This post covers *Vectors class 11 Physics revision notes* – chapter 4 with concepts, formulas, applications, numerical & Questions. These revision notes are good for CBSE, ISC, UPSC, and other exams. This covers the grade 12 Vector Physics syllabus of some international boards as well. Here we have covered Vector fundamentals & types, Laws for vector addition, dot and cross products, & relative velocity of rain with respect to a moving man (with numerical).

There are physical quantities that require both magnitude and direction for their complete description. A simple example of a vector is velocity. The statement that the velocity of a train is 100 Km/hour does not make much sense unless we also tell the direction in which the train is moving.

Force is another such quantity. We must specify not only the magnitude of the force but also the direction in which the force is applied. Such quantities are called vectors. A vector quantity has both magnitude and direction. Some examples of vector quantities in mechanics are displacement, acceleration, momentum, angular momentum, and torque, etc.

A vector is represented by a line with an arrow indicating its direction. Take vector AB in Fig. 1.

The length of the line represents its magnitude on some scale. The arrow indicates its direction.

[ All our posts on Vector are clubbed here: **Vector Physics tutorials**. Read these for a better understanding ]

- Classification of Vectors (with Diagram)
- Addition of Vectors:
- Numerical problems – solve using Parallelogram Law & Triangle Law
- Multiplication of vectors – dot and cross, formulas, rules, numerical
- Numerical Problems on Dot Product and Cross Product
- Relative velocity of Rain with respect to a Moving Man – Vector physics numericals | Applications of Vector Subtraction

**Classification of Vectors (with Diagram)**

**Equal Vector**

Two vectors are said to be equal if their magnitudes are equal and they point in the same direction.

Three vectors **A, B,** and C showed in the Figure are **equal**.

We say **A = B = C**.

But **D** is not equal to **A**.

**Negative of a Vector**

A vector (here D in the figure above) which has the same magnitude as A but has opposite direction, is called negative of **A**, or –**A**. Thus, **D** = –**A**

**Null Vector or Zero Vector**

A vector is said to be a zero or null vector if the magnitude of the vector is zero, i.e., the starting point and the endpoint of the Vector are the same.

**Unit Vector**

A vector is said to be a Unit vector if its magnitude is 1 unit and it has a specified direction. A unit vector is a dimensionless vector having a magnitude of exactly 1. Unit vectors are used to specify a given direction and have no other physical significance. They are used solely as a convenience in describing a direction in space.

It has neither units nor dimensions.

As an example, we can write vector **A** as A **n** where a cap(^) on **n** denotes a unit vector in the direction of **A**. A unit vector has been introduced to take care of the direction of the vector; the magnitude has been taken care of by A.

The unit vectors along coordinate axes are of particular importance. We shall use the symbols **i, j, k** to represent unit vectors pointing in the positive *x*, *y*, and *z *directions, respectively. The “^ hats” on the symbols are a standard notation for unit vectors, but many times just a bold font(like **i**) is popularly used without using the hat.

The unit vectors **i, j, k** form a set of mutually perpendicular vectors in a right-handed coordinate system, as shown in Figure. The magnitude of each unit vector equals 1;

that is, | **i** | = | **j** | = | **k** | = 1.

Unit vector along the x-axis is denoted by** i**, along the y-axis by** j**, and along the z-axis by **k**.

Using this notation, vector **A**, whose components along x and y axes are respectively A_{x} and A_{y}, can be written as A = A_{x} **i **+ A _{y} ** j**

**Addition of Vectors:**

Vectors of the same kind only can be added. For example, two forces can be added or two velocities can be added. But a force and a velocity cannot be added. Here we will discuss the **Triangle Law** and the **Parallelogram Law** to add vectors.

**Triangle Law**

**If two vectors are represented in magnitude and direction by the two sides of a triangle taken in order, the resultant is represented by the third side of the triangle taken in the opposite order. This is called the triangle law of vectors.**

The sum of two or more vectors is called the resultant vector. In Fig. above, pr is the resultant of A and B

**Parallelogram Law ( formulas & Angle)**

**If a parallelogram can be drawn so that two vectors can be placed with their tails connected as the two adjacent sides of the parallelogram with an angle θ between them, then the diagonal of the parallelogram represents their resultant vector or the vector sum.**

Let **A** and **B** be the two vectors and let θ be the angle between them as shown in Fig. above. To calculate the vector sum, we complete the parallelogram. Here side PQ represents vector **A**, side PS represents **B** and the diagonal PR represents the resultant vector **R. **Here,α angle is the angle the resultant makes with base vector and the angle denotes the direction of the resultant or the vector sum.

**Numerical** problems – solve using **Parallelogram Law** & Triangle Law

*1) *

*Vector A has a magnitude of 30 and it lies in such a way that it makes an angle of 30 degrees with another vector B of magnitude 40. What is the vector sum or resultant of A and B?*

Solution: Let, **R** be the vector sum of **A** and **B**.

|R| = √[30^{2} + 40^{2} + 2.30.40 cos 30] = **67.66**

If R makes an angle α with A then tanα = (B sinθ)/(A + Bcosθ)

here, A = 30, B = 40, θ = 30.

tanα = 40 sin 30 / (30 + 40 cos 30) = 20/(30 + 34.64) = .31

α = arctan (.31) = **17.22 degree**

Answer:Resultant is **67.66 at 17.22 degree from A**

**2)**

**A and B are two forces where |A| = 3 N and |B| = 4 N.**

**Can you draw the vector sum of these two forces using the triangle law?Also, find out its magnitude and direction using some suitable formula.**

**Solution:** Using Triangle law, vectors A and B are drawn in the **tail to tip** way to draw 2 sides of the triangle. R, the 3^{rd} side of the triangle (in red) shows the vector sum of A and B.

The magnitude of the vector sum can be calculated using the Pythagoras theorem in this case and B is perpendicular to A. So the magnitude of the vector sum: |R| = √(3^{2} + 4^{2}) = 5 N

If it makes an angle α with A, then tanα = 4/3

α = arctan (4/3) = 53.13

So, the vector sum is 5 N at 53.13 degrees from A

**The next section covers the Multiplication of vectors – dot and cross, formulas, rules, and numerical problems.**

## Multiplication of vectors – dot and cross, formulas, rules, numerical

### Dot product

**Scalar multiplication of two vectors yields a scalar product. Scalar multiplication is also known as the dot product.**

**Formula of Dot Product**

**The scalar product or dot product of 2 vectors A and B is expressed by the following equation:A.B = AB cos φ, where φ is the angle between the vectors, A is the magnitude of vector A and B is the magnitude of vector B.The scalar product is also called the dot product because of the dot notation that indicates it.**

**Rules** of Dot product or scalar product

1) Dot product is a scalar, it is commutative: **A.B **= **B.A **= ABcosθ.

2) It is also distributive: **A.(B + C) = A.B + A.C**.

3)

**i **.** j = j **.** i = 0j **.

**k =k**.

**j =0**

**k**.

**i =i . k =0**

**i **.** i = 1j **.

**j = 1**

**k**.

**k = 1**

4)

If, **A** = (A_{x} **i** + A_{y} **j** + A_{z} **k**) and **B** = (B_{x} **i** + B_{y} **j** + B_{z} **k**)

Then find **A.B**

**A** . **B** = (A_{x} **i** + A_{y} **j** + A_{z} **k**) . (B_{x} **i** + B_{y} **j** + B_{z} **k**)

= (A_{x} **i . **B_{x} **i** + A_{x} **i .** B_{y} **j** + A_{x} **i .** B_{z} **k**)**+ **(A

_{y}

**j .**B

_{x}

**i**+ A

_{y}

**j .**B

_{y}

**j**+ A

_{y}

**j .**B

_{z}

**k**)

**+**(A

_{z}

**k .**B

_{x}

**i**+ A

_{z}

**k .**B

_{y}

**j**+ A

_{z}

**k .**B

_{z}

**k**)

= (A_{x }B_{x} **i .** **i** + A_{x} B_{y} **i .** **j** + A_{x} B_{z }**i .** **k**)**+ **(A

_{y}B

_{x}

**j .**

**i**+ A

_{y}B

_{y }

**j .**

**j**+ A

_{y}B

_{z }

**j .**

**k**)

**+**(A

_{z}B

_{x }

**k .**

**i**+ A

_{z}B

_{y }

**k .**

**j**+ A

_{z}B

_{z }

**k .**

**k**)

**=**A

_{x }B

_{x }+ A

_{y}B

_{y }+ A

_{z}B

_{z}

**Vector Product**

We define the vector product to be a vector quantity with a direction perpendicular to this plane (that is, perpendicular to both **A **and **B**) and a magnitude equal to *AB *sinφ.

**Formula** of Vector Product or Cross Product

if **C **=**AxB**, then *C*= *AB sinφ *……………….. (equation 1)

[ Here, A and B are magnitudes of vectors **A **and **B **respectively.*C *is the magnitude of the vector **C**.

And, **C**= (cross) product of **A **and **B** ]

The direction of the product vector C =A × B is given by the right-hand rule. If the right hand is held so that the curling fingers point from A to B through the smaller angle between the two, then the thumb stretched at right angles to fingers will point in the direction of C.

**Rules** of Vector Product or Cross Product

1) Direction of vector **BxA**is opposite to that of the vector **AxB**. This means that **the vector product is not commutative**.

2)**i**x** j = kj **x

**k =i**

**k**x

**i =j**

**j **x** i = – k****k **x** j = – i****i **x** k = – j**

**i **x** i = 0j **x

**j = 0**

**k**x

**k = 0**

3) If, **A** = (A_{x} **i** + A_{y} **j** + A_{z} **k**) and **B** = (B_{x} **i** + B_{y} **j** + B_{z} **k**)

Then find **A X B**

**A** X **B** = (A_{x} **i** + A_{y} **j** + A_{z} **k**) x (B_{x} **i** + B_{y} **j** + B_{z} **k**)

= (A_{x} **i **xB_{x} **i** + A_{x} **i **x B_{y} **j** + A_{x} **i **x B_{z} **k**)**+ **(A

_{y}

**j**x B

_{x}

**i**+ A

_{y}

**j**x B

_{y}

**j**+ A

_{y}

**j**x B

_{z}

**k**)

**+**(A

_{z}

**k**x B

_{x}

**i**+ A

_{z}

**k**x B

_{y}

**j**+ A

_{z}

**k**x B

_{z}

**k**)

= (A_{x }B_{x} **i **x **i** + A_{x} B_{y} **i **x **j** + A_{x} B_{z }**i **x **k**)**+ **(A

_{y}B

_{x}

**j**x

**i**+ A

_{y}B

_{y }

**j**x

**j**+ A

_{y}B

_{z }

**j**x

**k**)

**+**(A

_{z}B

_{x }

**k**x

**i**+ A

_{z}B

_{y }

**k**x

**j**+ A

_{z}B

_{z }

**k**x

**k**)

= 0 + A

_{x}B

_{y}

**k**+ A

_{x}B

_{z}(-

**j**) + A

_{y}B

_{x}(-

**k**) + 0 + A

_{y}B

_{z}

**i**+ A

_{z}B

_{x}

**j**+ A

_{z}B

_{y}(-

**i**) + 0

= (A

_{y}B

_{z}– A

_{z}B

_{y})

**i**+ (A

_{z}B

_{x}– A

_{x}B

_{z})

**j**+ (A

_{x}B

_{y }– A

_{y}B

_{x})

**k**

**Numerical** Problems on Dot Product and Cross Product

**1) **Find out the dot product of vector **A **and** B **where** A**** = **4**i ****+ **5 **j ****+ 2k****, ****B**** = **6**i ****– **4 **j ****+ 3k****.**

Dot product D = A_{x }B_{x }+ A_{y} B_{y} + A_{z} B_{z} = 4.6 + 5(-4) + 2.3 = 24 – 20 + 6 = 10

2) Find out the cross product of vector **A **and** B **where** A**** = **4**i ****+ **5 **j ****+ 2k****, ****B**** = **6**i ****– **4 **j ****+ 3k****.**

Cross Product C = (A_{y} B_{z} – A_{z} B_{y} ) **i** + (A_{z} B_{x} – A_{x} B_{z}) **j** + (A_{x} B_{y }– A_{y} B_{x}) **k**

=(5.3 – 2(-4)) i + (2.6 – 4.3) j + (4(-4)-5.6) k

= 23 i + 0j -46k = 23i – 46k

**In the next section, **we have discussed and solved numerical problems related to the **Relative velocity of Rain with respect to a Moving Man** with the help of **Vector subtraction**.

## Relative velocity of **Rain with respect to a Moving Man – Vector physics numericals** | Applications of Vector Subtraction

### Rain drop and moving man – relative velocity (Problem number 1)

Rain is falling vertically with a speed of 35 m/s. A man rides a bicycle with a speed of 12 m/s in the east to west direction. What is the direction in which he should hold his umbrella?

Solution:

Velocity of rain = V_{r} = 35 m/s

Velocity of man = V_{m} = 12 m/s east to west

Relative velocity of rain with respect to the man =V = V_{r} – V_{m}

=> V = V_{r} + (– V_{m})

In the diagram, we have reversed the direction of the man to draw – V_{m}

Now, Let this relative velocity V makes an angle θ with the vertical.

tan θ = V_{m} / V_{r} = 12/35

θ = arctan (12/35) = 18.9 degrees with the vertical towards the west.

The man has to hold his umbrella at 18.9 degrees with the vertical.

**Rain drop and moving man – relative velocity (Problem number 2)**

Rain is falling vertically with a speed of *x* m/s. A man rides a bicycle with a speed of 12 m/s in the east to west direction. What is the value of x if the direction in which he holds his umbrella is 21 degrees with the vertical?

Solution:

Velocity of rain = V_{r} = x m/s

Velocity of man = V_{m} = 12 m/s east to west

Relative velocity of rain with respect to the man =V = V_{r} – V_{m}

=> V=_{} x + (– V_{m})

In the diagram we have reversed the direction of the man to draw – V_{m}

If this relative velocityV makes an angle θ = 21 degree with the vertical.

tan 21 = V_{m} / V_{r} = 12/x

=> 0.38 =12/x

x=12/0.38=31.57 m/s

Rain is falling vertically with a speed of *31.57* m/s

**Rain drop and moving man – relative velocity (Problem number 3)**

Rain is falling vertically with a speed of *40* m/s. A man rides a bicycle with a speed of 12 m/s in the east to west direction. What will be the magnitude of the relative velocity of rain with respect to the man? (i.e. What would be the apparent velocity of rain to the cyclist?)

Solution:

Velocity of rain = V_{r} = 40 m/s

Velocity of man = V_{m} = 12 m/s east to west

Relative velocity of rain with respect to the man =V = V_{r} – V_{m}

=> V=_{} V_{r} + (– V_{m})

In the diagram we have reversed the direction of the man to draw – V_{m}

Now V = √( V_{r}^{2} + V_{m}^{2}) = √( 40^{2} + 12^{2}) = 41.76 m/s

Hence, the apparent velocity of rain to the cyclist=41.76 m/s

Note: All our posts on Vector are clubbed here: **Vector Physics tutorials**. Read these for a better understanding