Informative line

### Preview Of Vectors

Learn examples of scalar and vector quantities as a physics quantities, concept of unit and displacement vector. Practice analytical method of vector addition using triangle law.

# Physical Quantities

• In physics we deal with those quantities which can be measured by comparing with some standard of same quantity. Such quantities are called physical quantities.
• All physical quantities can be classified into two categories:-

1. Scalars

2. Vectors

### 1. Scalars

• Those quantities that are fully described by a magnitude alone are called scalars. For scalars, the complete information is obtained by asking 'how much'.

For example:-

(3) How many apples are there in the bag?

### 2. Vectors

• Those quantities that are fully described by both a magnitude and direction are called vectors. For vectors the complete information is obtained by asking 'how much' and 'in which direction'. e.g. $$\vec {OA}$$  is a vector as shown in figure.

Examples:-

(1) For complete specification of a force, we need to know the magnitude as well as the direction of the force.

(2) For complete description of a point in space, both its distance and its direction from a fixed point (reference point) are needed.

Note:-

Assigning a direction to a scalar quantity will not convert it into a vector quantity.

e.g. Current

### Notation of vectors

• Assume that we have two quantities, A and B, where A is a vector quantity and B is a scalar quantity.
• To avoid the repetition of stating that A is a vector and B is a scalar, we use alphabet with arrow for quantity A i.e. $$\vec A$$ and simply an alphabet for quantity B i.e. $$B$$ .

#### Which of the following is not a vector quantity?

A 20 m towards east

B 10 m towards right

C Position of particle from origin

×

A vector quantity is one in which both magnitude and direction are specified.

Height of something can be completely specified by magnitude only, so it is not a vector quantity.

$$\therefore$$ Option D is correct.

### Which of the following is not a vector quantity?

A

20 m towards east

.

B

10 m towards right

C

Position of particle from origin

D

Option D is Correct

# Concept of Unit Vector

• A vector quantity needs two parameters for complete specification.

(1) Magnitude

(2) Direction

• In Physics, direction is denoted by a unit vector.

### Note:-

1. Unit vector is dimensionless and has magnitude '1'.
2. Unit vector is used to specify only direction and has no other physical significance.

e.g. :  Let A is a vector quantity

$$\underbrace{20\,m}_{magnitude}\,\,\,\,\, \underbrace{\text{towards east}}_{Unit \,\,vector}$$

Vector = magnitude × unit vector

### Representation

• Vectors are represented as $$\vec A ,\, \vec B, \,\vec C$$.
• Unit vectors are represented as  $$\hat A ,\, \hat B, \,\hat C$$.

$$\hat A$$ means towards $$\vec A$$

$$\hat B$$ means towards $$\vec B$$

• Magnitude of  $$\vec A$$ is represented as A or $$|\vec A|$$

Hence,

$$\underbrace{\vec A}_{Vector} =\,\,\,\underbrace{|\vec A|}_{Magnitude}\,\, \underbrace{\hat A}_{direction} \,\,\text{or}\,\, A.\hat A$$

### Some standard directions

$$\text {towards (+) }x- axis \,\,\,\,\hat i$$

$$\text {towards (+) }y- axis \,\,\,\,\hat j$$

$$\text {towards (+) }z- axis \,\,\,\,\hat k$$

#### A man runs from east to west with a speed of 10 m/s. Find the direction (unit vector) of velocity?

A East to West

B West to East

C North to South

D South to North

×

Direction of velocity (unit vector) is from east to west.

### A man runs from east to west with a speed of 10 m/s. Find the direction (unit vector) of velocity?

A

East to West

.

B

West to East

C

North to South

D

South to North

Option A is Correct

# Displacement Vector

• Displacement vector is known as the change in position vector of a particle.
• Suppose a particle, at point P, moves to point Q at some instant of time.
• The position vectors of P and Q are $$\vec r_1 \,\,\,\text {and}\,\,\,\vec r_2$$ respectively with respect to the origin 'O'.
• The displacement vector $$\vec{PQ}$$ represents the position of Q with respect to P as shown in figure.

$$\vec{PQ} = \vec {\Delta r} = \text {Displacement vector}$$

• Displacement vector is independent of choice of origin.

#### Choose the correct representation of displacement vector.

A

B

C

D

×

The position vectors of P and Q with respect to the origin 'O' are $$\vec r _1 \,\,\,\text {and }\,\,\vec r_2$$  respectively.

The displacement vector $$\vec{PQ}$$ represents the position of Q  with respect to P as shown in option B.

The representation of displacement vector $$\vec{PQ},$$ starts from point P and ends at point Q.

Hence, option (B) is correct.

### Choose the correct representation of displacement vector.

A
B
C
D

Option B is Correct

#### Consider a car travels 2 km westward and then 3 km southward. Calculate total displacement.

A $$\sqrt{13}\;km,\, tan^{-1}(1.5)$$ south of west

B $$5\,km,\, tan^{-1}(1.5)$$ south of west

C $$2\,km,\, tan^{-1}(2)$$ south of west

D $$6\,km,\, tan^{-1}(3)$$ south of west

×

Using Pythagoras theorem

$$d= \sqrt{(2)^2+ (3)^2}$$

$$d= \sqrt{13} \,km$$

$$tan\,\theta = \dfrac{3}{2}$$

$$\theta = tan^{-1} (1.5)$$

The horizontal displacement of $$\sqrt{13}\;km$$ from south of west by an angle $$\theta = tan ^{-1} (1.5)$$

Hence, option (A) is correct.

### Consider a car travels 2 km westward and then 3 km southward. Calculate total displacement.

A

$$\sqrt{13}\;km,\, tan^{-1}(1.5)$$ south of west

.

B

$$5\,km,\, tan^{-1}(1.5)$$ south of west

C

$$2\,km,\, tan^{-1}(2)$$ south of west

D

$$6\,km,\, tan^{-1}(3)$$ south of west

Option A is Correct

# Coordinates in Unit Vector Form

• Let us assume, a man starts walking from origin. He moves 4 m towards $$x - \text {axis}$$  and then moves 3 m towards $$y - \text {axis}$$. Vectorially, this process can be written as: -

Displacement of man $$= (4\,\hat i+3 \, \hat j) \,m$$

Here (+) sign doesn't mean addition; rather it represents continuity. We can interpret it as: -

$$4\,m \,\hat i \text{(towards }x- axis) + 3\,m\,\hat j \,\,(\text {towards } y- axis)$$

Graphically,

We can say that $$\vec{OA} = (4\,\hat i) \,m$$

$$\vec{AB} = (3\,\hat j) \,m$$

Hence, displacement = $$\vec{OA} +\vec{AB}= (4\,\hat i+3\,\hat j) \,m$$

#### A fly flies 3 m right, then 4 m up. Again it flies 5 m left and 3 m down. Find the displacement of the fly?

A $$(\hat i - \hat j)\,m$$

B $$(7\,\hat i - 2\,\hat j)\,m$$

C $$(2\,\hat i + \,\hat j)\,m$$

D $$(-2\,\hat i +\,\hat j)\,m$$

×

Given:   $$3\,m \,\,\,\,\,\text {right} = (3\,\hat i) \,m$$

$$4\,m \,\,\,\,\,\text {up} = (4\,\hat j) \,m$$

$$5\,m \,\,\,\,\,\text {left} = 5\,(\,-\hat i) \,m$$

$$3\,m \,\,\,\,\,\text {down} = 3\,(-\hat j) \,m$$

Displacement  $$=3\,\hat i +4\,\hat j + 5 (-\hat i) + 3 (-\hat j)$$

$$= (-2\,\hat i+\hat j )\,m$$

$$\therefore$$ Option (D) is correct.

### A fly flies 3 m right, then 4 m up. Again it flies 5 m left and 3 m down. Find the displacement of the fly?

A

$$(\hat i - \hat j)\,m$$

.

B

$$(7\,\hat i - 2\,\hat j)\,m$$

C

$$(2\,\hat i + \,\hat j)\,m$$

D

$$(-2\,\hat i +\,\hat j)\,m$$

Option D is Correct

# Triangle Law of Vector Addition

• Robin and Sara start from same point A (Park) to reach B (Shopping Mall).
• Robin reaches directly to B (Shopping Mall) but Sara goes to C (Grocery Store) before reaching to B (Shopping mall).
• Both of them start from same point and reach the same point, but through different paths. However, result is same for both.

$$\text{ Result of Robin = Result of Sara}\\ \\\\\,\,\text{A to B = A to C then C to B} \\ \vec{AB} = \vec {AC} + \vec {CB}$$

• Here, resultant means vector joining initial and final positions (independent of path).
• A to C  then C to B  is equivalent to A to B

$$\vec{AC} + \vec {CB} = \vec {AB}$$

Here, B is turning point.

Similarly,

$$\vec {AC} + \vec {CD} = \vec {AD}$$

This is known as triangle law of vector addition.

### To determine which side is the resultant of other two sides

Step 1 :   Observe the arrows carefully.

Step 2 :  Check which two sides are continuous or in same order.

Step 3 : Select the side which is in opposite order.

Step 4 :  Selected side is the resultant of other two sides taken in same order.

Note:

• Vector showing the path from B to A is $$\vec{BA}\,\,\&\,\, not \,\,\vec{AB}$$.
• $$\vec{AB}$$ represents position vector of B w.r.t  A.

Using triangle rule, we can conclude that: -

$$\vec{AB} =\vec{OB} -\vec{OA}$$

$$\vec{AB}=$$ Position vector of B – Position vector of A

#### For the figure shown, write one vector as a sum of other two.

A $$\vec {AB} =\vec {BC} +\vec {BA}$$

B $$\vec {AC} =\vec {AB} +\vec {BC}$$

C $$\vec {BA} =\vec {BC} +\vec {AC}$$

D $$\vec {BC} =\vec {BA} +\vec {AC}$$

×

$$\vec {BA}\,\,\text{and }\vec {AC}$$ are continuous.

$$\vec {BC}$$ is in opposite order.

$$\vec{BA} + \vec{AC} = \vec {BC}$$

### For the figure shown, write one vector as a sum of other two.

A

$$\vec {AB} =\vec {BC} +\vec {BA}$$

.

B

$$\vec {AC} =\vec {AB} +\vec {BC}$$

C

$$\vec {BA} =\vec {BC} +\vec {AC}$$

D

$$\vec {BC} =\vec {BA} +\vec {AC}$$

Option D is Correct

# Calculation of Displacement Vector

• Displacement vector is known as the change in position vector of a particle.
• Suppose a particle, at point P, moves to point Q at some instant of time.
• The position vectors of P and Q are $$\vec r_1 \,\,\,\text {and}\,\,\,\vec r_2$$ respectively, with respect to the origin 'O'.
• The displacement vector $$\vec{PQ}$$ represents the position of Q with respect to P as shown in figure.

$$\vec{PQ} = \vec {\Delta r} = \text {Displacement vector}$$

• Displacement vector is independent of choice of origin.
•           $$\vec{PQ} = \vec{OQ}-\vec{OP}$$

Here,    $$\Delta \vec r + \vec r_P = \vec r_Q$$

where $$\vec r_P \,\,\,and \,\,\, \vec r_Q$$ are two vectors

$$\Delta r \to$$ displacement vector

$$\vec {PQ} = \vec r _Q - \vec r _P$$

$$\vec {PQ} =$$ Position vector of  Q – Position vector of P

#### Calculate the displacement vector $$\vec{PQ}$$.

A $$4\,\hat i - 3\,\hat j$$

B $$5\,\hat i - 6\,\hat j$$

C $$\,\hat i+ 3\,\hat j$$

D $$4\,\hat i - 4\,\hat j$$

×

Let $$\vec r _1 =$$ Position vector of point P $$= 4\,\hat i + 3\,\hat j$$

$$\vec r _2 =$$ Position vector of point $$Q= 5\,\hat i + 6\,\hat j$$

Displacement vector

$$\vec{PQ}= \vec r_2 - \vec r_1$$

$$= (5-4) \hat i + (6-3)\hat j$$

$$= \hat i+3\,\hat j$$

Hence, option (C) is correct.

### Calculate the displacement vector $$\vec{PQ}$$.

A

$$4\,\hat i - 3\,\hat j$$

.

B

$$5\,\hat i - 6\,\hat j$$

C

$$\,\hat i+ 3\,\hat j$$

D

$$4\,\hat i - 4\,\hat j$$

Option C is Correct

# Analytical Method of Vector Addition Using Triangle Law

• Let's consider three points in space i.e. O, A and B.

• From the figure, if we move from point O  to A and then point A to B, which is equivalent to move from point O to B.
• Let OA represented by $$\vec P$$ , AB represented by $$\vec Q$$ and OB represented  by $$\vec R$$, then we can say that  $$\vec R$$  is the resultant vector of both, $$\vec P \,\,\,\text {and}\,\,\,\vec Q$$.

• The angle between $$\vec P \,\,\,\text {and}\,\,\,\vec Q$$  is $$\theta$$, as shown in figure.

• Arrange this triangle in such a manner so that $$\vec P$$ represents OA side of $$\Delta OAB$$$$\vec Q$$ represents OB side of  $$\Delta OAB$$ as shown in figure.

$$x$$ component of AB = Q $$cos\,\theta$$

$$y$$ component of AB = Q $$sin\,\theta$$

where

$$R \,cos \,\alpha = P+Q \,cos \,\theta$$

$$R \,sin \,\alpha = Q \,sin \,\theta$$

Calculating  resultant vector |R|

$$|R| = \sqrt{(R\,cos\,\alpha)^2 +(R\,sin\,\alpha)^2 }$$

$$= \sqrt{(P+Q\,cos\,\theta)^2 + (Q\,sin \,\theta)^2}$$

calculating angle $$\alpha$$ which is form in between  $$\vec P \,\,\,\text {and}\,\,\,\vec R$$

$$tan\,\alpha =\dfrac{Q\,sin\,\theta}{P+Q\,cos\,\theta}$$

$$\alpha = tan^{-1} \left(\dfrac{Q\,sin\,\theta}{P+Q\,cos\,\theta}\right)$$

#### An airplane is flying 30 km/hr in horizontal direction and flew 40 km/hr in vertical direction. What is the resultant velocity of plane?

A 30 km /hr, $$tan^{-1} \,2/3$$ with the horizontal

B 40 km /hr, $$tan^{-1} \,5/3$$ with the horizontal

C 20 km /hr, $$tan^{-1} \,1/3$$ with the vertical

D 50 km /hr, $$tan^{-1} \,4/3$$ with the horizontal

×

Let R be the resultant velocity and $$\alpha$$ is the angle of resultant with the horizontal.

$$|R| = \sqrt{(30)^2 + (40)^2}$$

$$|R| = \sqrt{900+1600}$$

$$|R| = 50 \,km/hr$$

$$tan \,\alpha = \dfrac{40}{30}$$

$$\alpha = tan^{-1} \dfrac{4}{3}$$  with the horizontal

Hence, option (D) is correct.

### An airplane is flying 30 km/hr in horizontal direction and flew 40 km/hr in vertical direction. What is the resultant velocity of plane?

A

30 km /hr, $$tan^{-1} \,2/3$$ with the horizontal

.

B

40 km /hr, $$tan^{-1} \,5/3$$ with the horizontal

C

20 km /hr, $$tan^{-1} \,1/3$$ with the vertical

D

50 km /hr, $$tan^{-1} \,4/3$$ with the horizontal

Option D is Correct