# 10.8 Vectors  (Page 4/22)

 Page 4 / 22
$\begin{array}{l}{v}_{1}=⟨2-0,0-0⟩\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=⟨2,0⟩\hfill \end{array}$

We also have v 2 with initial point $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}\left(0,\text{\hspace{0.17em}}3\right).\text{\hspace{0.17em}}$

$\begin{array}{l}{v}_{2}=⟨0-0,3-0⟩\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=⟨0,3⟩\hfill \end{array}$

Therefore, the position vector is

$\begin{array}{l}v=⟨2+0,3+0⟩\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=⟨2,3⟩\hfill \end{array}$

Using the Pythagorean Theorem, the magnitude of v 1 is 2, and the magnitude of v 2 is 3. To find the magnitude of v , use the formula with the position vector.

$\begin{array}{l}\hfill \\ \begin{array}{l}|v|=\sqrt{|{v}_{1}{|}^{2}+|{v}_{2}{|}^{2}}\hfill \\ \begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{{2}^{2}+{3}^{2}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{13}\hfill \end{array}\hfill \end{array}\hfill \end{array}$

The magnitude of v is $\text{\hspace{0.17em}}\sqrt{13}.\text{\hspace{0.17em}}$ To find the direction, we use the tangent function $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{y}{x}.$

$\begin{array}{l}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{{v}_{2}}{{v}_{1}}\hfill \\ \mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{3}{2}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta ={\mathrm{tan}}^{-1}\left(\frac{3}{2}\right)=56.3°\hfill \end{array}$

Thus, the magnitude of $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\sqrt{13}\text{\hspace{0.17em}}$ and the direction is $\text{\hspace{0.17em}}{56.3}^{\circ }$ off the horizontal.

## Finding the components of the vector

Find the components of the vector $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ with initial point $\text{\hspace{0.17em}}\left(3,2\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}\left(7,4\right).$

First find the standard position.

$\begin{array}{l}v=⟨7-3,4-2⟩\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=⟨4,2⟩\hfill \end{array}$

See the illustration in [link] .

The horizontal component is ${v}_{1}$ $=⟨4,0⟩\text{\hspace{0.17em}}$ and the vertical component is $\text{\hspace{0.17em}}{v}_{2}$ $=⟨0,2⟩.$

## Finding the unit vector in the direction of v

In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector    . We can then preserve the direction of the original vector while simplifying calculations.

Unit vectors are defined in terms of components. The horizontal unit vector is written as $i$ $=⟨1,0⟩$ and is directed along the positive horizontal axis. The vertical unit vector is written as $j$ $=⟨0,1⟩$ and is directed along the positive vertical axis. See [link] .

## The unit vectors

If $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ is a nonzero vector, then $\text{\hspace{0.17em}}\frac{v}{|v|}\text{\hspace{0.17em}}$ is a unit vector in the direction of $\text{\hspace{0.17em}}v.\text{\hspace{0.17em}}$ Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar.

## Finding the unit vector in the direction of v

Find a unit vector in the same direction as $v$ $=⟨-5,12⟩.$

First, we will find the magnitude.

$\begin{array}{l}|v|=\sqrt{{\left(-5\right)}^{2}+{\left(12\right)}^{2}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{25+144}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{169}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=13\hfill \end{array}$

Then we divide each component by $\text{\hspace{0.17em}}|v|,\text{\hspace{0.17em}}$ which gives a unit vector in the same direction as v :

$\frac{v}{|v|}=-\frac{5}{13}i+\frac{12}{13}j$

or, in component form

$\frac{v}{|v|}=⟨-\frac{5}{13},\frac{12}{13}⟩$

Verify that the magnitude of the unit vector equals 1. The magnitude of $\text{\hspace{0.17em}}-\frac{5}{13}i+\frac{12}{13}j\text{\hspace{0.17em}}$ is given as

The vector u $=\frac{5}{13}$ i $+\frac{12}{13}$ j is the unit vector in the same direction as v $=⟨-5,12⟩.$

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

So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j .

## Vectors in the rectangular plane

Given a vector $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ with initial point $\text{\hspace{0.17em}}P=\left({x}_{1},{y}_{1}\right)\text{\hspace{0.17em}}$ and terminal point $Q=\left({x}_{2},{y}_{2}\right),$ v is written as

$v=\left({x}_{2}-{x}_{1}\right)i+\left({y}_{1}-{y}_{2}\right)j$

The position vector from $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(a,b\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}\left({x}_{2}-{x}_{1}\right)=a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left({y}_{2}-{y}_{1}\right)=b,\text{\hspace{0.17em}}$ is written as v = a i + b j . This vector sum is called a linear combination of the vectors i and j .

The magnitude of v = a i + b j is given as $\text{\hspace{0.17em}}|v|=\sqrt{{a}^{2}+{b}^{2}}.\text{\hspace{0.17em}}$ See [link] .

stock therom F=(x2+y2) i-2xy J jaha x=a y=o y=b
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was  1350  bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after  3  hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as
how do I attempted a trig number as a starter