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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:-

(1) What is your height?

(2) What is your mass?

(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. \(B\) .

Illustration Questions

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 Your height

×

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

Your height

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

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

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

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

Illustration Questions

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.

image

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

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

Illustration Questions

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? 

image
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

Illustration Questions

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.

image
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

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.

Illustration Questions

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.

image

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 image
B image
C image
D image

Option B is Correct

Illustration Questions

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 

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

\(d= \sqrt{13} \,km\)

image

\(tan\,\theta = \dfrac{3}{2}\)

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

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

image

Hence, option (A) is correct. 

image

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

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 

Illustration Questions

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 \(Q= 5\,\hat i + 6\,\hat j\)

image

Displacement vector 

\(\vec{PQ}= \vec r_2 - \vec r_1\)

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

\(= \hat i+3\,\hat j\)

image

Hence, option (C) is correct.

image

Calculate the displacement vector \(\vec{PQ}\).

image
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  \(\vec P \,\,\,\text {and}\,\,\,\vec R\) 

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

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

Illustration Questions

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.

image

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

\(|R| = \sqrt{900+1600}\)

\(|R| = 50 \,km/hr\)

\(tan \,\alpha = \dfrac{40}{30}\)

\(\alpha = tan^{-1} \dfrac{4}{3}\)  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

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