# 8.8 Vectors  (Page 5/22)

 Page 5 / 22

## 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.

I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
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Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert