Vectors are used to represent forces in physics, handle 2D and 3D manipulation with computer graphics, and to calculate the forces acting on materials in textiles. You also deal with vector quantities in your everyday life, from driving your car down the road to planning the shortest route to get to uni. Use this resource to gain a basic understanding of vectors.
Vectors and scalars
A vector is a measurable quantity with direction. Examples of vectors are:
velocity – the speed at which an object is travelling in a given direction
force – an action that causes an object to change its velocity
displacement – the distance of an object from its original position
acceleration – the change in speed over time.
On the other hand, a scalar quantity has magnitude only, i.e. no direction. Examples include:
heat – the amount of thermal energy transferred or transformed within a system or between systems
mass – the amount of matter in an object
time – the amount of time that has passed
volume – the amount of space occupied by a substance.
Representing vectors
Vectors in three-dimensional space are defined by three directions that are perpendicular to each other. They can be denoted as bold letters or by using a vector arrow: \(\vec{a}\) or \(\vec{b}\) or \(\vec{c}\). In the following figure, the vector is \(\overrightarrow{PQ}\) as the arrow points from point P to point Q. The modern idea of vectors appeared late in the 19th century when Josiah Willard Gibbs and Oliver Heaviside (of USA and Britain, respectively) independently developed vector analysis to express the new laws of electromagnetism discovered by the Scottish physicist James Clerk Maxwell.
Vectors can also be defined along horizontal or vertical axes — that is, in two directions. A vector in the opposite direction from \(\vec{a}\) is denoted by \(-\vec{a}\).
Components of a vector
In the diagram, the vector \(\vec{r}\) is represented by \(\overrightarrow{OP}\) where \(P\) is the point \((x,y,z)\). This is called a position vector, which indicates where the vector is relative to the origin \(O(0,0,0)\).
If \(\vec{i}\), \(\vec{j}\), and \(\vec{k}\) are vectors parallel to the positive directions of the \(x\)-, \(y\)- and \(z\)-axis, respectively, then:
\(x\vec{i}\) is a vector of length \(x\) in the direction of the \(x\)-axis
\(y\vec{i}\) is a vector of length \(y\) in the direction of the \(y\)-axis
\(z\vec{k}\) is a vector of length \(z\) in the direction of the \(z\)-axis.
\(\overrightarrow{OP}\) is then the position vector \(x\vec{i}+y\vec{j}+z\vec{k}\), and \(x\), \(y\) and \(z\) are called the components of the vector.
The notation \((x,y,z)\) is used to denote the vector \(\vec{r}=(x\vec{i}+y\vec{j}+z\vec{k})\), as well as the coordinates of a point \(P(x,y,z)\).
Directed line segment
In the previous diagram, we defined vector \(P\) which extended from the origin O to point \(P(x,y,z)\). We can also define vectors extending from two points in space, such as from point \(P(x_{1},y_{1},z_{1})\) to point \(Q(x_{2},y_{2},z_{2})\).
\(\overrightarrow{PQ}\) is called a directed line segment or geometric vector. This vector is found by subtracting the coordinates of \(P\) (the initial point) from the coordinates of \(Q\) (the final point).
Let's consider the points \(P(3,4,1)\) and \(Q(5,6,-1)\).
When we apply the general formula:
\[\begin{align*} PQ & = (5-3)i+(6-4)+(-1-1)k\\
& = 2i+2j-2k
\end{align*}\]
Therefore, the directed line segment \(\overrightarrow{PQ}\) is represented by the vector \(2\vec{i}+2\vec{j}-2\vec{k}\) or \((2,2,-2)\). This is also used to represent ANY directed line segment with the same length and direction as \(\overrightarrow{PQ}\).
The directed line segment \(\overrightarrow{QP}\) has the same length as \(\overrightarrow{PQ}\) but is in the opposite direction.
\[\begin{align*} \overrightarrow{QP} & = -\overrightarrow{PQ}\\
& = -(2\vec{i}+2\vec{j}-2\vec{k})\\
& = -2\vec{i}-2\vec{j}+2\vec{k}\quad\textrm{or}\quad(-2,-2,2)
\end{align*}\]
Your turn – defining directed line segments and position vectors
Given the points \(A(3,0,4)\), \(B(-2,4,3)\), and \(C(1,-5,0)\), find:
\(\overrightarrow{AB}\)
\(\overrightarrow{AC}\)
\(\overrightarrow{CB}\)
\(\overrightarrow{BC}\)
\(\overrightarrow{CA}\)
Look at your answers for questions b and e, and c and d. Describe what you notice.
Define the position vectors of the points \(A\), \(B\) and \(C\).
\((-5,4,-1)\)
\((-2,-5,-4)\)
\((-3,9,3)\)
\((3,-9,-3)\)
\((2,5,4)\)
They are in opposite directions.
\(\overrightarrow{OA}=3\vec{i}+4\vec{k}\), \(\overrightarrow{OB}=2\vec{i}+4\vec{j}+3\vec{k}\), and \(\overrightarrow{OC}=\vec{i}-5\vec{j}\)
Adding and subtracting vectors
Vectors may be added or subtracted by adding or subtracting their corresponding components. This is done graphically using the triangle law.
Let's consider two vectors: \(\vec{a}\) extends from point \(P\) to point \(Q\), and \(\vec{b}\) extends from point \(Q\) to point \(R\).
To add vectors \(\vec{a}\) and \(\vec{b}\), we place the tail of vector \(\vec{b}\) at the head of vector \(\vec{a}\) (point Q). The vector sum, \(\vec{a}+\vec{b}\), is the vector \(\overrightarrow{PR}\), from the tail of vector \(\vec{a}\) to the head of \(\vec{b}\).
To subtract \(\vec{b}\) from \(\vec{a}\), we reverse the direction of \(\vec{b}\) to give \(-\vec{b}\), then place them head to tail like we did with addition. \(\vec{b}\) is now extending from point \(R\) to point \(Q\).
\(\vec{a}\) and \(-\vec{b}\) are added together. This means that \(\overrightarrow{PR}\) is the same as \(\vec{a}-\vec{b}\).
Given \(\vec{a}=(2,1,1)\), \(\vec{b}=(1,3,-3)\) and \(\vec{c}=(0,3,-2)\), find:
\(\vec{a}+\vec{b}\)
\(\vec{a}+\vec{c}\)
\(\vec{c}-\vec{b}\)
\(\vec{a}-\vec{b}\)
\((3,4,-2\))
\((2,4,-1\))
\((-1,0,1\))
\((1,-2,4\))
Magnitude of a vector
Once again, let's consider vector \(P(x,y,z)\) where \(\vec{P}=x\vec{i}+y\vec{j}+z\vec{k}\). The length or magnitude of \(\vec{P}\) is denoted by \(\left|\vec{P}\right|\) or \('\vec{P}'\), where:
\[\left|\vec{P}\right|=\sqrt{x^{2}+y^{2}+z^{2}}\]
Take, for example, the vector \(2\vec{i}+3\vec{j}-5\vec{k}\). Its length is:
\[\sqrt{2^{2}+3^{2}+(-5)^{2}}=\sqrt{38}\]
Unit vector
Any vector with a magnitude of \(1\) is called a unit vector. If \(\vec{P}\) is any vector, then a unit vector parallel to \(\vec{P}\) is written \(\hat{P}\) (read as P “hat”). The “hat” symbolises a unit vector.
The unit vector \(\hat{P}\) equals vector \(\vec{P}\) divided by its magnitude \(\left|\vec{P}\right|\).
\[\hat{P} = \frac{\vec{P}}{\left|\vec{P}\right|}\]
This can be rearranged to give:
\[\vec{P} = \left|\vec{P}\right|\hat{P}\]
The rearranged equation shows that vector \(P\) is the magnitude of the vector by its unit vector.
You should remember unit vectors \(\vec{i}\), \(\vec{j}\) and \(\vec{k}\), which are parallel to the \(x\)-, \(y\)- and \(z\)-axes, respectively.
Example 1 – finding unit vectors
If \(\overrightarrow{PQ}\) is the line \(2\vec{i}-5\vec{j}+\vec{k}\), find the unit vector parallel to \(\overrightarrow{PQ}\).
To find the unit vector, divide the vector by its magnitude. Let's calculate the magnitude first.
\[\begin{align*} \overrightarrow{PQ} & = \sqrt{2^{2}+\left(-5\right)^{2}+1^{2}}\\
& = \sqrt{30}
\end{align*}\] The unit vector parallel to \(\overrightarrow{PQ}\) is therefore:
\[\begin{align*} \overrightarrow{PQ} & = \frac{\overrightarrow{PQ}}{\left|\overrightarrow{PQ}\right|}\\
& = \frac{(2\vec{i}-5\vec{j}+\vec{k})}{\sqrt{30}}\\
& = \frac{1}{\sqrt{30}}(2\vec{i}-5\vec{j}+\vec{k})
\end{align*}\]
If \(\vec{a}=(1,2,3)\), find the unit vector parallel to \(\vec{a}\).
\[\begin{align*} \hat{a} & = \frac{\vec{a}}{\left| \vec{a} \right|}\\
& = \frac{(1\vec{i}+2\vec{j}+3\vec{k})}{\sqrt{1^{2}+2^{2}+3^{2}}}\\
& = \frac{(1\vec{i}+2\vec{j}+3\vec{k})}{\sqrt{14}}\\
& = \frac{1}{\sqrt{14}}(1\vec{i}+2\vec{j}+3\vec{k}
\end{align*}\]
Your turn – finding unit vectors
Find the length of the vectors.
\((3,-1,-1)\)
\((0,2,4)\)
\((0,-2,0)\)
Given the points \(A(3,0,4)\), \(B(0,4,3)\) and \(C(1,-5,0)\), find the unit vectors parallel to:
\(\overrightarrow{BA}\)
\(\overrightarrow{CB}\)
\(\overrightarrow{AC}\)
\(\sqrt{11}\)
\(\sqrt{20}=2\sqrt{5}\)
\(2\)
\(\dfrac{1}{\sqrt{26}}(3,-4,1)\)
\(\dfrac{1}{\sqrt{91}}(-1,9,3)\)
\(\dfrac{1}{\sqrt{45}}(-2,-5,-4)\)
\(\dfrac{1}{3\sqrt{5}}(-2,-5,-4)\)
Multiplying vectors by a scalar
A scalar is a number that you multiply by to change the magnitude of a value. To multiply a vector by a scalar, \(m\), you just multiply each component by \(m\).
\[m\vec{P}=mx\vec{i}+my\vec{j}+mz\vec{k}\]
The result is a vector of length \(m\times\left|\vec{P}\right|\).
If \(m>0\), the resultant vector is in the same direction as \(\vec{P}\).
If \(m<0\), the resultant vector is in the opposite direction as \(\vec{P}\).
Two vectors, \(\vec{a}\) and \(\vec{b}\), are said to be parallel if and ONLY if \(\vec{a}=k\vec{b}\) where \(k\) is a real constant.
Example – multiplying vectors by a scalar
Multiply \(\vec{a}=(3\vec{i}+\vec{j}-2\vec{k})\) by \(7\) and show that \(\left|7\vec{a}\right|=7\left|\vec{a}\right|\).
To multiply by a scalar, multiply each component of the vector by the scalar.
\[\begin{align*} 7\vec{a} & = 7(3\vec{i}+\vec{j}-2\vec{k})\\
& = 21\vec{i}+7\vec{j}-14\vec{k}
\end{align*}\]
