Definition of unit vector | what is a vector math

• table of content
  
• Vector
• Magnitude of vector
• Unit vector
• Vector opreations
• Cosine of vector
• Dotproduct of vector
• Cross product of vector
• Application of vector

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defination of vector

vector is an object that has both a magnitude & a direction
represent with "→"

what is dimensions ?

we know that(w.k.t) three dimensions are normally use.x-axis,y-axis,z-axis

firest is x also called x-axis

second y also called y-axis

third is z also called z-axis

represent of axis

if a ∈ R² the two dimensional vector a = (x,Y)

if a ∈ R³ then three dimensional vector a = (x, y, z)

represent axis in vector

x axis represent as I̅ or cap Î

Y axis represent as J̅ or cap j

Z axis represent as z̅ or cap k

conver simple vector to unit vector

1) A̅ = ( -3,2,-7)

step 1: find all axis in this case x axis is -3, y-axis is 2, z-axis -7

step 2: we know that x-axis represent with I̅ so x denote as -3I̅,same as y and z axis 2J̅ and -7K̅

step 3:final answer is

answer:- A̅ = -3i̅+2j̅-7k̅

2) A̅ = (4,2,1)
Answer:- A̅ = 4I̅+2J̅+1k̅

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Magnitude of vector

we know that vector has a magnitude and direction both. we have understand magnitude

how to find magnitude of vector ?

the magnitude of vector is length of vector
the magnitude of vector A is denoted as |A| same ase other alphabet and equation

formula

For two dimensional vector a

a̅ = (x, y)

Magnitude of a̅

|a̅| = `\sqrt{x^2+y^2}`

For three dimensional vector a

a̅ = (x, y, z)

Magnitude of a̅

|a̅| = `\sqrt{x^2+y^2+z^2}`

calculate methode
Case : 1 : If two vector given find magnitude?

Example : 1 : a̅ = (1, 2, 3) & b̅ = (4, 5, 6) find magnitude of |`\overline{ab}`|.

step 1: define a value of x₁, y₁, z₁ and x₂, y₂, z₂.

step 2:
`overline{ab}=`
`\sqrt{(x_2-x_1)^2\+(y_2-y_1)^2+(z_2-z_1)^2}`

Note :x₁, y₁, z₁ and x₂, y₂, z₂ is a element of a different value. Is a value element is constant.
(Number and Alphabets)

answer:-
`overline{ab}=`
`\sqrt{(x_2-x_1)^2\+(y_2-y_1)^2+(z_2-z_1)^2}`
`=\sqrt{(4-1)^2\+\(5-2)^2\+\(6-3)^2}`
`=\sqrt{(3)^2\+\(3)^{2\}+\\(3)^{2\}}`
`=\sqrt{9\+\9\+9}`
`=\sqrt{27}`
` =\sqrt{9\times\3}`
`=3\sqrt3`

Prectice Sum:

1) A̅ = (4,2,1) & b̅ = (4, 5, 6) find magnitude of |`\overline{ab}`|.
Answer:- A̅ = `\sqrt{34}`

2) A̅ = (11,12,8) & b̅ = (11, 12, 13) find magnitude of |`\overline{ab}`|.
Answer:- A̅ = 5

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unit vector

a vector That has a magnitude of 1 is unit vector

is also known as direction vector

definition:- give vector divided by it's magnitude is called unit vector.

formula of unit vector: `\frac\wedge a` = `\frac{\overline a}{\|\overline a\|}`

how to find unit vector:

Example : 1 : a̅ = (2, 0, 3)

step 1: defind vector axis element x, y, z.
x in this case a=2, b=0, c=3

step 2: We know that unit vector formula `\frac\wedge a` = `\frac{\overline a}{\|\overline a\|}` (Find given vector magnitude)
a̅ = (2, 0, 3) but `|\overline a\|` find magnitude first.

`|\overline a\|`= `\sqrt{x^2+y^2+z^2}`
=`\sqrt{(2)^2+(0)^2+\left(3\right)^2}`
=`\sqrt{4+0+9}`
= `\sqrt{13}`

step 3: Put value in unit vector formula:
`\frac\wedge a` = `\frac{\overline a}{\|\overline a\|}`=
`=\frac{\left(2,0,3\right)}{\sqrt13}`
answer:- `\frac\wedge a` = `\frac1{\sqrt{13}}`(2,0,3)

Prectice Sum:

Reprsent Of Vector in the form of unit vector i, j, k & final magnitude c.

1) b̅=(2,-3,1)
answer:- 2i̅-3j̅+1k̅
2) c̅=(-3,2,4)
answer:- -3i̅+2j̅+4k̅

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Vector operations

Addition of a vector:

Example: 1 : a̅ = (x, y, z), b̅=(p, q, r)

a̅ + b̅ = (x+p, y+q, z+r)

Example: 2 :Multiplication of vector by scalar a̅ = (x, y, z) find 3a̅.

Multiplied by 3.
Multiplied all element with 3.
answer:- 3a̅ =(3x̅, 3y̅, 3z̅)

Prectice Sum:

1)a̅ = (11, 20, 15), b̅=(-20, 9, -5)
answer:- a̅ + b̅ = (-9, 29, 10)
2) a̅ = (-9, 12, 19), b̅=(-4, 9, 6)
answer:- a̅ + b̅ = (-13, 21, 25)

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Direction cosine of a Vector

direct line op passing thrdirect line op passing through the origin makes `\alpha`,`\beta`,`\gamma` angles with
the axis x,y and z.

angle between x axis and line op is `\alpha`(alpha)

angle between y axis and line op is `\beta`(beta)

angle between z axis and line op is `\gamma`(gamma)

`\alpha` is also known as l

`\beta` is also known as m

`\gamma` is also known as n

cosine = (l,m,n)

formula :

Let's suppose given a=(x,y,z) find cosine of vector

cosine = `\frac{\ x}{\sqrt{x^2+y^2+z^2}}`,`\frac{\ y}{\sqrt{x^2+y^2+z^2}}`,`\frac{\ z}{\sqrt{x^2+y^2+z^2}}`

How to find cosine of a vector find?

find cosine of vector a̅ =(x,y,z)

Step 1:

cos`\alpha`=`\frac{\ x}{\sqrt{x^2+y^2+z^2}}`= l

cos`\beta`=`\frac{\ y}{\sqrt{x^2+y^2+z^2}}`=m

cos`\gamma`=`\frac{\ z}{\sqrt{x^2+y^2+z^2}}`=n

Practice sums:

1) a̅ = (1, 2, 3), b̅=(-1, 2, 3), c̅=(-1, 2,-3) then find direction cosion of 2a̅ + 3b̅ - c̅ ?
answer:-
cos`\alpha`=`\frac0{\sqrt{388}}`, cos`\beta`=`\frac8{\sqrt{388}}`, cos`\gamma`=`\frac{18}{\sqrt{388}}`

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Dot product

Dot product is multiplication of two vector.

formula of dot product:(X₁ • X₂ + Y₁ • Y₂ + Z₁ • Z₂)

how to Dot product of two vector:

Example: 1 : a̅ = (X₁, Y₁, Z₁), b̅=(X₂, Y₂, Z₂) Find Dot product of vector a̅ • b̅.

step 1: Defind X₁, Y₁, Z₁ & X₂, Y₂, Z₂.

X₁=X₁ X₂=X₂
Y₁=Y₁ Y₂=Y₂
Z₁=Z₁ Z₂=Z₂

step 2:Put value of in form

(X₁ • X₂ + Y₁ • Y₂ + Z₁ • Z₂)

Prectice Sum:

1) a̅ = (1, 2, 1), b̅=(2, -3, 0) Find Dot product of vector a̅ • b̅.
answer:-a̅ • b̅ = -4
2) a̅ = (-4, 9, 6), b̅=(0, -2, 2) Find Dot product of vector a̅ • b̅.
answer:-a̅ • b̅ = -6

Some case in dot product & scalar product:

If x̅ & y̅(means to vector) are perpendicular vectors then x̅•y̅=0

cos`\theta` =`\frac{\overline a • b }\|\overline a\|\|\overline b\|` (Find angle between two vector.)

how to find angle between two vector:

Example: 1 : a̅ = (2, 1, 1), b̅=(1, 1, 2) then find angle between a̅ & b̅.

step 1:Defind a̅ • b̅

a̅ • b̅ =(2, 1, 1)•(1, 1, 2)
=(2)•(1)+(1)•(1)+(1)•(2)
=2+ 1+ 2
a̅ • b̅ =5

step 2:Defind |a̅|• |b̅|

|a̅|=`\sqrt{(2)^{2\}\+\(1)^2\+(1)^2}`
|a̅|=`\sqrt{4\+\1\+\1}`
|a̅|=`\sqrt6`

|b̅|=`\sqrt{(1)^{2\}\+\(1)^2\+(2)^2}`
|b̅|=`\sqrt{1\+\1\+\4}`
|b̅|=`\sqrt6`

step 3: Put value in angle between formula:

cos`\theta` =`\frac{\overline a • b }\|\overline a\|\|\overline b\|`

=`\frac5{\sqrt6\cdot\sqrt6}`

cos`\theta` =`\frac5\6`

`\theta`=`\cos^{-1}\frac 5\6`

Prectice Sum:

1) x̅= (1, -1, 0), y̅=(1, 0, 1) then find angle between x̅ & y̅.
answer:- cos`\theta =\frac1\2` and `\theta`=`\cos^{-1}\frac 1\2`,`\theta`=60°

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Cross product or vector product

how to find Cross product two vector:

Example: 1 :a̅ = (X₁, Y₁, Z₁), b̅=(X₂, Y₂, Z₂)

step 1:Defind a̅ × b̅ =   

step 2:Solve determinant

answer:- i̅ (Y₁•Z₂ - Z₂•Y₁)-j̅ (X₁•Z₂ - Z₁•X₂) k̅(Y₂•X₁ - X₂•Y₁)

Note:

1 Row is fix all sum.

Second Raw elements of vector A.

Third Raw elements of vector B.

Prectice Sum:

1) x̅= (1, 2, 3), y̅=(2, 3, 1) then find x̅ × y̅.
answer:- x̅ × y̅ = (-7, 5, -1)

Case: 1: find angle between two vector.

formula

sin`\theta`= `\frac{|\overline a ×\overline b\|}{|\overline a\||\overline b\|}`

step 1:Find a̅ × b̅ (Determinant method)
step 2:Solve Determinant (Scroll up & Show step : 2)
step 3:Find magnitude of step 2 equation.
step 4:Find Each vector seprate magnitude.
step 5:Put value in formula.

Prectice Sum:

1)Prove that the angle between vector x̅= (1, 1, -1), y̅=(2, -2, 1) `\sin^{-1}\sqrt{\frac{26}{27}}`.
answer:-`\theta`= `\sin^{-1}\sqrt{\frac{26}{27}}`

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Application of Dot product

how to find work done:

formula:

W = F•D

Example: 1 :The force is applied 3i̅ -j̅ +2k̅ and i̅ +3j̅ -k̅ are actiny an a pertical. The pertical moves form the point 2i̅ +3j̅ +k̅ to the point 5i̅ +2j̅ +3k̅. Then find the work done.

step 1:Defind F.
F = F1 + F2
=(3, -1, 2)+(1, 3, -1)
=(3+1), (-1+3), (2+(1))
F =(4, 2, 1)

step 2:Defind d.
d = B - A
=(5, 2, 3)-(2, 3, 1)
=(3, -1, 2)

step 3:Defind work.
Work = F•D
=(4, 2, 1)•(3, -1, 2)
=(4)(3) + (2)(-1) + (1)(2)
Work =12 Units

Prectice Sum:

1)The force is applied i̅ +2j̅ -3k̅ and i̅ -j̅ +2k̅ are actiny an a pertical. The pertical moves form the point (3, 1, 2) to the point (1, 3, -1). Then find the work done.
answer:-Work =1 Units

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