**Class – 11 Physics, Chapter – 5, Vector Analysis**

Vector a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity’s magnitude.Although a vector has magnitude and direction, it does not have position. That is, as long as its length is not changed, a vector is not altered if it is displaced parallel to itself.

In contrast to vectors, ordinary quantities that have a magnitude but not a direction are called **scalar**. For example, **displacement**, **velocity**, and **acceleration** are vector quantities, while speed (the magnitude of velocity), time, and mass are scalars.

To qualify as a vector, a quantity having magnitude and direction must also obey certain rules of combination.

Although vectors are mathematically simple and extremely useful in discussing physics, they were not developed in their modern form until late in the 19th century, when **Josiah Willard Gibbs** and **Oliver Heaviside** (of the United States and England, respectively) each applied vector analysis in order to help express the new laws of electromagnetism, proposed by **James Clerk Maxwell**

**Vector Notation :** vectors are conventionally written as boldface letters or arrow headed letter as given below .

The above vector can be represented by a bold letter i.e **A **or arrow headed letter

**Important Terms :**

**1 ) Collinear Vectors : **

** **

Two or more **vectors** that lie on the same line or on a parallel line to this, are called **collinear vectors**. Two **collinear vector** may point in either same or opposite direction. But, they cannot be inclined at some angle from each other for sure. Angle between collinear vectors is either zero degree or 180 degree .

**2 ) Parallel Vectors :**

**Vectors** are **parallel** if they have the same direction. Both components of one **vector** must be in the same ratio to the corresponding components of the **parallel vector**. Angle between parallel vectors is zero degree

**3 ) Anti-Parallel Vectors :**

Two vectors are said to be antiparallel, if they point in **exactly** opposite direction. The angle between two anti parallel vectors is 180 degree

**4 ) Unit Vector:**

The vector having unit magnitude is called as unit vector. It is used to denote direction of a vector. A **unit vector** is often denoted by a lowercase letter with a circumflex, or “hat”: (pronounced “i-hat”). The term direction **vector** is used to describe a **unit vector** being used to represent spatial direction. Any **vector** can become a **unit vector** by dividing it by the **vector’s** magnitude.

**5 ) Fixed Vectors:**

** Fixed vector** is that **vector** whose initial point or tail is **fixed**. It is also known as localised **vector**. For example, The initial point of a position **vector** is **fixed** at the origin of the coordinate axes. So, position **vector** is a **fixed** or localised vector.

**6 ) Free Vectors:**

A vector which is drawn parallel to a given vector through a specified point unlike free vector in space is called a localised vector. The effect of a force acting on a body depends not only on the magnitude & direction but also on its point of application & line of action.

**7 ) Co-Initial Vectors :**

When two vectors are drawn so they have the same initial (starting) point, then they are “co-initial”.

** **

**8 ) Coterminous Vectors:**

Coterminous vectors are those vectors which have common terminal point .

**9 ) Negative Vector :**

Two **vectors** are said to be **negative** if they have same magnitude, but opposite direction. Thus, the positive **vector** implies its direction, while if we put a **negative** sign in front of the **vector**, then it would indicate the exactly opposite direction of given positive **vector**.

**10 ) Position Vector :**

**Position vector**, straight line having one end fixed to a body and the other end attached to a moving point and used to describe the **position** of the point relative to the body. As the point moves, the **position vector** will change in length or in direction or in both length and direction.**Position vector**, straight line having one end fixed to a body and the other end attached to a moving point and used to describe the position of the point relative to the body. As the point moves, the position vector will change in length or in direction or in both length and direction. If drawn to some scale, the change in length will signify a change in the magnitude of the vector, while a change in direction will signify a rotation of the vector. Changes in magnitude and direction are the only changes that a position vector can experience, and the **velocity** of the point is defined as the time rate of change of the position vector.

For a point moving on a straight path, a position vector coinciding with the path is the most convenient; the velocity of the point is equal to the rate at which the magnitude of the vector changes with respect to time, and it will be a vector lying along the line. For a point moving on a circular path, a position vector coinciding with a radius of the circle is the most convenient; the velocity of the point is equal to the rate at which the direction of the vector changes with respect to time, and it will be a vector at right angles to the position vector. For a point moving on a non circular curved path, the position vector changes in both magnitude and direction; the velocity of the point is the sum of the two rates of change, one a vector along the position vector and the other a vector at right angles to it.

**11 ) Displacement Vector:**

A displacement is a vector whose length is the shortest distance from the initial to the final position of a point P. It quantifies both the distance and direction of an imaginary motion along a straight line from the initial position to the final position of the point.

**12 ) Equal Vectors :**

Equal vectors are vectors that have the same magnitude and the same direction. Equal vectors may start at different positions. Note that when the vectors are equal, the directed line segments are parallel. Equality Of Column Vectors. If two vectors are equal then their vector columns are equal.

**13 ) Null Vector : **

A null vector is a vector having magnitude equal to zero.It is represented by . A null vector has no direction or it may have any direction. Generally a null vector is either equal to resultant of two equal vectors acting in opposite directions or multiple vectors in different directions.

**14 ) Equality of Vectors: **

If two vectors look exactly alike with regard to length and direction, then they are exactly the same vector. Positioning of the vectors does not matter. In the picture here, and have a common length and direction, so they are equal and we can write .

**15 ) Co Planar Vectors: **

Vectors parallel to the same plane, or lie on the same plane are called **co planar vectors.**

**Condition of Vectors Coplanarity : **

- For 3-vectors: The three vectors are coplanar if their scalar triple product is zero.
- For 3-vectors: The three vectors are coplanar if they are linearly independent
- For n-vectors:n Vectors are coplanar if among them no more than two linearly independent vectors .

**Vector operation :**

- Addition of a vectors
- Subtraction of vectors
- Multiplication of vectors
- Resolution of a vector

**Addition of Vectors :**

A variety of mathematical operations can be performed with and upon vectors. One such operation is the addition of vectors. Two vectors can be added together to determine the result (or resultant). This process of adding two or more vectors has already been discussed in an earlier unit. Recall in our discussion of Newton’s laws of motion, that the *net force* experienced by an object was determined by computing the vector sum of all the individual forces acting upon that object. That is the net force was the result (or resultant) of adding up all the force vectors. During that unit, the rules for summing vectors (such as force vectors) were kept relatively simple. Observe the following summations of two force vectors :

These rules for summing vectors were applied to **free body diagrams** in order to determine the net force (i.e., the vector sum of all the individual forces). Sample applications are shown in the diagram below.

In this unit, the task of summing vectors will be extended to more complicated cases in which the vectors are directed in directions other than purely vertical and horizontal directions. For example, a vector directed up and to the right will be added to a vector directed up and to the left. The *vector sum* will be determined for the more complicated cases shown in the diagrams below.

There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors.

There basically three laws of vector addition

- Triangle law of addition
- Parallelogram law of vector addition
- Polygon law of vector addition.

**1) Triangle Law of Vectors Addition:**

It states that “ if two vectors can be represented both in magnitude and direction by two sides of a triangle taken in same order , then the resultant is represented completely, both in magnitude and direction , by the third side of the triangle taken in the reverse order”.

In the above figure, “AB” and “BC” are the two sides of ABC, taken in same direction represent two vectors **a **and **b . **Here , the bold letters indicate vector quantity . **c** is the resultant vector represented by side “AC” taken in reverse order .

**2) Parallelogram Law of Vector Addition :**

It states that “ if two vectors acting simultaneously at apoint can be represented both in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point , then the resultant is represented completely both in magnitude and direction by the diagonal of a parallelogram passing through that point”.

In the above figure , “OA” and “OB” are the two adjacent sides of a parallelogram OACB drawn from same point O representing two vectors **P **and **Q ** acting simultaneously at a point O . According to parallelogram law of vector addition , diagonal “OC” is the resultant vector **R** passing through same point O.

**3) Polygon Law of Vector Addition : **

**Polygon law of vector addition** states that “if a number of **vectors** can be represented in magnitude and direction by the sides of a **polygon** taken in the same order, then their resultant is represented in magnitude and direction by the closing side of the **polygon** taken in the opposite order”.

In triangle OKL, the vectors **A** and **B** are represented by the sides **OK **and **KL** taken in the same order. Therefore, from the triangle law of vector addition, the closing side **OL** taken in the opposite order represents the resultant of vectors ** OK ** and ** KL**

Thus, **OK + KL = OL** … (1)

By applying the triangle law of vector addition to the triangle OLM. It shows that the side OM is the resultant of vectors OL and LM i.e **OL + LM = OM**

Using eg (1), we get,

**OK + KL + LM = OM ……………………… ( 2 ) **

Similarly applying the triangle law of vector addition to the triangle OMN, we get,

**OM + MN = ON ……………….. ( 3 )**

Using equation ( 2) , we get

**OK + KL + MN = ON ………………………. ( 4 )**

Now, **OK = A, **

**KL = B,**

**LM = C,**

And **MN = D,**

Hence, the equation ( 4 ) becomes,

**ON = OK + KL + LM + MN **

**R = A + B+ C + D**

**Analytical method to calculate the magnitude and direction of resultant vectors of two vectors**

**1) By Using Parallelogram Law of Vector Addition:**

**Let two vectors P and Q represented by two adjacent sides of a parallelogram AB and AD drawn from same point A . θ is the angle between the two vectors .**

**R** is the resultant vector represented by diagonal AC passing through point A.

Applying pythagoras theorem in the right angled triangle AEC,

AC^{2}=AE^{2}+EC^{2}——————————-(1)

**magnitude of vector P**=AB=DC—————————(2)

**magnitude of vector Q**=AD=BC——————————(3)

(Unbold letter may be treated as the magnitude of the vector )

Again in right angled triangle BEC ,

cosθ = base / hypotenuse

= BE / BC

Therefore, BE = BCcosθ

= Qcosθ [by using equation (3)]

And sinθ = perpendicular / hypotenuse

= CE / BC

Therefore, CE = BCsinθ

= Qsinθ ———(4) [by using equation (3)]

AE = AB + BE

= P + Qcosθ——————————(5)

Now, putting the values of AE, AC and EC in the equation (1) we get

AC^{2} = AE^{2} + EC^{2}

⇒ R^{2} =(P+cosθ)^{2} + Q^{2}sin^{2}θ

⇒ R^{2} =P^{2} + Q^{2}cos^{2}θ + 2PQcosθ + Q^{2}sin^{2}θ

⇒ R^{2} =P^{2} + Q^{2}(cos^{2}θ + sin^{2}θ) + 2PQcosθ

Putting cos^{2}θ+sin^{2}θ = 1 in the above equation, we get

⇒ R^{2} = P^{2 }+ Q^{2 }+ 2PQcosθ

⇒ **R = (P**^{2}**+Q**^{2}**+2PQcosθ) ^{1/2 }**

The above expression is the magnitude of the resultant vector .

Now let us find the direction of **R**.

Let Ф is the angle made by **R** with the vector **P**.

In the **Δ**AEC,

tanФ = perpendicular / base

= CE / AE

= Qsinθ / ( P + Qcosθ ) [by using the equations (4) and (5) ]

**Ф = tan ^{-1}**

**{ Qsin**

**θ / ( P + Qcosθ ) }**

**2) By Using the Triangle Law of Addition :**

Let two vectors **P **and **Q **are the two vectors represented by two sides OA and OB of a **Δ**OBC taken in the same order and θ is the angle between them.

Let us extend OA to point C and BC is the perpendicular drop from point B on OC.

Applying pythagoras theorem in the right angled triangle OCB,

OB^{2 }= BC^{2 }+ OC^{2}——————————–(1)

OA = magnitude of **P **= P

AB = magnitude of **Q **= Q

OB = magnitude of resultant vector **R **= R

Again in the right angled triangle ABC,

cosθ = AC / AB

AC = ABcosθ

= Qcosθ

And sinθ = BC / AB

BC = ABsinθ

= Qsinθ

OC = OA + AC

= P + Qcosθ

Now putting the values of OC, BC and OA in equation (1), we get

OB^{2 }= OC^{2 }+ BC^{2}

⇒ R^{2} = ( P + Qcosθ )^{2 }+ Q^{2}sin^{2}θ

⇒ R^{2} = P^{2 }+ Q^{2}cos^{2}θ + 2PQcosθ + Q^{2}sin^{2}θ

⇒ R^{2 }= P^{2 }+ Q^{2}( cos^{2}θ + sin^{2}θ ) + 2PQcosθ

Putting cos^{2}θ + sin^{2}θ = 1 in the above equation, we get

⇒ R^{2 }= P^{2 }+ Q^{2 }+ 2PQcosθ

⇒ **R = ( P**^{2 }**+ Q**^{2 }**+ 2PQcosθ ) ^{1/2}**

The above expression is the magnitude of the resultant vector.

Now let us find the direction of **R**.

Let Ф is the angle made by **R** with the vector **P**.

Let Ф is the angle made by **R** with the vector **P**.

In the **Δ**OCB,

tanФ = perpendicular / base

= BC / OC

= Qsinθ / ( P+ Qcosθ )

**Ф = tan ^{-1}**

**{ Qsin**

**θ / ( P + Qcosθ ) }**

**Questions on the above discussion **

**1 ) Calculate the angle between a two dyne and a three dyne force so that their sum is four dyne.**

**Solution :**

** **Let **P** = 2 dyne

**= 3 dyne **

Resultant ( **R **) = 4 dyne

θ = Angle between **A **and **B **

R = ( P^{2 }+ Q^{2 }+ 2PQcosθ )^{1/2}

squaring both sides we get,

R^{2} = ( P^{2 }+ Q^{2 }+ 2PQcosθ )

⇒ 4^{2} = 2^{2} + 3^{2} + 2 *2 * 3 cosθ

⇒ 16 – 13 = 12 cosθ

⇒ 3 / 12 = cosθ

⇒ θ = cos ^{-1}( 1 / 4 )

= 75.522 degree

Therefore, angle between 2 dyne and 3 dyne forces is 75.522 degree.

**2 ) A motorboat is racing towards north at 25 km/hr and the water current in that region is 10 km/hr in the direction of 60 degree east of south. Find the resultant velocity of the boat.**

**Solution : **

velocity of motorboat ( **P **) = 25 km / hr towards north

velocity of water current ( **Q** ) = 10 km / hr in the direction of 60 degree east of south

Angle between **P **and **Q ** ( θ ) = 90 + ( 90 – 60 ) = 120 degree

we know that,

R = ( P^{2 }+ Q^{2 }+ 2PQcosθ )^{1/2}

Squaring both sides we get,

R^{2} = ( P^{2 }+ Q^{2 }+ 2PQcosθ )

= 25^{2} + 10^{2} + 2 * 25 * 10 * cos120

= 625 + 100 + ( – 250 )

= 725 – 250

= 475

R = 21.79 km / hr

α = tan^{-1}{ Psinθ / ( Q + Pcosθ ) }

= tan^{-1} { 25 * sin120 / ( 10 + 25 * cos 120 ) }

= 87.36 degree

**3 ) Resultant of two equal vectors a and b inclined at an angle θ is c. Calculate θ.**

**Given : magnitude of a = magnitude of b = magnitude of c**

**Solution : **Since vectors **a **and **b **are equal vectors

magnitude of ** a = ** magnitude of **b = ** magnitude of c = x ( say )

as we know that

c^{2} = ( a^{2} + b^{2} + 2abcosθ )

⇒ x^{2} = x^{2} + x^{2} + 2x^{2}cosθ

⇒ x^{2} + 2x^{2}cosθ = 0

⇒ x^{2} = – 2x^{2}cosθ

⇒ cosθ = – 1 / 2

⇒ cosθ = cos120

⇒ θ = 120

**4 ) Two vectors acting in opposite direction have a resultant of 10 units. If they act at right angles to each other, the resultant is 50 units. Calculate the magnitudes of the vectors.**

**Solution : **

Let two vectors are **P** and **Q**

when the vectors are opposite in direction ( angle between the vector is 180 degree ), the magnitude of resultant vector is given by

R = ( P^{2 }+ Q^{2 }+ 2PQcosθ )^{1/2}

10 = ( P^{2} + Q^{2} + 2PQcos180 ) ^{1/2}

10 = ( P^{2} + Q^{2} – 2PQ)^{1/2}

10 = ( P – Q ) ——————————————– ( 1 )

when the vectors act at right angles to each other then the magnitude of the resultant is given by

r = ( P^{2} + Q^{2} + 2PQcos90 ) ^{1/2}

50 = ( P^{2} + Q^{2} )^{1/2 }( ∵ cos90 = 0 )

squaring both sides we get

2500 = P^{2} + Q^{2}

2500 = ( P – Q )^{2} + 2PQ

putting the value of ( P – Q ) we get

2500 = 100 + 2PQ

2500 – 100 = 2PQ

2PQ = 2400

Now squaring both sides of equation ( 1 ) we get

100 = ( P – Q )^{2}

100 = ( P + Q )^{2} – 4PQ

100 = ( P + Q )^{2} – 4800

100 + 4800 = ( P + Q )^{2}

4900 = ( P + Q ) ^{2}

70 = ( P + Q ) —————————- ( 2 )

On solving equations ( 1 ) & ( 2 ) we get

P = 40 units

Q = 30 units

The required magnitude of vectors are 40 units and 30 units.

**5 ) Resultant of two vectors which have equal magnitudes and which act at right angle is 1414 dyne. Calculate the magnitude of each force.**

**solution :**

let two vectors are **P** and **Q**

Given that P = Q

resultant vector = 1414 dyne

( P^{2 }+ Q^{2 }+ 2PQcos90 )^{1/2 }=1414

squaring both sides we get

P^{2} + P^{2} + 0 = 1414*1414

2P^{2} = 1414*1414

P = √{(1414*1414 )/2 } = 1414/√2 = 999.85 ≃ 1000 dyne

The magnitude of the givenequal vectors are 1000 dyne.

**6 ) Two forces of 5 kgf are acting at an angle of 120 degree****. Calculate the magnitude and direction of the resultant force.**

**Solution :**

Let two forces are represented by P and Q.

P = Q = 5 kg-f

angle between P and Q = 120 degree

magnitude of resultant of given vectors ( R ) = ( P^{2 }+ Q^{2 }+ 2PQcos120 )^{1/2}

= { 5^{2} + 5^{2} + 2 * 5 * 5 * ( – 0.5 ) }^{1/2}

= 5 kg – f

angle made by **R** with Q = tan^{-1} { 5sin120 / ( 5 + 5cos120 ) } = 60 degree.

**7 ) A body is simultaneously given two velocities, one 10 m/se due east and other 20 m/sec due north-west. Calculate the resultant velocity.**

**Solution :**

Let P = 10 m/sec due east

Q = 20 m/sec due north – west

angle between them = 90 + 45 = 135 degree

resultant velocity = ( P^{2 }+ Q^{2 }+ 2PQcos135 )^{1/2}

= ( 10^{2} + 20^{2} +2*10*20cos135 )^{1/2}

= ( 100 + 400 + 400 cos135 )^{1/2}

= 14.74 m/sec

**8 ) At what angle do the forces (A+B) act ( A – B ) so that the magnitude of the resultant is ( 3A**^{2 }**+ B ^{2}**

**)**

^{1/2 }**?**

**Solution : **

Let P = ( A + B )

Q = ( P – Q )

θ = angle between P and Q

Resultant ( R ) = ( 3A^{2 }+ B^{2} )^{1/2}

( P^{2 }+ Q^{2 }+ 2PQcosθ ) = ( 3A^{2 }+ B^{2} )

( A + B )^{2} + ( A – B )^{2} + 2*( A + B ) * ( A – B ) cosθ = 3A^{2} + B^{2}

2A^{2} + 2B^{2} + 2 ( A^{2} – B^{2} ) cosθ = 3A^{2} + B^{2}

2 ( A^{2} – B^{2} ) cosθ = 3A^{2} + B^{2} – 2A^{2} – 2B^{2}

cosθ = ( A^{2} – B^{2} ) / 2( A^{2} – B^{2} )

cosθ = 1 / 2

cosθ = cos60

θ = 60

**∴ **required angle is 60 degree.

**9 ) A man walks 4 m towards east and then turns 60 degree**** to the north of east and again walks 4 m. calculate net displacement.**

**Solution : **

Let P = 4 m due east

Q = 4 m making 60 degree north of east

angle between P and Q = ( 90 + 60 ) degree = 150 degree

magnitude of net displacement ( R ) = ( P^{2 }+ Q^{2 }+ 2PQcosθ )^{1/2}

= ( 4^{2} + 4^{2} + 2*4*4cos150 )^{1/2}

= ( 16 + 16 – 27.713 )^{1/2}

= 2.071 m

angle made by R with P = tan^{-1} { ( Psin150 ) / ( Q + Pcos150 ) } = tan^{-1} { (4*sin150 ) / ( 4 + 4cos150 ) } = 20.6 degree.

**10 ) Two forces whose magnitude are in the ratio 3:5 give a resultant a 35 N. If the angle of the inclination be 60 degree****, calculate the magnitude of each force.**

**Solution : **

Let two forces are represented by **P** and **Q**

P **:** Q = 3 **:** 5

P / Q = 3 / 5

P = 3k

Q = 5k

where k is non zero constant

angle between P and Q = 60 degree

Resultant = 35 N

( P^{2 }+ Q^{2 }+ 2PQcosθ ) = 35*35

9k^{2} + 25k^{2} + 30k^{2}cos60 = 1225

31k^{2} + 15k^{2} = 1225

46k^{2} = 1225

k^{2} = 1225 / 46

k = 5.16

k ≃ 5

**∴** the magnitudes of given vectors are ( 3*5 ) = 15 N and ( 5*5 ) = 25 N

**11 ) The greatest and the least resultant of two forces acting at a point is 10 N and 6 N. if each force is increased by 3 N, find the resultant of new forces when acting at a point at an angle of 90 degree**** with each other.**

**Solution : **

Let P and Q represent two vectors.

greatest resultant = 10 N

( P + Q ) = 10 ——————————————- ( 1 )

least resultant = 6 N

( P – Q ) = 6 ————————— ( 2 )

adding equation ( 1 ) and ( 2 ) we get

2P = 16

P = 8 N

putting the value of P in equation ( 1 ) we get

Q = 10 – 8 = 2 N

Now, after adding the magnitude of new forces are

A = P + 3 = 8 + 3 = 11 N

B = Q + 3 = 2 + 3 = 5 N

angle between A and B = 90 degree

required resultant = ( A^{2} + B^{2} + 2ABcos90 )^{1/2}

= ( 11^{2} + 5^{2} )^{1/2}

= ( 121 + 25 )^{1/2}

= 12.08 N

**Subtraction of vectors : **

A – B, then, is the same thing as A + (–B). For instance, let’s take the two **vectors** A and B: To **subtract** B from A, take a **vector** of the same magnitude as B, but pointing in the opposite direction, and add that **vector** to A, using either the tip-to-tail method or the parallelogram method.

**Questions on above discussion**

**1 ) Find the change in velocity of a yacht if it changes its velocity from 5 m/sec due north to 3 m/sec due west .**

**Solution : **

let **P** = 5 m/sec due north

**Q** = 3 m/sec

angle between P and Q = 90 degree

required change in velocity = **√ **( P^{2} + Q^{2} ) = **√**( 5^{2} + 3^{2} ) = 5.8 m

**2 ) Find the change in velocity of a tennis ball if it approaches a tennis racket at 30 m/sec and leaves the racket in the opposite direction at 40 m/sec.**

**Solution : **

Let **P **= 30 m/sec

**Q** = 40 m/ sec

change in velocity = P – Q = 30 – 40 = -10 m

here negative sign indicates that the direction of change in velocity is opposite to the initial velocity ( **P** )

**3 ) Show that, for any two vectors A and B, the vector (B-A) is equal to -(A-B).**

**Solution : **

Let **A** = A_{1} **î **+ A_{2} **ĵ** + A_{3}**k̂ **_{}

and **B** = B_{1} **î**+ B_{2} **ĵ**+ B_{3}**k̂**

( **B** – **A** ) = B_{1} **î**+ B_{2} **ĵ**+ B_{3}** k̂ **– A

_{1}

**î –**A

_{2}

**ĵ**– A

_{3}

**k̂**( **B** – **A** ) = – ( A_{1} **î **+ A_{2} **ĵ** + A_{3}** k̂ **– B

_{1}

**î –**B

_{2}

**ĵ –**B

_{3}

**) = – (**

**k̂****A**–

**B**)

( **B** – **A** ) = – ( **A** – **B** )

**( Proved )**

**4 ) Two people are pushing a disabled car. One exerts a force of 200 N east, the other a force of 150 N east. What is the net force exerted on the car? (Assume friction to be negligible.)**

**Solution : **

one force = 200 N due east

other force = 150 N due east

angle between these forces is zero that is these forces are in the same direction. So, the net force is given by

net force = one force + other force

= 200 + 150 = 350 N in the previous direction

**5 ) An airplane heads due north at 100 m/s through a 30 m/s cross wind blowing from the east to the west. Determine the resultant velocity of the airplane (relative to due north).**

**6 ) A mountain climbing expedition establishes a base camp and two intermediate camps, A and B. Camp A is 11,200 m east of and 3,200 m above base camp. Camp B is 8,400 m east of and 1,700 m higher than Camp A. Determine the displacement between base camp and Camp B.**

**7 ) Find the difference of vectors a = {1; 2} and b = {4; 8}.**

**8 ) Find the difference of vectors a = {1; 2; 5} and b = {4; 8; 1}.**

**Multiplication of vectors:**

There are two ways of multiplying two vectors

- Scalar product or dot product
- Vector product or cross product

**Scalar product OR Dot product : **

Algebraically , the scalar product of two vectors is the product of the magnitudes of each vector and the cosine of the angle between the two vectors.

**a.b **= abcosθ

a = magnitude of **a**

b = magnitude of **b**

(here bold letter represents vector while unbold letter represent vector’s magnitude )

θ = angle between the vectors **a **and **b**

Properties of scalar products :

**Commutative**: …**u · v**=**v · u**- Distributive over vector
**addition**: - Bilinear:
- Scalar multiplication:
- Not
**associative**because the dot product between a scalar (a ⋅ b) and a vector (c) is not defined, which means that the expressions involved in the**associative**property, (a ⋅ b) ⋅ c or a ⋅ (b ⋅ c) are both ill-defined. … **u · v**= |**u**||**v**| cos θ**u · v**=**v · u****u · v**= 0 when**u**and**v**are orthogonal.**0 · 0**= 0- |
**v**|2 =**v · v** - a (
**u·v**) = (a**u**)**·****v** - (a
**u**+ b**v**)**·****w**= (a**u**)**·****w**+ (b**v**)**·****w**

**There are several applications of scalar product. Most important ones are discussed below:**

- The scalar product can be used in order determine if two vectors are perpendicular. If their scalar product is zero, then they would be perpendicular.
- Similarly, scalar product is used to find if two vectors are parallel or not. When the scalar product comes out to be equal to the product of their magnitudes, both are parallel.
- The scalar product can be applied to calculate the angle between two vectors if scalar product is given.
- This product is also used to determine the type of angle between two vectors. If the scalar product of two vectors is greater than zero, then the angle between them is acute. And if the scalar product is less than zero, then the angle is obtuse.
- Many physical quantities are defined using the scalar product of two vectors, such as magnetic flux is said to be the scalar product of area vectors and magnetic field, and work done is equal to the scalar product of force and displacement vectors.

**In rectangular component form the scalar product is as given below :**

**A = **A_{1}**i**+A_{2}**j**+A_{3}**k**

**B **= B_{1}**i**+B_{2}**j**+B_{3}**k**

**A.B **= (A_{1}**i**+A_{2}**j**+A_{3}**k**)**.**(B_{1}**i**+B_{2}**j**+B_{3}**k**)

=A_{1}B_{1}(**i.i**)+A_{1}B** _{2}**(

**i.j**)+A

_{1}B

_{3}(

**i.k**)+A

_{2}B

_{1}(

**j.i**)+A

_{2}B

_{2}(

**j.j**)+A

_{2}B

_{3}(

**j.k**)+A

_{3}B

_{1}(

**k.i**)+A

_{3}B

_{2}(

**k.j**)+A

_{3}B

_{3}(

**k.k**)

=A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3}

[ **i.i**=1, **i.j**=**j.i**=**i.k**=**k.i**=**k.j**=**j.k**=0 ]

**Vector product OR cross product :**

The **cross product**, also called the **vector product**, is an operation on two vectors. The **cross product** of two vectors produces a third **vector** which is perpendicular to the plane in which the first two lie. That is, for the **cross** of two vectors, A and B, we place A and B so that their tails are at a common point.

Geometrically, the cross product of two vectors is the area of the parallelogram between them.

The symbol used to represent this operation is a large diagonal cross (×), which is where the name “cross product” comes from. Since this product has magnitude and direction, it is also known as the vector product.

**A** × **B** = *AB* sin θ **n̂**

The vector n̂ (“n hat”) is a unit vector perpendicular to the plane formed by the two vectors. The direction of n̂ is determined by the right hand rule, which will be discussed shortly.

**Properties of cross product :**

1 ) The cross product is distributive…

**A** × (**B** + **C**) = (**A** × **B**) + (**A** × **C**)

but not commutative…

**A** × **B** = −**B** × **A**

Reversing the order of cross multiplication reverses the direction of the product.

2 ) Since two identical vectors produce a degenerate parallelogram with no area, the cross product of any vector with itself is zero…**A** × **A** = 0

Applying this corollary to the unit vectors means that the cross product of any unit vector with itself is zero.

**î** × **î** = **ĵ** × **ĵ** = **k̂** × **k̂** = (1)(1)(sin 0°) = 0

3 ) It should be apparent that the cross product of any unit vector with any other will have a magnitude of one. (The sine of 90° is one, after all.)

4 )** Cross product of two** non-zero **vectors** a and b is equal to zero if and only if the **vectors** are collinear.

**Direction of cross product :**

The **right hand rule** for cross multiplication relates the direction of the two vectors with the direction of their product. Since cross multiplication is not commutative, the order of operations is important.

- Hold your right hand flat with your thumb perpendicular to your fingers. Do not bend your thumb at anytime.
- Point your fingers in the direction of the first vector.
- Orient your palm so that when you fold your fingers they point in the direction of the second vector.
- Your thumb is now pointing in the direction of the cross product.

A right-handed coordinate system, which is the usual coordinate system used in physics and mathematics, is one in which any cyclic product of the three coordinate axes is positive and any anti cyclic product is negative. Imagine a clock with the three letters x-y-z on it instead of the usual twelve numbers. Any product of these three letters that runs around the clock in the same direction as the sequence x-y-z is cyclic and positive. Any product that runs in the opposite direction is anti cyclic and negative.

**The cross product of a cyclic pair**

**of unit vectors is positive.**

**The cross product of an anti cyclic pair**

**of unit vectors is negative.**

Using this knowledge we can derive a formula for the cross product of any two vectors in rectangular form. The resulting product looks like it’s going to be a terrible mess, and it is!

**A × B = (Ax î + Ay ĵ + Az k̂) × (Bx î + By ĵ + Bz k̂)**

The product of two tri – nomials has nine terms.

**A ×**

**B = (**

**Ax î × Bx î ) + ( Ax î × By ĵ ) + ( Ax î × Bz k̂ )****+**

**(**

**Ay ĵ × Bx î ) + (****Ay ĵ × By ĵ ) + (****Ay ĵ × Bz k̂ )****+ (**** Az k̂ × Bx î ) + ( Az k̂ × By ĵ ) + ( Az k̂ × Bz k̂ )**

As we know that the cross product of two perpendiculars vectors is unity.

**ĵ × î = – k̂ , î × ĵ = k̂, k̂ × î = ĵ , î × k̂ = – ĵ, k̂ × ĵ = -î, ĵ × k̂ = î**

Therefore, three terms will be cancelled and the cross product of two same unit vectors is zero i.e

**ĵ × ĵ = 0, î × î = 0, k̂ × k̂ = 0**

**A × B = Ax By k̂ – Ax Bz ĵ – Ay Bx k̂ + Ay Bz î + Az Bx ĵ – Az By î**

** = ( Ay Bz – Az By ) î + (Az Bx – Ax Bz ) ĵ + ( AxBy _ Ay Bx ) k̂**

This expression is also achieved by expanding of determinant of the components of the above two vectors.

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