# 8.8 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.

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
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris