\mathbf{鈫�}}{\mathbf{x}}$ in ${\mathit{R}}^{\mathbf{n}}$. Compare this with Exercise $\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{.}\mathbf{43}\mathit{c}$. This vector role="math" localid="1660390639527" $\stackrel{\mathbf{鈫�}}{\mathbf{u}}$ is called the cross product of ${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}$,${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{3}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{n}}$ , written as$\stackrel{\mathbf{鈫�}}{\mathbf{u}}\mathbf{=}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{3}}\mathbf{脳}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{n}}\mathbf{.}$In other words, the cross product is defined by the fact that$\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{脳}\left({\stackrel{鈫�}{v}}_2脳{\stackrel{鈫�}{v}}_3脳...脳{\stackrel{鈫�}{v}}_n\right)$$\mathbf{=}\mathit{d}\mathit{e}\mathit{t}\mathbf{\left[}\begin{array}{ccccc}\mathbf{1}& \mathbf{|}& \mathbf{|}& & \mathbf{|}\\ \stackrel{\mathbf{鈫�}}{\mathbf{x}}& {\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}& {\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{3}}& \mathbf{.}\mathbf{.}\mathbf{.}& {\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{n}}\\ \mathbf{|}& \mathbf{|}& \mathbf{|}& & \mathbf{|}\end{array}\mathbf{\right]}$ for all role="math" localid="1660390666262" $\stackrel{\mathbf{鈫�}}{\mathbf{x}}$ in role="math" localid="1660390669928" ${\mathit{R}}^{\mathbf{n}}$. Note that the cross product in role="math" localid="1660390676371" ${\mathit{R}}^{\mathbf{n}}$ is defined for role="math" localid="1660390680958" $\mathit{n}\mathbf{-}\mathbf{1}$ vectors only. (For example, you cannot form the cross product of just two vectors in role="math" localid="1660390685393" ${\mathit{R}}^{\mathbf{4}}$.) Since the$\mathit{i}$th component of a vector role="math" localid="1660390689594" $\stackrel{\mathbf{鈫�}}{\mathbf{w}}$ is ${\stackrel{\mathbf{鈫�}}{\mathbf{e}}}_{\mathbf{i}}\mathbf{.}\stackrel{\mathbf{鈫�}}{\mathbf{w}}$, we can find the cross product by components as follows: role="math" localid="1660390696637" $\mathit{i}$ th component of ${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{3}}\mathbf{脳}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{n}}$.

a. When is ${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{3}}\mathbf{脳}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{n}}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{0}}$? Give your answer in terms of linear independence.

b. Find ${\stackrel{\mathbf{鈫�}}{\mathbf{e}}}_{\mathbf{2}}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{e}}}_{\mathbf{3}}\mathbf{脳}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{e}}}_{\mathbf{n}}$.

c. Show that${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{3}}\mathbf{脳}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{n}}$is orthogonal to all the vectors ${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{i}}$, for $\mathit{i}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{n}$.

d. What is the relationship between ${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{3}}\mathbf{脳}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{n}}$ and ${\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{3}}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{脳}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{脳}{\stackrel{\mathbf{鈫�}}{\mathbf{v}}}_{\mathbf{n}}$? (We swap the first two factors.)

e. Express det$\left[{\stackrel{鈫�}{v}}_2脳{\stackrel{鈫�}{v}}_3脳...脳{\stackrel{鈫�}{v}}_n{\stackrel{鈫�}{v}}_2{\stackrel{鈫�}{v}}_3...{\stackrel{鈫�}{v}}_n\right]$ in terms of $||{\stackrel{鈫�}{v}}_2脳{\stackrel{鈫�}{v}}_3脳...脳{\stackrel{鈫�}{v}}_n||$.
f. How do we know that the cross product of two vectors in ${\mathit{R}}^{\mathbf{3}}$, as defined here, is the same as the standard cross product in ${\mathit{R}}^{\mathbf{3}}$? See Definition A.9 of the Appendix.

Step by step solution

01

Step by Step Solution: Step 1: Give your answer in terms of linear independence

a. The equation $\stackrel{鈫�}{v_2}脳\stackrel{鈫�}{v_3}脳鈥�脳\stackrel{鈫�}{v_n}=0$applies if and only if the given determinant is 0 for any given $\stackrel{鈫�}{x}$, which is if and only if${\stackrel{鈫�}{v}}_2,{\stackrel{鈫�}{v}}_3,鈥�,\stackrel{鈫�}{v_n}$ are linearly dependent.

02

To find e→2×e→3×...×e→n

b. To find$\stackrel{鈫�}{e_2}脳\stackrel{鈫�}{e_3}脳鈥�脳{\stackrel{鈫�}{e}}_n$, we have to find the vector$\stackrel{鈫�}{u}鈭�{\mathrm{鈩�}}^n$such that

role="math" localid="1660389198561" $\det \left[\begin{array}{ccccc}\stackrel{鈫�}{x}& \stackrel{鈫�}{e_2}& \stackrel{鈫�}{e_3}& ...& \stackrel{鈫�}{e_n}\end{array}\right]=\stackrel{鈫�}{x}路\stackrel{鈫�}{u},鈭赌\stackrel{鈫�}{x}鈭�{\mathrm{鈩�}}^n$

This gives us $\stackrel{鈫�}{x}路\stackrel{鈫�}{u}=x_1$

So,

$\stackrel{鈫�}{u}=\left[\begin{array}{c}1\\ 0\\ 鈰�\\ 0\end{array}\right]$

$\stackrel{鈫�}{u}=\stackrel{鈫�}{e_1}$

03

To show that v→2×v→3×...×v→n.

c. For $i=2,鈥�,n$, if $\stackrel{鈫�}{x}={\stackrel{鈫�}{v}}_i$, then the first and the $i$-th column of the given matrix are the same, which means the determinant is 0 , giving us

${\stackrel{鈫�}{v}}_i路\left(\stackrel{鈫�}{v_2}脳{\stackrel{鈫�}{v}}_3脳鈥�{\stackrel{鈫�}{v}}_n\right)=0,鈭赌i=2,鈥�,n$

04

To find the relationship.

d. We compute,

$\det \left[\begin{array}{ccccc}\stackrel{鈫�}{x}& \stackrel{鈫�}{v_2}& \stackrel{鈫�}{v_3}& ...& {\stackrel{鈫�}{v}}_n\end{array}\right]=-\det \left[\begin{array}{ccccc}\stackrel{鈫�}{x}& \stackrel{鈫�}{v_3}& \stackrel{鈫�}{v_2}& ...& {\stackrel{鈫�}{v}}_n\end{array}\right]$

which gives us,

$\stackrel{鈫�}{x}路\left(\stackrel{鈫�}{v_2}脳\stackrel{鈫�}{v_3}脳鈥�脳{\stackrel{鈫�}{v}}_n\right)=-\stackrel{鈫�}{x}路\left(\stackrel{鈫�}{v_3}脳\stackrel{鈫�}{v_2}脳鈥�脳{\stackrel{鈫�}{v}}_n\right)$

Thus,

${\stackrel{鈫�}{v}}_3脳{\stackrel{鈫�}{v}}_2脳鈥�脳\stackrel{鈫�}{v_n}=-\stackrel{鈫�}{v_2}脳{\stackrel{鈫�}{v}}_3脳鈥�脳{\stackrel{鈫�}{v}}_n$

05

To express v→2×v→3×...×v→nv→2v→3...v→n

e. To express we get,

$$\begin{gathered} \det \left[\begin{array}{cc}\stackrel{鈫�}{v_2}脳\stackrel{鈫�}{v_3}脳鈥�脳{\stackrel{鈫�}{v}}_n& {\stackrel{鈫�}{v}}_2{\stackrel{鈫�}{v}}_3鈥�\stackrel{鈫�}{v_n}\end{array}\right]=\left(\stackrel{鈫�}{v_2}脳\stackrel{鈫�}{v_3}脳鈥�脳\stackrel{鈫�}{v_n}\right)路\left(\stackrel{鈫�}{v_2}脳\stackrel{鈫�}{v_3}脳鈥�脳\stackrel{鈫�}{v_n}\right) \\ \det \left[\begin{array}{cc}\stackrel{鈫�}{v_2}脳\stackrel{鈫�}{v_3}脳鈥�脳{\stackrel{鈫�}{v}}_n& {\stackrel{鈫�}{v}}_2{\stackrel{鈫�}{v}}_3鈥�\stackrel{鈫�}{v_n}\end{array}\right]={\left|\left|\stackrel{鈫�}{v_2}脳\stackrel{鈫�}{v_3}脳鈥�脳\stackrel{鈫�}{v_n}\right|\right|}^2 \end{gathered}$$

06

Step 6: How do calculate the cross product of two vectors

f) If we were to find the cross product of two vectors $\stackrel{鈫�}{v},\stackrel{鈫�}{w}鈭�{\mathrm{鈩�}}^3$, as defined here, we would have to solve:

$$\begin{gathered} \left|\begin{array}{ccc}{x}_1& v_1& w_1\\ x_2& v_2& w_2\\ x_3& v_3& w_3\end{array}\right|=\left(x_1,x_2,x_3\right)路\left(u_1,u_2,u_3\right),鈭赌\left(x_1,x_2,x_3\right)鈭�{\mathrm{鈩�}}^3 \\ \left(v_2{w}_3-v_3{w}_2\right)x_1-\left(v_1{w}_3-v_3{w}_1\right)x_2+\left(v_1{w}_2-v_2{w}_1\right)x_3=u_1{x}_1+u_2{x}_2+u_3{x}_3 \\ {u}_1=v_2{w}_3-v_3{w}_2,u_2=v_3{w}_1-v_1{w}_3,u_3=v_1{w}_2-v_2{w}_1 \end{gathered}$$

So, $\stackrel{鈫�}{v}脳\stackrel{鈫�}{w}=\left(v_2{w}_3-v_3{w}_2,v_3{w}_1-v_1{w}_3,v_1{w}_2-v_2{w}_1\right).$

However, if we were to calculate their cross product by the standard definition, we would compute:

$$\begin{gathered} \stackrel{鈫�}{v}脳\stackrel{鈫�}{w}=\left|\begin{array}{ccc}\stackrel{鈫�}{e_1}& \stackrel{鈫�}{e_2}& \stackrel{鈫�}{e_3}\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}\right| \\ \stackrel{鈫�}{v}脳\stackrel{鈫�}{w}=\left(v_2{w}_3-w_2{v}_3,v_3{w}_1-v_1{w}_3,v_1{w}_2-v_2{w}_1\right), \end{gathered}$$

This is the same.

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