Chapter 10.2, Problem 61E

To determine

To find: The terminal point of the vector $\text{v}=\langle 4,\frac{3}{2},-\frac{1}{4}\rangle$ .

Answer to Problem 61E

The terminal point of the vector $\text{v}=\langle 4,\frac{3}{2},-\frac{1}{4}\rangle$ is $\left(6,\frac{5}{2},-\frac{7}{4}\right)$ .

Explanation of Solution

Given information:

The vector is $\text{v}=\langle 4,\frac{3}{2},-\frac{1}{4}\rangle$ and initial point of vector is $\left(2,1,-\frac{3}{2}\right)$ .

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 $2$ , $1$ , $-\frac{3}{2}$ for $p_1$ , $p_2$ , $p_3$ and $\langle 4,\frac{3}{2},-\frac{1}{4}\rangle$ for $\text{v}$ in the formula.

  $\begin{array}{c}v=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle \\ \langle 4,\frac{3}{2},-\frac{1}{4}\rangle =\langle q_1-2,q_2-1,q_3-\left(-\frac{3}{2}\right)\rangle \\ \langle 4,\frac{3}{2},-\frac{1}{4}\rangle =\langle q_1-2,q_2-1,q_3+\frac{3}{2}\rangle \end{array}$

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

  $\begin{array}{c}{q}_1-2=4\\ q_2-1=\frac{3}{2}\\ q_3+\frac{3}{2}=-\frac{1}{4}\end{array}$

Solving all the equations gives the values $q_1=6$ , $q_2=\frac{5}{2}$ and $q_3=-\frac{7}{4}$ .

Therefore, the terminal point of the vector $\text{v}=\langle 4,\frac{3}{2},-\frac{1}{4}\rangle$ is $\left(6,\frac{5}{2},-\frac{7}{4}\right)$ .