A vector describes a movement from one point to another. It is a mathematical quantity having both Magnitude & Direction. The length of the segment of the directed line is called the magnitude of the vector and the angle at which the vector is inclined shows the direction of the vector.
The beginning point of a vector is called as “Tail” and the end side (having arrow) is called as “Head.”
Examples & Representation
Velocity, Acceleration, Force, Increase/Decrease in Temperature etc.
A vector between two points A and B is given as \(\overrightarrow{AB}\), or vector a.
Understanding more about Vectors
Breaking a vector into its x and y components is the most common way for solving vectors.
A vector “a” is inclined with horizontal having an angle equal to \(\theta\).
This given vector “a” can be broken down into two components i.e. a_{x }and a_{y}.
The component a_{x }is called as “Horizontal component” whose value is \(a \cos \theta\).
The component a_{y }is called as “Vertical component” whose value is \(a \sin \theta\).
Example Given vector V, having magnitude of 10 units & inclined at \(60^{\circ}\). Break down the given vector into its two component.
Solution \(\overrightarrow{v} \), having magnitude(V) = 10 units and \(\theta = 60^{\circ}\) Horizontal component (V_{x}) = \(V \cos \theta\) V_{x }= \(10\; \cos 60^{\circ}\) V_{x }= \(10 \times 0.5\) V_{x }= 5 units Now, Vertical component(V_{y}) = \(V \sin \theta\) V_{y }= \(10\; \sin 60^{\circ}\) V_{y }= \(10 \times \frac{\sqrt{3}}{2}\) V_{y }= \(10 \sqrt{3}\) units 
Magnitude of a Vector
The magnitude of a vector is shown by vertical lines on both the sides of the given vector.
\(\left  a \right \)
Mathematically, the magnitude of a vector is calculated by the help of “Pythagoras Theorem,” i.e.
\(\left  a \right  = \sqrt{x^{2}+y^{2}}\),
Example Find the magnitude of vector a (3,4).
Solution Given \( \overrightarrow{a} \)= (3,4) \(\left  a \right = \sqrt{x^{2}+y^{2}}\) \(\left  a \right = \sqrt{3^{2}+4^{2}}\) \(\Rightarrow \left  a \right = \sqrt{9 + 16} = \sqrt{25} \) Therefore, \(\left  a \right = 5\) 
Operation on Vector–
Vector operation such as Addition, Subtraction, Multiplication etc. can be done easily.
1. Addition of Vectors
The two vectors a and b can be added giving the sum to be a + b. This requires joining them head to tail.
We can translate the vector b till its tail meets the head of a. The line segment that is directed from the tail of vector a to the head of vector b is the vector “a + b”.
Characteristics of Vector Addition
 Commutative Law the order of addition does not matter, i.e, a + b = b + a
 Associative law the sum of three vectors has nothing to do with which pair of the vectors is added at the beginning.
i.e. (a + b) + c = a + (b + c)
2. Subtraction of Vectors
Before going to the operation it is necessary to know about reverse vector(a).
A reverse vector (a) which is opposite of a has similar magnitude as a but pointed in opposite direction.
First, we find the reverse vector.
Then add them as the usual addition.
Such as if we wanna find vector b – a
Then, b – a = b + (a)
3. Multiplication of Vectors

Scalar Multiplication
Multiplication of a vector by a scalar quantity is called “Scaling.”
In this type of multiplication, only the magnitude of a vector is changed not the direction.

Vector Multiplication
It is of two types “Cross product” and “Dot product.”
Cross Product
The cross product of two vectors results in a vector quantity. It is represented by a cross sign between two vectors.
\(a \times b\) 
Mathematical value of a cross product
\(a \times b = \left  a \right  \left  b \right \sin \theta \;\hat{n}\) 
where, \(\left  a \right \) is the magnitude of vector a.
\(\left  a \right \) is the magnitude of vector b.
\(\theta\) is the angle between two vectors a & b.
and \(\hat{n}\) is a unit vector showing the direction of the multiplication of two vectors.
Dot product
The dot product of two vectors always result in scalar quantity, i.e. it has only magnitude and no direction. It is represented by a dot in between two vectors.
\(a.b\) 
Mathematical value
\(a . b = \left  a \right  \left  b \right \cos \theta\) 
Example Find the scalar and vector multiplication of two vectors a and b given by \(3\hat{i}1\hat{j}+2\hat{k}\) and \(1\hat{i}+2\hat{j}+3\hat{k}\) respectively.
Solution Given vector a (3,1,2) and vector b (1,2,3) Vector product ( or Cross product) = \(\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix}\) \(\vec{a} \times \vec{b} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \hat{i} – \begin{vmatrix} 3 & 2\\ 1 & 3 \end{vmatrix}\hat{j} + \begin{vmatrix} 3 & 1\\ 1 & 2 \end{vmatrix}\hat{k}\) \(\vec{a} \times \vec{b} = 1 \hat{i} – (7)\hat{j} + (5) \hat{k}\) \(\vec{a} \times \vec{b} \) = \( 1 \hat{i} 7 \hat{j} – 5 \hat{k}\) Now, \(\left  \vec{a} \times \vec{b} \right  = \sqrt{(1)^{2}+ (7)^{2}+ (5)^{2}}\) \(\left  \vec{a} \times \vec{b} \right \) = \(\sqrt{75} = 5\sqrt{3}\) Now finding magnitudes of vector a and b \(\left  \vec{a} \right  = \sqrt{(3)^{2}+(1)^{2}+(2)^{2}}\) \(\left  \vec{a} \right \) = \(\sqrt{9+1+4} =\sqrt{14}\) \(\left  \vec{b} \right  = \sqrt{(1)^{2}+(2)^{2}+(3)^{2}}\) \(\left  \vec{b} \right \) = \(\sqrt{1+4+9} =\sqrt{14}\) \(\sin \theta\) = \(\frac{\left  \vec{a} \times \vec{b} \right  }{\left  \vec{a} \right  \times \left  \vec{b} \right  }\) \(\sin \theta = \frac{5\sqrt{3} }{\sqrt{14} \times \sqrt{14}}\) \(\sin \theta = \frac{5\sqrt{3} }{14}\) Or, \(\theta\) = \(\sin^{1}\left ( \frac{5\sqrt{3} }{14} \right )\) Thus the Vector product is equal to \(1 \hat{i} 7 \hat{j} – 5 \hat{k} \) Scalar product (or Dot product) = \(\vec{a}.\vec{b}= \left  a \right \left  b \right  \cos \theta\) Where \(\theta\) is the angle between the vectors. But we don’t know the angle between the vectors thus another method of multiplication can be used. \(\vec{a}.\vec{b}= (3\hat{i}1\hat{j}+2\hat{k}).(1\hat{i}2\hat{j}+ 3\hat{k})\) \(\vec{a}.\vec{b}= 3(\hat{i}.\hat{i})+2(\hat{j}.\hat{j})+6(\hat{k}.\hat{k})\) \(\vec{a}.\vec{b}= 3+2+6\) \(\vec{a}.\vec{b}= 11\) 
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