Types of Vectors

In Mathematics and Physics, vectors are quantities that have both magnitude and direction. They are essential in representing physical quantities like force, velocity, and displacement. Understanding the types of vectors is important for solving problems in mechanics, electromagnetism, and geometry.

1.0What Are Vectors?

A vector is represented by an arrow where:

• The length of the arrow shows the magnitude.
• The direction of the arrow shows the direction of the vector.

Vectors are usually denoted in bold (e.g., A, B) or with an arrow above the letter (e.g., \vec{A}, \vec{B} ).

2.0Types of Vectors

Vectors are classified based on their properties and applications. Below are the main types of vectors you should know:

1. Zero Vector

• A vector with zero magnitude and no specific direction.
• Denoted as:

$\vec{0}=0$

• Example: The displacement of a stationary object.

2. Unit Vector

• A vector with magnitude equal to 1, used to specify direction.
• Denoted as $\hat{i},\hat{j}$ and $\hat{k}$ in 3D space.
• Example:

$\hat{i}=(1,0,0),\hat{j}=(0,1,0),\hat{k}=(0,0,1)$

3. Equal Vectors

• Vectors having the same magnitude and direction, regardless of their initial point.
• Example:

$\vec{A}=(3,4),\vec{B}=(3,4)then\vec{A}=\vec{B}$

4. Position Vector

• A vector that represents the position of a point with respect to the origin.
• Example: The position vector of point P(x, y, z) is:

$\overrightarrow{OP}=x\hat{i}+y\hat{j}+z\hat{k}$

5. Zero Vector vs. Null Vector

• A Zero Vector is the vector with magnitude zero.
• A Null Vector is sometimes used interchangeably but typically refers to a zero vector in context.

6. Collinear Vectors

• Vectors lying along the same line or parallel lines.
• Example:

$\vec{A}=(2,4),\vec{B}=(1,2)$

Since $\vec{A}=2\vec{B}$, are collinear vectors.

7. Coplanar Vectors

• Vectors lying in the same plane.
• Example: Vectors $\vec{A},\vec{B}$ and $\vec{c}$ are coplanar vectors if:

$\vec{A}.(\vec{B}\times \vec{C})=0$

3.0Types of Vectors with Examples Summary Table

4.0Solved Examples on Types of Vectors

Example 1: Given $\vec{A}=3\hat{i}+4\hat{j}$ and $\vec{B}=-6\hat{i}-8\hat{j}$ check whether $\vec{A}$ and $\vec{B}$ are collinear.

Solution:

Two vectors are collinear if one is a scalar multiple of the other.
Here,

$\vec{B}=-2\vec{A}$

Thus, $\vec{A}and\vec{B}$ are collinear.

Example 2: Find the position vector of point P(2, -3, 5).

Solution:

The position vector \vec{OP} is:

$\overrightarrow{OP}=2\hat{i}-3\hat{j}+5\hat{k}$

So,

$\overrightarrow{OP}=2\hat{i}-3\hat{j}+5\hat{k}$

Example 3: Determine whether the vectors $\vec{A}=2\hat{i}+\hat{j}+\hat{k},\vec{B}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{C}=3\hat{i}+5\hat{j}+7\hat{k}$ are coplanar.

Solution:

Vectors are coplanar if:

$\vec{A}.(\vec{B}\times \vec{C})=0$

First, compute $\vec{B}\times \vec{C}$:

$\vec{B}\times \vec{C}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 3& 5& 7\end{array}\right|$

$=\hat{i}(2\cdot 7-3\cdot 5)-\hat{j}(1\cdot 7-3\cdot 3)+\hat{k}(1\cdot 5-2\cdot 3)$

$=\hat{i}(14-15)-\hat{j}(7-9)+\hat{k}(5-6)$

$=-\hat{i}+2\hat{j}-\hat{k}$

Now compute $\vec{A}.(\vec{B}\times \vec{C})$:

$(2\hat{i}+\hat{j}+\hat{k})\cdot (-\hat{i}+2\hat{j}-\hat{k})$

$=(2)(-1)+(1)(2)+(1)(-1)$

$=-2+2-1=-1$

Since $\vec{A}\cdot (\vec{B}\times \vec{C})\ne 0$_,  are not coplanar vectors.

Example 4: Find the unit vector in the direction of $\vec{A}=3\hat{i}-4\hat{j}+12\hat{k}$

Solution:
Step 1: Find the magnitude of $\vec{A}$:

$|\vec{A}|=\sqrt{3^2+(-4{)}^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13$

Step 2: Unit vector $\hat{A}$ in the direction of $\vec{A}$:

$\hat{A}=\frac{\vec{A}}{|\vec{A}|}=\frac{3\hat{i}-4\hat{j}+12\hat{k}}{13}$

Final Answer:

$\hat{A}=\frac{3}{13}\hat{i}-\frac{4}{13}\hat{j}+\frac{12}{13}\hat{k}$

Example 5: Given

$\vec{A}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{B}=4\hat{i}-\hat{j}+2\hat{k}$  find $\vec{A}\cdot \vec{B}$

Solution:

Dot product formula:

$\vec{A}\cdot \vec{B}=(1)(4)+(2)(-1)+(3)(2)=4-2+6=8$

Final Answer: 8 

Example 6: If

$$\begin{gathered} \vec{A}=2\hat{i}+3\hat{j}+\lambda \hat{k}and \\ vecB=-\hat{i}+4\hat{j}+2\hat{k} \end{gathered}$$

are perpendicular, find \lambda.

Solution:

Perpendicular vectors:

$\vec{A}\cdot \vec{B}=0$

Compute the dot product:

$(2)(-1)+(3)(4)+(\lambda )(2)=-2+12+2\lambda =0$

Simplify:

$10+2\lambda =0⟹\lambda =-5$

Final Answer: −5 

Example 7: Check whether the following vectors are coplanar:

$\vec{A}=\hat{i}+2\hat{j}+3\hat{k},\vec{B}=2\hat{i}+4\hat{j}+6\hat{k},\vec{C}=3\hat{i}+6\hat{j}+9\hat{k}$

Solution:

Vectors are coplanar if:

$\vec{A}\cdot (\vec{B}\times \vec{C})=0$

First, compute $\vec{B}\times \vec{C}$:

$\vec{B}\times \vec{C}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 4& 6\\ 3& 6& 9\end{array}\right|$

$=\hat{i}(4\times 9-6\times 6)-\hat{j}(2\times 9-6\times 3)+\hat{k}(2\times 6-4\times 3)$

$=\hat{i}(36-36)-\hat{j}(18-18)+\hat{k}(12-12)=\vec{0}$

Thus,

$\vec{A}\cdot (\vec{B}\times \vec{C})=\vec{A}\cdot \vec{0}=0$

So, the vectors are coplanar.

5.0Practice Questions on Types of Vectors

1. Given $\vec{A}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{B}=2\hat{i}+4\hat{j}+6\hat{k}$ determine if $\vec{A}$ and $\vec{B}$ are collinear.
2. Find the position vector of point Q(-1, 4, 2).
3. Are the vectors $\vec{P}=3\hat{i}-\hat{j}+2\hat{k},\vec{Q}=-\hat{i}+2\hat{j}-\hat{k},and\vec{R}=2\hat{i}+\hat{j}+\hat{k}$ coplanar?
4. If $\vec{A}=4\hat{i}+3\hat{j}$ and $\vec{B}=-2\hat{i}-1.5\hat{j}$  find the scalar multiple relationship between them.

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