3 vectors and scalar

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.


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.


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.


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


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.


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


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.


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


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.


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.


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.


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.


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.


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


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



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



Components

(Continued)

Third Quadrant~A = �4i� 5jUse the Pythagorean theorem todetermine the magnitude:

j ~Aj =p(�4)2 + (�5)2 = 6:403


� = tan�1(�5�4) = 51:34o

Use the tangent of the angle withrespect to the X axis: � = 180o + 51:34o = 231:34o



Components

(Continued)

Fourth Quadrant~A = +4i� 5jUse the Pythagorean theorem todetermine the magnitude:

j ~Aj =p42 + (�5)2 = 6:403


� = tan�1(�54) = �51:34o

Use the tangent of the angle withrespect to the X axis: � = 360o � 51:34o = 308:66o


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.


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.


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


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.


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


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.


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:


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


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


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


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


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.


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 = ~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


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:


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:


B) Next �nd the angles between the two vectors:


18:00 = 5:00p40:00 sin(�)

18:00

5:00p40:00

= sin(�)

� = sin�1 (0:5692)

� = 34:47o

3 vectors and scalar

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Transcript of 3 vectors and scalar