Types Of Vectors

A quantity that has magnitude as well as a direction in space and follows the triangle law of addition is called a vector quantity, e.g., velocity, force, displacement, etc. A vector is represented by a directed line segment (an arrow). In real life, we use vectors for tracking objects that are in motion, and localization of places and things.

This Story also Contains

1. What are Vectors?
2. Different types of Vectors
3. Properties Of Vector
4. Applications of Vectors
5. Solved Examples Based on Types Of Vectors

In this article, we will cover the concept of Types Of Vectors. This topic falls under the broader category of Vector Algebra, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of seven questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2014, two in 2019, and one in 2021.

What are Vectors?

Vectors are geometrical entities that have magnitude and direction. A vector can be represented by a line with an arrow pointing towards its direction, and its length represents the magnitude of the vector.

Different types of Vectors

The vectors are named differently as types of vectors based on their properties such as magnitude, direction, and their relationship with other vectors.

Zero or Null Vector

A vector whose initial and terminal points coincide is called a zero vector (or null vector) and it is denoted as $\overrightarrow{0}$. The magnitude of the zero vector is zero and the direction of the zero vector is indeterminate. A zero vector cannot be assigned a definite direction as it has zero magnitude or, alternatively, it may be regarded as having any direction

It can also be denoted by $\overrightarrow{A A}$ or $\overrightarrow{B B}$ etc.

Unit Vector

A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. Unit vectors are denoted by small letters with a cap on them

The unit vector in the direction of a given vector $\vec{a}$ is denoted by $\widehat{\mathbf{a}}$ read as "a cap". Thus, $|\widehat{\mathbf{a}}|=1$.

$
\hat{\mathbf{a}}=\frac{\overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|}
$

Coinitial Vectors

Two or more vectors having the same initial point are called coinitial vectors.

Collinear Vector

Two or more vectors are said to be collinear (or Parallel) if they are parallel to the same line, irrespective of their magnitudes and directions

Non-collinear Vector

Two vectors acting in different directions are called non-collinear vectors. Non-collinear vectors are often called independent vectors.

Coplanar Vector

Two parallel vectors or non-collinear vectors are always coplanar or two vectors a and b in different directions determine a unique plane in space.

Like and Unlike Vectors

Vectors are said to be like when they have the same direction and unlike when they have opposite directions.

Both like and unlike vectors are parallel to each other.

Equal Vectors

Two vectors $\vec{a}$ and $\vec{b}$ are said to be equal, if they have the same magnitude and same direction regardless of the positions of their initial points, and written as $\vec{a}=\vec{b}$.

Negative of a Vector

The vector has the same magnitude as that of a given vector (say $\overrightarrow{\mathbf{a}}$) but has an opposite direction.

It is denoted by $-\overrightarrow{\mathbf{a}}$.
Thus if $\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{AB}}$ then, $-\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{BA}}$

Coplanar Vector

A system of vectors is said to be coplanar if they lie on the same plane.

Note: 2 vectors are always coplanar.

Free Vector

Vectors whose initial points are not specified are called free vectors.

Localized vector

A vector drawn parallel to a given vector, but through a specified point as the initial point, is called a localized vector.

Parallel Vector

Two or more vectors are said to be parallel if they have the same support or parallel support. Parallel vectors may have equal or unequal magnitudes and their directions may be same or opposite.

Properties Of Vector

• The Dot Product of two vectors is a scalar and lies in the plane of the two vectors.

• The cross product of two vectors is a vector, which is perpendicular to the plane containing these two vectors.

• The addition of vectors is commutative and associative.

Applications of Vectors

Some of the important applications of vectors in real life are listed below:

• The direction in which the force is applied to move the object can be found using vectors.

• To understand how gravity applies as a force on a vertically moving body.

• The motion of a body confined to a plane can be obtained using vectors.

• Vectors help in defining the force applied on a body simultaneously in the three dimensions.

• Vectors are used in the field of Engineering, to check if the force is much stronger than the structure and if it will sustain, or collapse.

• In various oscillators, vectors are used.

• Vectors also have their applications in ‘Quantum Mechanics’.

• The velocity of liquid flow in a pipe can be determined in terms of the vector field - for example, fluid mechanics.

• We may also observe them everywhere in general relativity.

• Vectors are used in various wave propagations such as vibration propagation, sound propagation, AC

Recommended Video Based on Types of Vectors

Solved Examples Based on Types Of Vectors

Example 1: If vectors $\overrightarrow{a_1}=x \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{a_2}=\hat{i}+y \hat{j}+z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i}+y \hat{j}+z \hat{k}$ is: [JEE MAINS 2021]

Solution

$\vec{a}_1$ and $\vec{a}_2$ are collinear so $\frac{x}{1}=\frac{-1}{y}=\frac{1}{z}$ unit vector in direction of $x \hat{i}+y \hat{j}+z \hat{k}= \pm \frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$

Hence, the answer is $\frac{1}{\sqrt{3}}(\hat{i}-\widehat{j}+\hat{k})$

Example 2: Let $\vec{a}=i+2 j+4 k, \vec{j}=i+\lambda j+4 k$ and $\vec{e}=2 i+4 j+\left(\lambda^2-1\right) k$ be coplanar vectors. Then the non-zero vector $\bar{a} \times \vec{c}$ is: [JEE MAINS 2019]

Solution: Coplanar vectors -A given number of vectors are called coplanar if their line segments are parallel to the same plane. Two vectors are always coplanar.

For three coplanar vectors,

$[\vec{a} \vec{b} \vec{c}]=0$

$\vec{a}, \vec{b}$ and $\vec{c}$ are three coplanar vectors.

$
\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+4 \hat{k} \\
& \vec{b}=\hat{i}+\lambda \hat{j}+4 \hat{k} \\
& \begin{aligned}
& \vec{c}=2 \hat{i}+4 \hat{j}+\left(\lambda^2-1\right) \hat{k} \\
& {[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{lll}
1 & 2 & 4 \\
1 & \lambda & 4 \\
2 & 4 & \lambda^2-1
\end{array}\right| } \\
& \qquad=\lambda\left(\lambda^2-1\right)-16-2\left(\lambda^2-1-8\right)+4(4-2 \lambda) \\
& \begin{aligned}
\Rightarrow[\vec{a} \vec{b} \vec{c}]=(\lambda-3)(\lambda+3)(\lambda-2)
\end{aligned} \\
& \text { for } \lambda= \pm 3 \vec{c}=2 \vec{a} \Rightarrow \vec{a} \times \vec{c}=0 \\
& \text { for } \lambda=2, \\
& \vec{a} \times \vec{c}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 4 \\
2 & 4 & 3
\end{array}\right|=-10 \hat{i}+5 \hat{j}
\end{aligned}
\end{aligned}
$

Hence, the answer is $-10 \hat{i}+5 \hat{j}$

Example 3: Let $\vec{\alpha}=3 \hat{i}+\hat{j}$ and $\vec{\beta}=2 \hat{i}-\hat{j}+3 \hat{k}$ If $\vec{\beta}=\overrightarrow{\beta_1}-\overrightarrow{\beta_2}$ where$\vec{\beta}_1$ is parallel to $\vec{\alpha}$ and $\overrightarrow{\beta_2}$ is perpendicular to $\vec{\alpha}$, then $\overrightarrow{\beta_1} \times \overrightarrow{\beta_2}$ is equal to : [JEE MAINS 2019]

Solution: $\vec{\alpha}=3 \hat{i}+\hat{j}$

$\vec{\beta}_1$ is parallel to $\vec{\alpha}$, so $\vec{\beta}_1=k \vec{\alpha}$

$
\vec{\beta}_1=k(3 \hat{i}+\hat{j})
$

and

$
\begin{aligned}
& \vec{\beta}_2=\vec{\beta}_1-\vec{\beta} \\
& \vec{\beta}_2=k \vec{\alpha}-\vec{\beta}=k(3 \hat{i}+\hat{j})-(2 \hat{i}-\hat{j}+3 \hat{j})
\end{aligned}
$

$\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$, so $\vec{\beta}_2 \cdot \vec{\alpha}=0 \Rightarrow(3 k-2) \cdot 3+(k+1)=0$

$
\begin{aligned}
& k=\frac{1}{2} \\
& \therefore \overrightarrow{\beta_1} \times \overrightarrow{\beta_2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
\frac{3}{2} & \frac{1}{2} & 0 \\
\frac{-1}{2} & \frac{3}{2} & -3
\end{array}\right| \\
& =-\frac{3}{2} \hat{i}+\frac{9}{2} \hat{j}+\frac{5}{2} k \\
& =\frac{1}{2}(-3 \hat{i}+9 \hat{j}+5 \hat{k})
\end{aligned}
$
Hence, the answer is $\frac{1}{2}(-3 \hat{i}+9 \hat{j}+5 \hat{k})$

$
\frac{1}{2}(-3 \hat{i}+9 \hat{j}+5 \hat{k})
$
Example 4: If $\hat{x}, \dot{y}$, and $\tilde{z}$ are three unit vectors in three-dimensional space, then the minimum value of $|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2$ is:
Solution: Unit vector - A vector of unit magnitude in the direction of a given vector $\vec{a}$ is called a unit vector. It is denoted by $\hat{a}$.
Scalar Product of two vectors $-\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}$

Hence the minimum value is 3 .

$
\begin{aligned}
& \vec{a} \cdot \vec{a}=a^2=|\vec{a}|^2 \\
& |\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2=|\hat{x}|^2+|\hat{y}|^2+2 \hat{x} \hat{y}+|\hat{y}|^2+|\hat{z}|^2+2 \hat{y} \hat{z}+|\hat{x}|^2+|\hat{z}|^2+2 \hat{z} \hat{x} \\
& =6+2(\hat{x} \hat{y}+\hat{y} \hat{z}+\hat{z} \hat{x}) \\
& =6+\left[(\hat{x}+\hat{y}+\hat{z})^2-\left(\hat{x}^2+\hat{y}^2+\hat{z}^2\right)\right] \\
& |\hat{x}|^2=|\hat{y}|^2=|\hat{z}|^2=1 \\
& =6+\left[(\hat{x}+\hat{y}+\hat{z})^2-\left(|\hat{x}|^2+|\hat{y}|^2+|\hat{z}|^2\right)\right] \\
& =6+\left[(\hat{x}+\hat{y}+\hat{z})^2-(1+1+1)\right] \\
& =3+(\hat{x}+\hat{y}+\hat{z})^2 \\
& \geq 3
\end{aligned}
$

Hence the minimum value is 3.

Example 5: The magnitude of the vector from point $A$ with position vector $3 \hat{i}-4 \hat{j}+\hat{k}$ to point $B$ with position vector $\hat{i}+2 \hat{j}+4 \hat{k}$.
Solution: Magnitude of a Vector - The length of the directed line segment $\overrightarrow{A B}$ is called its magnitude.
It is denoted by $|\overrightarrow{A B}|$

$
\begin{aligned}
& \overrightarrow{O A}=3 \hat{i}-4 \hat{j}+\hat{k} ; \overrightarrow{O B}=\hat{i}+2 \hat{j}+4 \hat{k} \\
& \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=-2 \hat{i}+6 \hat{j}+3 \hat{k} \\
& \therefore|\overrightarrow{A B}|=\sqrt{(-2)^2+(6)^2+(3)^2}=7
\end{aligned}
$
Hence, the answer is 7 units