# 10.8 Vectors  (Page 5/22)

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## Writing a vector in terms of i And j

Given a vector $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ with initial point $\text{\hspace{0.17em}}P=\left(2,-6\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}Q=\left(-6,6\right),\text{\hspace{0.17em}}$ write the vector in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j.$

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

$\begin{array}{l}v=\left({x}_{2}-{x}_{1}\right)i+\left({y}_{2}-{y}_{1}\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(-6-2\right)i+\left(6-\left(-6\right)\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-8i+12j\hfill \end{array}$

## Writing a vector in terms of i And j Using initial and terminal points

Given initial point $\text{\hspace{0.17em}}{P}_{1}=\left(-1,3\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}{P}_{2}=\left(2,7\right),\text{\hspace{0.17em}}$ write the vector $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j.\text{\hspace{0.17em}}$

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

$\begin{array}{l}v=\left({x}_{2}-{x}_{1}\right)i+\left({y}_{2}-{y}_{1}\right)j\hfill \\ v=\left(2-\left(-1\right)\right)i+\left(7-3\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=3i+4j\hfill \end{array}$

Write the vector $\text{\hspace{0.17em}}u\text{\hspace{0.17em}}$ with initial point $\text{\hspace{0.17em}}P=\left(-1,6\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}Q=\left(7,-5\right)\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j.$

$u=8i-11j$

## Performing operations on vectors in terms of i And j

When vectors are written in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j,\text{\hspace{0.17em}}$ we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.

## Adding and subtracting vectors in rectangular coordinates

Given v = a i + b j and u = c i + d j , then

$\begin{array}{c}v+u=\left(a+c\right)i+\left(b+d\right)j\\ v-u=\left(a-c\right)i+\left(b-d\right)j\end{array}$

## Finding the sum of the vectors

Find the sum of $\text{\hspace{0.17em}}{v}_{1}=2i-3j\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{v}_{2}=4i+5j.$

According to the formula, we have

$\begin{array}{l}{v}_{1}+{v}_{2}=\left(2+4\right)i+\left(-3+5\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6i+2j\hfill \end{array}$

## Calculating the component form of a vector: direction

We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}j.\text{\hspace{0.17em}}$ For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.

Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with $\text{\hspace{0.17em}}|v|\text{\hspace{0.17em}}$ replacing $\text{\hspace{0.17em}}r.$

## Vector components in terms of magnitude and direction

Given a position vector $\text{\hspace{0.17em}}v=⟨x,y⟩\text{\hspace{0.17em}}$ and a direction angle $\text{\hspace{0.17em}}\theta ,$

$\begin{array}{lll}\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{x}{|v|}\hfill & \text{and}\begin{array}{cc}& \end{array}\hfill & \mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{y}{|v|}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=|v|\mathrm{cos}\text{\hspace{0.17em}}\theta \begin{array}{cc}& \end{array}\hfill & \hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=|v|\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \end{array}$

Thus, $\text{\hspace{0.17em}}v=xi+yj=|v|\mathrm{cos}\text{\hspace{0.17em}}\theta i+|v|\mathrm{sin}\text{\hspace{0.17em}}\theta j,\text{\hspace{0.17em}}$ and magnitude is expressed as $\text{\hspace{0.17em}}|v|=\sqrt{{x}^{2}+{y}^{2}}.$

## Writing a vector in terms of magnitude and direction

Write a vector with length 7 at an angle of 135° to the positive x -axis in terms of magnitude and direction.

Using the conversion formulas $\text{\hspace{0.17em}}x=|v|\mathrm{cos}\text{\hspace{0.17em}}\theta i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=|v|\mathrm{sin}\text{\hspace{0.17em}}\theta j,\text{\hspace{0.17em}}$ we find that

$\begin{array}{l}x=7\mathrm{cos}\left(135°\right)i\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{7\sqrt{2}}{2}\hfill \\ y=7\mathrm{sin}\left(135°\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{7\sqrt{2}}{2}\hfill \end{array}$

This vector can be written as $\text{\hspace{0.17em}}v=7\mathrm{cos}\left(135°\right)i+7\mathrm{sin}\left(135°\right)j\text{\hspace{0.17em}}$ or simplified as

$v=-\frac{7\sqrt{2}}{2}i+\frac{7\sqrt{2}}{2}j$

A vector travels from the origin to the point $\text{\hspace{0.17em}}\left(3,5\right).\text{\hspace{0.17em}}$ Write the vector in terms of magnitude and direction.

$v=\sqrt{34}\mathrm{cos}\left(59°\right)i+\sqrt{34}\mathrm{sin}\left(59°\right)j$

Magnitude = $\text{\hspace{0.17em}}\sqrt{34}$

$\theta ={\mathrm{tan}}^{-1}\left(\frac{5}{3}\right)=59.04°$

## Finding the dot product of two vectors

As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product . We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
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