Chapter 10.2, Problem 60E

To determine

To find: The terminal point of the vector $\text{v}=\langle -2,15,-8\rangle$ .

Answer to Problem 60E

The terminal point of the vector $\text{v}=\langle -2,15,-8\rangle$ is $\left(3,9,-8\right)$ .

Explanation of Solution

Given information:

The vector is $\text{v}=\langle -2,15,-8\rangle$ and the initial point of the vector is $\left(5,-6,0\right)$ .

Calculation:

Use the formula for component form of vector $\text{v}$ with initial point $\langle p_1,p_2,p_3\rangle$ and terminal point $\langle q_1,q_2,q_3\rangle$ .

  $v=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle$

Substitute $5$ , $-6$ , $0$ for $p_1$ , $p_2$ , $p_3$ and $\langle -2,15,-8\rangle$ for $\text{v}$ in the formula.

  $\begin{array}{c}v=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle \\ \langle -2,15,-8\rangle =\langle q_1-5,q_2-\left(-6\right),q_3-0\rangle \\ \langle -2,15,-8\rangle =\langle q_1-5,q_2+6,q_3\rangle \end{array}$

From the above equation, the following three equations can be extracted as given below.

  $\begin{array}{c}{q}_1-5=-2\\ q_2+6=15\\ q_3=-8\end{array}$

Solving all the equations gives the values $q_1=3$ , $q_2=9$ and $q_3=-8$ .

Therefore, the terminal point of the vector $\text{v}$ is $\left(3,9,-8\right)$ .