Chapter 9, Problem 1T

(a)

To determine

The graph of vector u in coordinate plane with initial point $P\left(3,-1\right)$ and terminal point $Q\left(-3,9\right)$.

Explanation of Solution

The initial and terminal point of the vector u are $P\left(3,-1\right)$ and $Q\left(-3,9\right)$ respectively.

Figure (1)

Thus, the Figure (1) shows the graph of vector u in coordinate plane with initial point $P\left(3,-1\right)$ and terminal point $Q\left(-3,9\right)$.

(b)

To determine

The vector u in terms of i and j.

Answer to Problem 1T

The vector u in terms of i and j is $-6i+10j$.

Explanation of Solution

Given:

The initial and terminal point of the vector u are $P\left(3,-1\right)$ and $Q\left(-3,9\right)$ respectively.

Formula used: The vector v with initial point $P\left(x_1,y_1\right)$ and the terminal point $Q\left(x_2,y_2\right)$ is given below,

$v=\langle x_2-x_1,y_2-y_1\rangle$ (1)

The vector $v=\langle a_1,a_2\rangle$ is expressed in terms of i and j as given below,

$v=\langle a_1,a_2\rangle =a_{1}i+a_{2}j$ (2)

Calculation:

Substitute 3 for $x_1$, $-1$ for $y_1$, $-3$ for $x_2$ and 9 for $y_2$ in equation (1), to find the vector u with initial point $P\left(3,-1\right)$ and terminal point $Q\left(-3,9\right)$,

$\begin{array}{c}u=\langle -3-3,9-\left(-1\right)\rangle \\ =\langle -6,9+1\rangle \\ =\langle -6,10\rangle \end{array}$

Substitute $-6$ for $a_1$ and 10 for $a_2$ in equation (2), to express the vector u in terms of i and j,

$\begin{array}{c}u=\left(-6\right)i+\left(10\right)j\\ =-6i+10j\end{array}$

Thus, the vector u in terms of i and j is $-6i+10j$.

(c)

To determine

The length of vector u.

Answer to Problem 1T

The length of vector u is $2\sqrt{34}$.

Explanation of Solution

Given:

From part (b), the vector u in terms of i and j is given below,

$u=-6i+10j$

Formula used: The length of the vector $v=a_{1}i+a_{2}j$ is given below,

$\left|v\right|=\sqrt{a_1^2+a_2^2}$ (3)

Calculation:

The vector u is $-6i+10j$.

Substitute $-6$ for $a_1$ and 10 for $a_2$ in equation (3), to find the length of the vector u,

$\begin{array}{c}\left|u\right|=\sqrt{{\left(-6\right)}^2+{\left(10\right)}^2}\\ =\sqrt{36+100}\\ =\sqrt{136}\\ =2\sqrt{34}\end{array}$

Thus, the length of vector u is $2\sqrt{34}$.