Chapter 11, Problem 14STP

To determine

To find: which of the following vectors have a dot product of -4.

1. $\begin{array}{l}u=\langle 0,4,0\rangle \\ v=\langle 0,0,-1\rangle \end{array}$
2. $\begin{array}{l}u=\langle -2,1,0\rangle \\ v=\langle -2,-1,0\rangle \end{array}$
3. $\begin{array}{l}u=\langle -1,0,1\rangle \\ v=\langle 2,0,-2\rangle \end{array}$
4. $\begin{array}{l}u=\langle 1,5,0\rangle \\ v=\langle -1,1,0\rangle \end{array}$

Answer to Problem 14STP

(C) $\begin{array}{l}u=\langle -1,0,1\rangle \\ v=\langle 2,0,-2\rangle \end{array}$ have a dot product of -4.

Explanation of Solution

Given information:

Formula used:

  $\begin{array}{l}\text{if}u=<u_1,u_2,u_3>\text{and}v=<v_1,v_2,v_3>\text{then}\\ u.v=(u_1\times v_1)+(u_2\times v_2)+(u_3\times v_3)\end{array}$

Calculation:

(A)  $\begin{array}{l}u=\langle 0,4,0\rangle \\ v=\langle 0,0,-1\rangle \\ u.v=0\times 0+4\times 0+0\times (-1)=0+0+0=0\end{array}$

(B)   $\begin{array}{l}u=\langle -2,1,0\rangle \\ v=\langle -2,-1,0\rangle \\ u.v=(-2)\times (-2)+1\times (-1)+0\times 0=4-1+0=3\end{array}$

(C)   $\begin{array}{l}u=\langle -1,0,1\rangle \\ v=\langle 2,0,-2\rangle \\ u.v=(-1)\times (2)+0\times 0+1\times (-2)=-2+0-2=-4\end{array}$

(D)  $\begin{array}{l}u=\langle 1,5,0\rangle \\ v=\langle -1,1,0\rangle \\ u.v=1\times (-1)+5\times 1+0\times 0=-1+5+0=4\end{array}$

Therefore, (C) $\begin{array}{l}u=\langle -1,0,1\rangle \\ v=\langle 2,0,-2\rangle \end{array}$ have a dot product of -4.