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c C. M. Jenkins, Dept of Physics, U. of South Alabama 1
Vectors
Some quantities require just a number to describe them.
Volume: This is a one liter coke...
Temperature: It is 93o outside...
Relative Humidity.... With 96% humidity!
These quantities are called scalars.For other quantities a number is not enough.
Some quantities need an number and a di-rection.
We agree on a point.
Then I walk a distance of 5 meters fromthat point.
My �nal location is anywhere on a circle of5 meters from the starting point.
These quantities are calledVectors.Examples of vector quantities:
Displacement (or location).
Velocity.
Acceleration.
Force.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 2
Vectors are represented as: ~A orA
Graphically, a vector is drawn as a line withan arrow head on it.The arrowhead end is called the head.The other end is called the tail.
Remember: the vector is described by a number and direction.
The number corresponds to the length of the vector and is calledthe magnitude.
The magnitude of ~A (the length of the vector) is representedby: A or j ~Aj.
The direction of the vector corresponds to the direction that thevector is pointing.
The direction of the vector is represented by the angle withrespect to the X axis.
Or the direction of vector ~A is represented by the unit vector:
A =~A
j ~Aj .
Vectors may be displaced (i.e. moved).
The vector is not changed if its length and direction are notchanged.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 3
Vector AdditionTriangle Method
Two vectors may be added.This operation is de�ned di�erently than the addition of two scalars.
We must add the vectors in a way that adds the lengths (i.e.) themagnitudes..
But we must also account for their directions...
Consider adding two vectors ~A and ~B.The result is the resultant vector ~R:
~R = ~A+ ~B
At �rst we de�ne vector additionby graphical methods.The �rst method is the Trianglemethod:
Draw vector ~A (i.e. from tail to head).
At the head of ~A, place the tail of ~B.
Draw vector ~B.
The resultant vector (~R) is determined by drawing a vector fromthe tail of ~A to the head of ~B.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 4
Vector AdditionParallelogram Method
The second method is the parallelogram method.This method gives the same result as the triangle method...
Consider adding two vectors ~A and~B.The result is the resultant vector~R:
~R = ~A+ ~B
Draw vector ~A (i.e. from tail to head).
At the head of ~A, place the tail of ~B.
Draw vector ~B.
Make copies of ~A and ~B.
Displace the copy of ~B until its tail is touching the tail of ~A.
Displace the copy of ~A until its tail is touching the head of thecopy of ~B.
The resultant vector ~R is drawn along the diagonal from the tailsof vectors ~A and ~B to the heads of ~A and ~B.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 5
Properties of addition of vectors
Two vectors may be added in any order: ~A + ~B = ~B + ~A, i.e.vector addition commutes.
Three or more vectors may be grouped in any order when added:~A+�~B + ~C
�=�~A+ ~B
�+ ~C , i.e. vector addition is associative.
Any number of vectors may be added together:
~R = ~A+ ~B + ~C + ~D
c C. M. Jenkins, Dept of Physics, U. of South Alabama 6
Inverse of a Vector
Scalar InverseNumbers have an additive inverse or inverse:
For the number a, there exist an additive inverse: -a.
A number a added to its additive inverse -a gives the identity(zero) as a result.
a + (-a) = 0.
Note this is how the operation of subtraction is de�ned. (i.e.just remove the parenthesis in the above example).
Vector InverseThe inverse of a vector is constructed by taking the vector and chang-ing its direction by 180o without changing its length.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 7
Vector Subtraction
We use the inverse vector and vec-tor addition to de�ne vector sub-traction.Suppose we wish to subtract ~B
from ~A to get the resultant vec-tor ~R:
~R = ~A� ~B
Draw vector ~A (i.e. from tail to head).
Construct the inverse of vector ~B.
At the head of ~A, place the tail of (- ~B).
Draw vector (- ~B).
The resultant vector (~R) is determined by drawing a vector fromthe tail of ~A to the head of ~B.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 8
Or construct a parallelogram:
~R = ~A+ ~B
Draw vector ~A (i.e. from tail to head).At the head of ~A, place the tail of ~B.Draw vector ~B.Make copies of ~A and ~B. Displace the copyof ~B until its tail is touching the tail of ~A.Displace the copy of ~A until its tail is touch-ing the head of the copy of ~B.The resultant vector ~R is drawn along the di-agonal from the head of vector ~B (original)heads of ~A (original).
This diagram may be simpli�ed by drawing ~A and~B with their tails at the same point.The resultant (~R = ~A� ~B) is drawn with its tail atthe head of ~B to the head of ~A.Note that these methods are as accurate as thegraphical tools (such as rulers and protractors) thatare used...We need numerical methods to calculate vector ad-
dition and subtraction.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 9
Coordinate Systems
Consider a world consisting of a plane.We want to describe the location of points in this plane.Coordinate Systems for that world needs:
A single point to measure allother points from: origin.
Two di�erent directions to mea-sure along.
In our case twomutually per-pendicular directions.
Lets call the \horizontal" di-rection (or axis) the \X" axis, the\vertical" direction the \Y" axis.
A rule to tell use the order thatthe directions position is commu-nicated.
In our case (X, Y).The location of (5,3) is indi-
cated in the Figure....This is called a Cartesian coordinate system.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 10
Components of Vectors
Let's consider a vector ~A that hasa magnitude of 7.0 with an angleof 50o with respect to the X axis.
Lets construct vector ~A with the
vector sum of two two vectors: ~AX
and ~AY .~AX is parallel to the X axis.~AY is parallel to the Y axis.The vectors ~AX and ~AY are calledthe components of the ~A.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 11
A Cartesian coordinate system allows us to use right triangles andtherefore trigonometry.
From the Figure you can identify:The right angle.The hypotenuse.Use the angle with respect to the X axis (50o).Then the length of ~AX is: AX = A cos(50o).And the length of ~AY is: AY = A sin(50o).These lengths are called the components of ~A.
The directions of these lengths are called the unit vectorsof ~A.
i means one unit along the X direction.j means one unit along the Y direction.
So ~AX (which is parallel to the X axis) is written as: ~AX = AX i =A cos(50o)i.And ~AY (which is parallel to the Y axis) is written as: ~AY = AY j =A sin(50o)j .The vector ~A = A cos(50o)i+ A sin(50o)j.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 12
Example: Components of Vectors
Back to our example of a vector
~A that has a magnitude of 7.0with an angle of 50o with respectto the X axis.Find the components of this vec-tor and write the the vector ~A incomponent form.The X component of the vector is:
Ax = 7:0 cos(50o) = 7:0(0:6427) = 4:500
The Y component of this vectoris:
Ay = 7:0 sin(50o) = 7:0(0:7660) = 5:362
Using the unit vectors:
~A = 4:500i+ 5:362j
c C. M. Jenkins, Dept of Physics, U. of South Alabama 13
Example: Magnitude & Direction of a Vector from
Components
Suppose a vector is given by its components.How do we �nd the magnitude of the vector:
Use the Pythagorean theorem: j ~Aj =qA2x + A2
y
How do we �nd the direction (i.e. the angle with respect to the X axis)of the vector?
Use the tangent of the angle with respect to the X axis: � =Tan�1(Ay
Ax
)Note that the angle with respect to the X axis becomes tricky if the
vector is not located in Quadrant I.First Quadrant~A = 4i+ 5jUse the Pythagorean theorem todetermine the magnitude:
j ~Aj =p42 + 52 = 6:403
Use the tangent of the angle withrespect to the X axis:
� = tan�1(5
4) = 51:34o
c C. M. Jenkins, Dept of Physics, U. of South Alabama 14
Example: Magnitude & Direction of a Vector from
Components
(Continued)
Second Quadrant~A = �4i+ 5jUse the Pythagorean theorem todetermine the magnitude:
j ~Aj =p(�4)2 + 52 = 6:403
The value of tangent returned byyour calculator is:
� = tan�1(5
�4) = �51:34o
Use the tangent of the angle withrespect to the X axis: � = 180o � 51:34o = 128:66o
c C. M. Jenkins, Dept of Physics, U. of South Alabama 15
Example: Magnitude & Direction of a Vector from
Components
(Continued)
Third Quadrant~A = �4i� 5jUse the Pythagorean theorem todetermine the magnitude:
j ~Aj =p(�4)2 + (�5)2 = 6:403
The value of tangent returned byyour calculator is:
� = tan�1(�5�4) = 51:34o
Use the tangent of the angle withrespect to the X axis: � = 180o + 51:34o = 231:34o
c C. M. Jenkins, Dept of Physics, U. of South Alabama 16
Example: Magnitude & Direction of a Vector from
Components
(Continued)
Fourth Quadrant~A = +4i� 5jUse the Pythagorean theorem todetermine the magnitude:
j ~Aj =p42 + (�5)2 = 6:403
The value of tangent returned byyour calculator is:
� = tan�1(�54) = �51:34o
Use the tangent of the angle withrespect to the X axis: � = 360o � 51:34o = 308:66o
c C. M. Jenkins, Dept of Physics, U. of South Alabama 17
Addition of Vectors
We want a numerical method to add two or more vectors.The rule for addition of vectors is simple:
To add vectors: algebraicly add the components (i.e.add the X components together, then add the Y com-ponents together).
Remember the result is a vector, which requires at least twonumbers to describe it.
These two numbers are:
A magnitude and direction.
Or an X component and Y component.
Suppose we want to add two vectors ~C =~A+ ~B:
~A = Axi + Ayj~B = Bxi + Byj
Using unit vectors:
~A = Axi + Ayj~B = Bxi + Byj~C = (Ax + Bx)i + (Ay +By)j
Vector addition using the components depicted graphically.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 18
Addition of Vectors
Example: Vector AdditionFind the resultant from the sum of ~A and ~B, where:
~A = (7i + 5j) m~B = (�17i + 9j) m
Just add the components....
~A = (7i + 5j) m
+ ~B = (�17i + 9j) m~C = (�10i + 14j) m
Find the magnitude and direction of the resultant vector (~C).Magnitude:
C =p(�10)2 + 142 = 17:20m
The angle with respect to the X axis:
� = tan�1(14
�10) = �54:46o
Since the resultant vector is in quadrant II, the angle is: � = 180o �54:46o = 125:54o.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 19
Vector Subtraction using Components
The numerical method to subtract vectors is very similar to adding twovectors.The rule for subtraction of two vectors is:
To subtract vectors: subtract the components(i.e. subtract the X components, then subtractthe Y components).
Remember the result is a vector, which re-quires at least two numbers to describe it.
These two numbers are:
A magnitude and direction.
Or an X component and Y component.
Suppose we want to subtract two vectors ~C = ~A� ~B:
~A = Axi + Ayj~B = Bxi + Byj
Using unit vectors:
~A = Axi + Ayj
� ( ~B = Bxi + Byj)~C = (Ax �Bx)i + (Ay � By)j
c C. M. Jenkins, Dept of Physics, U. of South Alabama 20
Example: Vector Addition
Find the resultant ~C = ~A� ~B, where:
~A = (7i + 5j) m
� ( ~B = (�17i + 9j) m )
Just subtract the components....
Using unit vectors:
~A = (7i + 5j) m
� ~B = (�17i + 9j) m~C = (24i � 4j) m
Find the magnitude and direction of the resultant vector (~C).Magnitude:
C =p(24)2 + (�4)2 = 24:33m
The angle with respect to the X axis:
� = tan�1(�424
) = �9:46o
Since the resultant vector is in quadrant IV, the angle is: � = 360o �9:46o = 350:53o.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 21
Example: A Vector Equation
A hiker walks along a trail in fourlegs. The �rst leg is 60 m north,the second leg is 130 m east, head-ing and distance of the third legis 80 m at an angle of 30o westof north. The four leg has an un-known length and heading. Thehiker ends up at a distance of 132.72m 13.06o west of north. Find thelength and heading of the fourthleg.Let the hike be represented by thevector equation:
~R = ~A+ ~B + ~C + ~D
where the unknown vector is ~D. Solving this equation for ~D:
~D = ~R � ~A� ~B � ~C
To calculate ~D, resolve all known vectors into components:
~A = (0i + 60j) m~B = (130i + 0j) m~C = (�80 sin(30)i + 80 cos(30)j) m~D = (Dxi + Dyj) m~R = (�132:72 sin(13:06)i + 132:72 cos(13:06)j) m
Or:~A = (0i + 60j) m~B = (130i + 0j) m~C = (�40i + 69:28j) m~D = (Dxi + Dyj) m~R = (�29:99i + 129:29j) m
c C. M. Jenkins, Dept of Physics, U. of South Alabama 22
Solving for the components of ~D:
~R = (�29:99i + 129:66j) m
� ( ~A = (0i + 60j) m )
� ( ~B = (130i + 0j) m )
� ( ~C = (�40i + 69:28j) m )~D = (�119:99x � 0:38j) m
So the vector describing the fourth leg is:
~D = (�119:99 x� 0:38 j) m
The length of ~D (i.e. the distance) is:
j ~Dj =p(�119:99)2 + (�0:38)2 = 119:99 m
The heading is:
� = tan�1(�0:38�119:99) = 0:181o
Both the \X" and \Y" compo-nents are negative.So this vector is in the third quad-rant.Or 0.181o South of West.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 23
Scalar or Dot Product
The scalar or dot product is a vector operation where two vectors aremultiplied and a scalar results.
The dot product is de�ned as:
~A � ~B = j ~Ajj ~Bj cos(�)
Where � is the angle between the two vectors.
the scalar product commutes: ~A� ~B = ~B � ~A.
The scalar product obeys the distributive prop-
erty: ~A �
�~B + ~C
�= ~A �
~B + ~A �~C
Note the dot product between two identicalunit vectors is:
i � i = jijjij cos(0)
i � i = 1
Note the dot product between two di�erent unit vectors is:
i � j = jijjjj cos(90)
i � j = 0
This gives us the general rule for a dot product of:
c C. M. Jenkins, Dept of Physics, U. of South Alabama 24
i � i = 1 j � j = 1 k � k = 1
i � j = 0 i � k = 0 j � k = 0
Geometrically, the dot product is the projection of on vector onto theother.
Or, howmuch of one vector is par-allel to the second.
The dot product between two vec-tors is easily calculated using com-ponents.
~A � ~B =�Axi+ Ayj + Azk
���Bxi+ Byj + Bzk
�
Just multiply out as a polynomial:
~A � ~B = AxBxi � i+AxBy i � j + AxBz i � k +AyBxj � i+ AyByj � j + AyBz j � k +AzBxk � i+ AzByk � j +AzBzk � k
Now apply the rules for the dot product between two unit vectors:
~A � ~B = AxBxi � i%1 + AxBy i � j%0 + AxBz i � k%0 +
AyBxj � i%0 + AyByj � j%1 +AyBzj � k%0 +
AzBxk � i%0 +AzByk � j%0 + AzBzk � k%1
c C. M. Jenkins, Dept of Physics, U. of South Alabama 25
This gives us the result
~A � ~B = AxBx +AyBy + AzBz
Note any vector \dotted" into itself gives the square of the magnitudeof that vector:
~A � ~A = AxAx +AyAy + AzAz
~A � ~A = A2x +A2
y +A2z
~A � ~A = j ~Aj2
c C. M. Jenkins, Dept of Physics, U. of South Alabama 26
Example: The Dot Product
Vector A: ~A = 3:00 i+ 4:00 j.Vector B: ~B = 6:00 i+ 2:00 j.
First �ne the dot product: ~A � ~B:~A � ~B =
�3:00 i+ 4:00 j
���6:00 i+ 2:00 j
�
~A � ~B = (3:00)(6:00)+ (4:00)(2:00)
~A � ~B = 26:00
Find the angle between vectors ~A and ~B.
~A � ~B = j ~Ajj ~Bj cos(�)So �rst �nd the magnitudes of vectors ~A and ~B.
First, vector A:
j ~Aj =qA2x + A2
y
j ~Aj =p(3:00)2 + (4:00)2
j ~Aj = 5:00
Next, vector B:
j ~Bj =qB2x +B2
y
j ~Bj =p(6:00)2 + (2:00)2
j ~Bj = p40:00
c C. M. Jenkins, Dept of Physics, U. of South Alabama 27
Using the de�nition of the dot product:
~A � ~B = j ~Ajj ~Bj cos(�)
26:00 = (5:00)p40:00 cos(�)
26:00
5:00p40:00
= cos(�)
� = cos�1 (0:8222)
� = 34:47o
c C. M. Jenkins, Dept of Physics, U. of South Alabama 28
Vector or Cross Product
The vector or cross product is a vector oper-ation where two vectors are multiplied and avector results.
~C = ~A� ~B
This vector is perpendicular to the plane de�ned by the two vec-tors that are multiplied.
The magnitude of the cross product is de�ned as:
j ~A� ~Bj = j ~Ajj ~Bj sin(�)
Where � is the angle between the two vectors.
The direction of the resultant vector ~C is determined by the right handrule.
Use your right hand.... Take the �ngers andpoint them in the direction of the �rst vectorin the cross product. Orient your hand so thatthe second vector points out of you palm. Thethumb points in the direction of the resultantvector ~C.
The cross product does not commute: ~A�~B = �
~B �~A.
This may be proven by use of of the right hand rule.
c C. M. Jenkins, Dept of Physics, U. of South Alabama 29
The magnitude of the cross product between the same unit vector is:
ji� ij = jijjij sin(0)
ji� ij = (1)(1)(0)
ji� ij = 0
so:
i� i = 0 j � j = 0 k � k = 0
Note, that this means: ~A�~A = 0 .
The cross product between two di�erent unitvector is:
ji� jj = jijjjj sin(90)
ji� jj = (1)(1)(1)
ji� jj = 1
The direction is determined by the right hand rule.For this example, the direction is k.From this example and the right hand rule, we can deduce the followingrelationships:
i� j = k k � i = j j � k = i
Suppose we have two vectors: ~A = Axi + Ayj + Azk and ~B = Bxi +Byj + Bzk
And we want to take the cross product of these two vectors:
c C. M. Jenkins, Dept of Physics, U. of South Alabama 30
~C = ~A� ~B
~A� ~B =�Axi+ Ayj + Azk
���Bxi+Byj +Bzk
�
Just multiply out as a polynomial (be careful of the order of multipli-cation) :
~A� ~B = AxBxi� i+AxBy i� j + AxBz i� k +
AyBxj � i+ AyByj � j + AyBzj � k +
AzBxk � i+ AzByk � j +AzBzk � k
Apply the rules for taking the cross product between unit vectors
~A� ~B = AxBxi� i%0 + AxBy i� j% k + AxBz i� k%�j +
AyBxj � i%�k +AyByj � j%0 +AyBzj � k% i +
AzBxk � i% j + AzByk � j% �i + AzBzk � k%0
so the result is:
~A� ~B = [AyBz �AzBy] i+ [AzBx � AxBz] j + [AxBy �AyBx] k
The cross product between these two vectors are more easily calculatedby using the determinant of a 3� 3 matrix:
~A� ~B = Deti j k
Ax Ay Az
Bx By Bz
=
[AyBz �AzBy] i+
[AxBz �AzBx] (�j)+[AxBy �AyBx] k
c C. M. Jenkins, Dept of Physics, U. of South Alabama 31
Example: The Cross Product
Vector A: ~A = 3:00i+ 4:00j + 0:00k.Vector B: ~B = 6:00i+ 2:00j + 0:00k.
A) First �nd the cross product: ~A� ~B:
~A� ~B = Deti j k
3:00 4:00 0:006:00 2:00 0:00
=
[(4:00)(0:00)� (0:00)(2:00)] i+
[(3:00)(0:00)� (0:00)(3:00)] (�j)+[(3:00)(2:00)� (4:00)(6:00)] k
So:~A� ~B = 0:00i+ 0:00j � 18:00k
B) Next �nd the angles between the two vectors:
j ~A� ~Bj = j ~Ajj ~Bj sin(�)
So �rst �nd the magnitudes of vectors ~A and ~B.
First, vector A:
j ~Aj =qA2x + A2
y
j ~Aj =p(3:00)2 + (4:00)2
j ~Aj = 5:00
Next, vector B:
j ~Bj =qB2x +B2
y
j ~Bj =p(6:00)2 + (2:00)2
j ~Bj = p40:00
Using the de�nition of the dot product:
c C. M. Jenkins, Dept of Physics, U. of South Alabama 32
B) Next �nd the angles between the two vectors:
j ~A� ~Bj = j ~Ajj ~Bj sin(�)
18:00 = 5:00p40:00 sin(�)
18:00
5:00p40:00
= sin(�)
� = sin�1 (0:5692)
� = 34:47o