Vector Concepts and Representations
Vectors are fundamental mathematical objects that possess both magnitude and direction, describing a translation. Hence they showcase both quantity and orientation in some space.
They can be written as a bold letter or with a vector sign on top, for example
$$\vec{v} \text{ OR } \mathbf{v}$$
Position and Displacement Vectors
Position vectors represent the location of a point in space relative to a fixed origin. For instance, if we consider a point P in three-dimensional space, its position vector $\vec{OP}$ would extend from the origin O to the point P.
Displacement vectors, on the other hand, describe the change in position from one point to another. If we have two points A and B, the displacement vector $\vec{AB}$ represents the directed path from A to B. Or from B to A would be $\vec{BA}$
Let the right red position vector be $\vec{OA}=\mathbf{a}$ and the left red position vector be $\vec{OB} = \mathbf{b}$. Then the blue vector goes from $\vec{OA}$ to $\vec{OB}$ which is $\vec{AB}$ and can be calculated as $\mathbf{b}- \mathbf{a}$
NoteA position vector is technically also a direction vector but referring to a specific point, that's why we notate $\mathbf{b} - \mathbf{a}$ where $\mathbf{a},\mathbf{b}$ are actual vectors, instead of using the position notation
Representation Using Directed Line Segments
Vectors are often visually represented as arrows, where the length of the arrow corresponds to the vector's magnitude, and the direction of the arrow indicates the vector's direction.
NoteThe same vector can be represented by any arrow with the same length and direction, regardless of its starting point. This is known as the "free vector" concept.
Base Vectors and Components
Base Vectors i, j, k
In three-dimensional space, we define three standard unit vectors:
- $i = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$
- $j = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$
- $k = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$
These vectors form a basis for 3D space, meaning any vector can be expressed as a linear combination of these base vectors.
Components of a Vector
A vector v can be expressed in terms of its components:
$\mathbf{v} = (v_1, v_2, v_3) = v_1\times i + v_2\times j + v_3\times k$
Where $v_1$, $v_2$, and $v_3$ are scalar values representing the vector's components along the x, y, and z axes respectively.
ExampleThe vector $\mathbf{v} = \begin{pmatrix} 3\\ -2 \\ 5 \end{pmatrix}$ can be written as:
$$v = 3i - 2j + 5k$$
This representation shows how much of each base vector contributes to the overall vector.
Magnitude and Unit
Since a vector goes in multiple directions, the actual length of the vector, or in other words the magnitude, can be calculated by taking the sum of the squares of each component. This is a simple property of Pythagoras, and for 2-dimensional or 3-dimensional, can be easily shown.
The magnitude is written as $||\mathbf{v}||$ and for a vector $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$ can be calculated as
$$||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2}$$
A vector that has a length of $1$ is called a unit vector. In fact a vector going one direction can be scaled into a unit vector in the same direction by
$$\mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||}$$
where $\mathbf{u}$ is a unit vector in the same direction as $\mathbf{v}$
ExampleConsider a vector $\mathbf{v} = \begin{pmatrix} 4 \\ 4 \\ 2 \end{pmatrix}$, then we can calculate the magnitude by
$$||\mathbf{v}|| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$$
We show how to turn this into a unit vector in the scalar multiplication section.
Vector Operations
Sum and Difference of Vectors
Vectors can be added or subtracted component-wise:
For vectors $\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$:
$$\mathbf{a} + \mathbf{b} = \begin{pmatrix} a_1+b_1 \\ a_2+b_2 \\ a_3+b_3 \end{pmatrix}$$
$$\mathbf{a} - \mathbf{b} = \begin{pmatrix} a_1-b_1 \\ a_2-b_2 \\ a_3-b_3 \end{pmatrix}$$
Geometrically, vector addition can be visualized using the parallelogram or tip-to-tail method.
Figure A above shows the tip-to-tail method. You have two vectors (in red) and to find their addition, you connect the tip of one to the tail of the other. Then drawing a vector from the tail of the 'behind' vector to the tip of the 'front' vector leaves you with their addition (in green).
Figure B similarly shows the parallelogram, where the two red vectors $\mathbf{a}$ are the same, and the two purple vectors $\mathbf{b}$ are the same. Hence $\mathbf{a}+\mathbf{b}$ would result in the diagonal of the parallelogram which is in green.
Similarly, you can backtrack to find how to subtract vectors. For example in Figure B, to subtract the purple vector $\mathbf{b}$ from the green one $\mathbf{a}+\mathbf{b}$, you can connect their tips and draw a vector from the green one $\mathbf{a}+\mathbf{b}$ to the purple one $\mathbf{b}$ which leaves you with the red one $\mathbf{a}$
And if you want to subtract the green one $\mathbf{a}+\mathbf{b}$ from the purple one $\mathbf{b}$, you can start by doing the same but now instead, drawing the vector from the purple one $\mathbf{b}$ to the green one $\mathbf{a}+\mathbf{b}$ which leaves you with a similar vector to the red one $\mathbf{a}$ but in the opposite direction!
The same can be done with Figure A of course.
Zero Vector and Negation
The zero vector, denoted as $\vec{0}$, is a vector with all components equal to zero: $\vec{0} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$. The position vector of the origin $\vec{OO}$ is the zero vector!
For any vector $\mathbf{v}$, its negation $-\mathbf{v}$ is a vector with the same magnitude but opposite direction:
If $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$, then $-\mathbf{v} = \begin{pmatrix} -v_1 \\ -v_2 \\ -v_3 \end{pmatrix}$
Scalar Multiplication
Multiplying a vector by a scalar k changes its magnitude, in other words, you scale the vector in the same direction if $k>0$ or in the opposite direction if $k<0$
$k\times v = \begin{pmatrix} k\times v_1 \\ k\times v_2 \\ k \times v_3 \end{pmatrix}$
ExampleFor a vector $\mathbf{v}$, you can show the result of the negative vector $-\mathbf{v}$ as we did above by taking $k=-1$
Further, let $\mathbf{v} = \begin{pmatrix} 4\\ 4\\ 2\end{pmatrix}$, then we know $||\mathbf{v}|| = 6$ from the magnitude section, hence to find the unit vector in the same direction, we do
$$\mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} = \frac{1}{||\mathbf{v}||}\mathbf{v}$$
which is just a scalar multiple of $\mathbf{v}$ with $k=\frac{1}{||\mathbf{v}||}$ so the unit vector become
$$\mathbf{u} = \begin{pmatrix} \frac{4}{6} \\ \frac{4}{6} \\ \frac{2}{6} \end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ \frac{2}{3}\\ \frac{1}{3}\end{pmatrix}$$
Parallel Vectors
Two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ are parallel if one is a scalar multiple of the other:
$\mathbf{a}$ is parallel to $\mathbf{b}$ if and only if $\mathbf{a} = k\mathbf{b}$ for some scalar $k ≠ 0$
ExampleA vector $\mathbf{a}=\begin{pmatrix} 4 \\ 3 \\ 1\end{pmatrix}$ is parellel to $\mathbf{b} = \begin{pmatrix} 8\\6 \\2 \end{pmatrix}$ because $2\mathbf{a} = \mathbf{b}$
Common MistakeVectors do not indicate position. Only position vectors indicate a specific position on the coordinate system. Don't confuse lines, points and vectors.
Vectors only show magnitude and direction but if you drew two vectors with the same length facing the same direction on different areas of the map, then they are the same vector!
ExampleLet's say you have a point $A=(1,3,4)$ with position vector $\vec{OA}=\begin{pmatrix} 1\\3 \\ 4\end{pmatrix}=\mathbf{a}$ and a point $B=(3,2,5)$ such that the position vector is $\vec{OB}=\begin{pmatrix}3 \\2 \\5 \end{pmatrix}=\mathbf{b}$. Then the following vectors can be calculated as
$$\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 3-1 \\2-3 \\5-4 \end{pmatrix} = \begin{pmatrix}2 \\-1 \\1 \end{pmatrix}$$
$$\vec{BA} = \mathbf{a} - \mathbf{b} = -(\mathbf{b} - \mathbf{a}) = -\begin{pmatrix} 2\\-1 \\1 \end{pmatrix} = \begin{pmatrix} -2\\1 \\-1 \end{pmatrix}$$
The distance between the two points $A$ and $B$ can be calculated by finding the magnitude (length) of the vector between them $\vec{AB}$ or $\vec{BA}$
Both lead to the same result since values are squared, hence the magnitude is always positive, showing the length only.
$$||\mathbf{b}-\mathbf{a}|| = ||\mathbf{a} - \mathbf{b}|| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$$
Hence the distance between $A$ and $B$ is $\sqrt{6}$