Chapter 9.5, Problem 37E

a.

To determine

Calculate triple products for the vectors.

  $u=\langle 0,1,1\rangle ,v=\langle 1,0,1\rangle ,w=\langle 1,1,0\rangle$

Answer to Problem 37E

  $u•(v\times w)=2$ , $u•(w\times v)=-2$ , $v•(u\times w)=-2$ , $v•(w\times u)=2$ , $w•(u\times v)=2$ , $w•(v\times u)=-2$ .

Explanation of Solution

Given information:

Three vectors $u,v\&w$ , are given calculate scalar triple product.

  $\begin{array}{l}u.(v\times w),u.(w\times v),v.(u\times w),\\ v.(w\times u),w.(u\times v),w.(v\times u)\end{array}$

Calculation:

Consider the following vectors

  $u=\langle 0,1,1\rangle ,v=\langle 1,0,1\rangle ,w=\langle 1,1,0\rangle$

Remember the form for the triple product

  $\begin{array}{l}a•(b\times c)=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|\\ u•(v\times w)=\left|\begin{array}{ccc}{u}_1& u_2& u_3\\ b_1& v_2& v_3\\ w_1& w_2& w_3\end{array}\right|Pluginforthevectorsofinterest\\ u•(v\times w)=\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|\\ u•(v\times w)=0(0-1)-1(0-1)+1(1)Simplify\\ u•(v\times w)=2\end{array}$

Put the second triple product of interest

  $\begin{array}{l}u•(v\times w)=\left|\begin{array}{ccc}{u}_1& u_2& u_3\\ w_1& w_2& w_3\\ v_1& v_2& v_3\end{array}\right|Pluginforthevectorsofinterest\\ u•(w\times v)=\left|\begin{array}{ccc}0& 1& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right|\\ u•(w\times v)=0(1-0)-1(1-0)+1(0-1)Simplify\\ u•(w\times v)=-2\\ \end{array}$

Put the third triple product of interest

  $\begin{array}{l}v•(u\times w)=\left|\begin{array}{ccc}{v}_1& v_2& v_3\\ u_1& u_2& u_3\\ w_1& w_2& w_3\end{array}\right|Pluginforthevectorsofinterest\\ v•(u\times w)=\left|\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 1& 1& 0\end{array}\right|\\ v•(u\times w)=1(0-1)-0(0-1)+1(0-1)Simplify\\ v•(u\times w)=-2\\ \end{array}$

Put the fourth triple product of interest

  $\begin{array}{l}v•(u\times w)=\left|\begin{array}{ccc}{v}_1& v_2& v_3\\ w_1& w_2& w_3\\ u_1& u_2& u_3\end{array}\right|Pluginforthevectorsofinterest\\ v•(w\times u)=\left|\begin{array}{ccc}1& 0& 1\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|\\ v•(w\times u)=1(1-0)-0(1-0)+1(1)Simplify\\ v•(w\times u)=2\\ \end{array}$

Put the fifth triple product of interest

  $\begin{array}{l}w•(u\times v)=\left|\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\right|\\ w•(u\times v)=1(1-0)-1(0-1)+0(0-1)Simplify\\ w•(u\times v)=2\end{array}$

Put the sixth triple product of interest

  $\begin{array}{l}w•(u\times v)=\left|\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& 1\end{array}\right|\\ w•(v\times u)=1(0-1)-1(1-0)+0(1-0)Simplify\\ w•(v\times u)=-2\end{array}$

b.

To determine

Make a conjecture about the relationships between six triple products.

Answer to Problem 37E

  $\begin{array}{l}u•(v\times w)=v•(w\times u)=w•(u\times v)=2\\ u•(w\times v)=v•(u\times w)=w•(v\times u)=-2\end{array}$

Explanation of Solution

Given information:

Three vectors $u,v\&w$ , are given calculate scalar triple product.

  $\begin{array}{l}u.(v\times w),u.(w\times v),v.(u\times w),\\ v.(w\times u),w.(u\times v),w.(v\times u)\end{array}$

Calculation:

On the basis of the following observation it seems as though

  $\begin{array}{l}u•(v\times w)=v•(w\times u)=w•(u\times v)=2\\ u•(w\times v)=v•(u\times w)=w•(v\times u)=-2\end{array}$

c.

To determine

Prove the conjecture you made in part (b).

Answer to Problem 37E

  $\begin{array}{l}u•(v\times w)=-u•(w\times u)\mathrm{Re}call that a \times b=-\left(b\times a\right)\\ v•\left(w\times u\right)=-v•\left(u\times w\right) \mathrm{Re}call that a\times b=-\left(b\times a\right) \\ w•\left(u\times v\right) =-w•\left(v\times u\right) \mathrm{Re}acll that a\times b=-\left(b\times a\right)\end{array}$

Explanation of Solution

Given information:

Three vectors $u,v\&w$ , are given calculate scalar triple product.

  $\begin{array}{l}u.(v\times w),u.(w\times v),v.(u\times w),\\ v.(w\times u),w.(u\times v),w.(v\times u)\end{array}$

Calculation:

Prove the conjecture of the relationship in the previous part.

  $\begin{array}{l}u•(v\times w)=-u•(w\times u)\mathrm{Re}call that a \times b=-\left(b\times a\right)\\ v•\left(w\times u\right)=-v•\left(u\times w\right) \mathrm{Re}call that a\times b=-\left(b\times a\right) \\ w•\left(u\times v\right) =-w•\left(v\times u\right) \mathrm{Re}acll that a\times b=-\left(b\times a\right)\end{array}$