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Vector Decomposition 1
We have already seen that writing a vector in polar form allows us to fully describe an arbitrary vector without the need to draw that vector.
Exam Tip
Even if a diagram is not necessary, it is always a good idea to draw a labelled diagram in your actual exam whenever possible so that you can more easily visualise the question!
For example, writing \(\vec{a} = \langle 7 \,\, \angle \,\, 120^{\circ}\rangle\) is completely equivalent to drawing this diagram.
\[\,\]Figure 1.7.1
There is in fact another, more useful way of writing down a vector.
To start, we first need to define what a unit vector is.
Key Point
A unit vector is any vector of magnitude \(1\) unit.
How does that help?
Well, consider our vector \(\vec{a}\) above. Recall that anytime we multiply a scalar by a vector, it changes the magnitude of the vector but either only flips or does not change the direction of that vector.
In this example, therefore, if there is a unit vector \(\vec{m}\) that is parallel to \(\vec{a}\), then we can simply write \(\vec{a}=7\vec{m}\)!
\[\,\]Figure 1.7.2
In general, we can write any vector \(\vec{a}\) as a scalar multiple of some unit vector \(\vec{m}\). For a given vector \(\vec{a}\), we typically write this scalar multiple simply as \(a\), i.e.
\[\vec{a} = a\vec{m}\]
Therefore, for our particular example, \(a=7\). We refer to these numbers in front of units vectors as the coefficients of the vector.
Key Point
In general, we can write any arbitrary vector \(\vec{a}\) as \(\vec{a} = a \vec{m}\), where \(a\) is a positive or negative scalar referred to as the vector coefficient and \(\vec{m}\) is a unit vector either parallel or antiparallel to \(\vec{a}\).
That does help a little, but not a lot. If a vector was stated in a question as \(\vec{a}=7\vec{m}\), and if the question told us that \(\vec{m}\) was a unit vector, then we would know that that vector has a magnitude of \(7\). We would also know that this vector was parallel to \(\vec{m}\). (If we instead have a vector \(\vec{b}=-5\vec{m}\), then that vector would instead be antiparallel to \(\vec{m}\).)
However, we wouldn’t know the direction of \(\vec{a}\) unless we were also told the direction of \(\vec{m}\)!
There are, however, two particular unit vectors that do help us in this regard as their directions are specifically defined.
They are so special, in fact, that they have their own names!
Key Point
\(\vec{i}\) and \(\vec{j}\) are unit vectors defined as follows:
- \(\vec{i}\) is a vector of magnitude \(1\) unit pointing in the positive \(x\) direction.
- \(\vec{j}\) is a vector of magnitude \(1\) unit pointing in the positive \(y\) direction.
\[\,\]Figure 1.7.3
Thus, in cases where we are told that a vector \(\vec{a}\) is a scalar multiple of either \(\vec{i}\) or \(\vec{j}\), then we immedialy know the direction of \(\vec{a}\) without having to draw it!
For example, this vector of magnitude \(2\) units can simply be written as \(2\vec{i}\).
\[\,\]Figure 1.7.4
In the opposite sense, the vector could instead be stated and we now know immediately how to draw it.
For example, if we are told that a vector is of the form \(\vec{b}=-3 \vec{j}\), we know that we can draw it as shown.
\[\,\]Figure 1.7.5
But wait, how does this help us with our original vector \(\vec{a}=\langle 7 \,\, \angle \,\, 120^{\circ}\rangle\)? It’s not parallel to \(\vec{i}\) or \(\vec{j}\)?
\[\,\]Figure 1.7.6
Well, let us now do something a bit odd with this vector and apply the Triangle Law backwards.
According to this law, this vector is exactly the same as the sum of these two vectors.
\[\,\]Figure 1.7.7
We can write this in equation form as
\[\vec{a} = \vec{a_x} +\vec{a_y} \tag{1.7.1}\]
This is what is known as decomposing a vector. We write an arbitrary vector as the sum of vectors which are parallel (or antiparallel) to axes.
Instead of adding vectors together to get a resultant vector like we did previously, we are now taking a vector and decomposing it into component vectors.
But what is the benefit of doing that? Have we not just introduced even more vectors and made things more complicated?!
Well, let’s look at one of these component vectors a little closer.
\[\,\]Figure 1.7.8
This vector lies entirely along the \(x\) axis, just like our unit vector \(\vec{i}\)!
\[\,\]Figure 1.7.9
Therefore, we can use the same approach as we just used. There must be some scalar (let’s call it \(a_x\)) such that
\[\vec{a_x} = a_x \vec{i}\]
and likewise for the other component vector
\[\vec{a_y} = a_y \vec{j}\]
If we now insert both of these into equation \((1.7.1)\) above, we can now write this arbitrary vector in what is known as component form, i.e. in terms of \(\vec{i}\) and \(\vec{j}\)!
\[\vec{a} = a_x\vec{i} +a_y\vec{j} \]
How do we find the coefficients \(a_x\) and \(a_y\)?
Well, to do that, we need to use some trigonometry!
First, notice that we’ll always have a right angled triangle when dealing with decomposed vectors.
\[\,\]Figure 1.7.10
Let’s look at this triangle more carefully by looking at just the magnitudes of the vectors so that we have a triangle that we are more accustomed to.
\[\,\]Figure 1.7.11
First, what is the angle \(A\)?
Well, as we are given the vector in polar form, we are given the argument \(\theta\) (\(= 120^\circ\) in this case) which we can use to find \(A\).
\[\,\]Figure 1.7.12
For our example, this angle \(A\) is therefore given by
\begin{align}A &=180^{\circ}-120^{\circ}\\&=60^{\circ}\end{align}
What about \(|\vec{a_x}|\), the length of the adjacent of this right-angled triangle?
Well, we know the length of the hypotenuse \(|\vec{a}|\) (\(=7\) in our example) and we now know the angle between the adjacent and the hypotenuse \(A\) (\(=60^{\circ}\) in our example).
\[\,\]Figure 1.7.13
Therefore, we want to use the trigonometric function that relates these two quantities to the adjacent, i.e. the cosine function.
\begin{align}\cos A &=\frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\&= \frac{|\vec{a_x}|}{|\vec{a}|}\end{align}
Rearranging this, we therefore get
\[|\vec{a_x}| = |\vec{a}|\cos A \]
For our particular example, we obtain
\begin{align}|\vec{a_x}| &= |\vec{a}|\cos A\\ &=7 \cos 60^{\circ}\\ &=(7)(0.5)\\ &=3.5\end{align}
Similarly, using the same approach for \(\vec{a_y}\), we obtain
\begin{align}\sin A &=\frac{\mbox{opposite}}{\mbox{hypotenuse}}\\&= \frac{|\vec{a_y}|}{|\vec{a}|}\end{align}
and therefore
\[|\vec{a_y}| = |\vec{a}|\sin A \]
Again, for our particular example, this gives
\begin{align}|\vec{a_y}| &= |\vec{a}|\sin A\\ &=7 \sin 60^{\circ}\\ &\approx(7)(0.866)\\ &=6.062\end{align}
Now that we know both magnitudes, we can easily write each component vector. If we now return to our triangle of vectors
\[\,\]Figure 1.7.14
we see therefore that
\begin{align} \vec{a}_x = -3.5\vec{i} && \vec{a}_y = 6.062\vec{j} \end{align}
(since \(\vec{a}_x\) is in the opposite direction to \(\vec{i}\) and \(\vec{a}_y\) is in the same direction as \(\vec{j}\)) and thus
\[\vec{a} = -3.5\vec{i}+6.062\vec{j}\]
Done! We now have created a different way of writing this vector. Writing this vector in polar form
\[\vec{a} = \langle 7 \,\, \angle \,\, 120^{\circ}\rangle\]
\[\,\]Figure 1.7.15
is completely equivalent to writing it in component form.
\[\vec{a} = -3.5\vec{i}+6.062\vec{j}\]
\[\,\]Figure 1.7.16
Cause of Confusion
One of the most common mistakes made by students in this subject is to assume that a magnitude like \(|\vec{a_x}|\) and a coefficient like \(a_x\) are the same thing (and similarly for \(|\vec{a_y}|\) and \(a_y\)).
Let’s quickly remind ourselves of what these four scalars refer to.
\(|\vec{a_x}|\) is defined as the magnitude of the vector \(\vec{a_x}\). As it is a magnitude, it is always a positive quantity as a magnitude (arrow length) is always positive. (An arrow cannot have a negative length!)
The coefficient \(a_x\), on the other hand, is simply defined as the number in front of the unit vector \(\vec{i}\). It could therefore be either positive or negative! If \(\vec{a_x}\) is in the same direction (parallel) to \(\vec{i}\), then \(a_x\) will be positive. If \(\vec{a_x}\) is instead in the opposite direction (antiparallel) to \(\vec{i}\), then \(a_x\) will be negative!
In our example above, we found that \(|\vec{a_x}|=3.5\). However, as the vector \(\vec{a_x}\) was pointing in the negative \(x\) direction, i.e. in the opposite direction to the unit vector \(\vec{i}\), we wrote that \(\vec{a_x} = -3.5 \vec{i}\) and therefore \(a_x=-3.5\). As \(\vec{a_y}\) was instead pointing in the positive \(y\) direction, we instead wrote this vector as \(\vec{a_y} = 6.062 \vec{j}\) and therefore \(a_y=6.062\).
Remember, while \(|\vec{a_x}|\) and \(a_x\) always have the same value, they may have different signs (and likewise for \(|\vec{a_y}|\) and \(a_y\)). It is incredibly common for students to lose marks here by getting the signs wrong, so be careful!
As we see, to convert a vector in polar form to a vector in component form, we need to perform a series of steps.
Any collection of such steps that gives us our desired result is referred to as an algorithm.
Key Point
An algorithm is a set of steps or instructions that are performed in order to solve a problem.
The algorithm to convert a vector in polar form to a vector in component form can therefore be summarised as follows.
Key Point
To convert a vector in polar form (\(\vec{a} = \langle |\vec{a}| \,\, \angle \,\, \theta\rangle\)) to a vector in component form (\(\vec{a} =a_x \vec{i} + a_y \vec{j}\)), we perform the following algorithm:
- Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
- Step 2: Use the argument \(\theta\) to find the relevant angle \(A\) in this triangle.
- Step 3: Calculate the magnitudes of the decomposed vectors using \(|\vec{a_x}| = |\vec{a}|\cos A \) and \(|\vec{a_y}| = |\vec{a}|\sin A \).
- Step 4: Calculate \(a_x\) and \(a_y\) using \(|\vec{a_x}|\) and \(|\vec{a_y}|\), i.e. be careful with signs!
- Step 5: Write \(\vec{a} =a_x \vec{i} + a_y \vec{j}\).
Exam Tip
In the actual exam, you do not need to write out the steps of any algorithm that is presented in these lessons.
Instead, you should only perform each step as stated and in the order that they are stated.
Guided Example 1.7.1
Write the vector \(\vec{a}=\langle 3 \,\, \angle \,\, 20^{\circ}\rangle\) in component form, correct to two decimal places.
We are being asked to convert a vector from polar form into component form and therefore shall follow the list of steps.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the angle \(\theta\) given in polar form is the same angle that is within the triangle, i.e. \(A=20^{\circ}\).
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{a_x}| = |\vec{a}|\cos A} \) and \(\boldsymbol{|\vec{a_y}| = |\vec{a}|\sin A} \).
\begin{align}|\vec{a_x}| &= |\vec{a}| \cos A \\&= 3\cos 20^{\circ} \\&\approx 2.82\end{align}
\begin{align}|\vec{a_y}| &= |\vec{a}| \sin A \\&= 3\sin 20^{\circ} \\&\approx 1.03 \end{align}
Step 4: Calculate \(\boldsymbol{a_x}\) and \(\boldsymbol{a_y}\) using \(\boldsymbol{|\vec{a_x}|}\) and \(\boldsymbol{|\vec{a_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{a_x}\) is in the same direction as \(\vec{i}\) and \(\vec{a_y}\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}a_x = 2.82 && a_y = 1.03\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{a} =a_x \vec{i} + a_y \vec{j}}\).
\[\vec{a}=2.82\vec{i} + 1.03\vec{j}\]
Guided Example 1.7.2
Write the vector \(\vec{a}=\langle 4 \,\, \angle \,\, 230^{\circ}\rangle\) in component form, correct to two decimal places.
We are being asked to convert a vector from polar form into component form and therefore shall follow the list of steps.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
The angle \(\theta\) given in polar form is the angle of the vector relative to the positive \(x\) axis. The angle \(A\), according to the above diagram, is therefore
\begin{align}A &= 270^{\circ}-230^{\circ}\\&=40^{\circ}\end{align}
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{a_x}| = |\vec{a}|\cos A} \) and \(\boldsymbol{|\vec{a_y}| = |\vec{a}|\sin A} \).
\begin{align}|\vec{a_x}| &= |\vec{a}| \cos A \\&= 4\cos 40^{\circ} \\&\approx 3.06\end{align}
\begin{align}|\vec{a_y}| &= |\vec{a}| \sin A \\&= 4\sin 40^{\circ} \\&\approx 2.57 \end{align}
\[\,\]
Step 4: Calculate \(\boldsymbol{a_x}\) and \(\boldsymbol{a_y}\) using \(\boldsymbol{|\vec{a_x}|}\) and \(\boldsymbol{|\vec{a_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{a_x}\) is in the opposite direction to \(\vec{i}\) and \(\vec{a_y}\) is in the opposite direction to \(\vec{j}\). Therefore
\begin{align}a_x = -3.06 && a_y = -2.57\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{a} =a_x \vec{i} + a_y \vec{j}}\).
\[\vec{a}=- 3.06\vec{i} -2.57\vec{j}\]
Guided Example 1.7.3
A vector \(\vec{p}\) has a magnitude of \(10\) units and subtends an angle \(\theta\) relative to the positive \(x\) axis, where \(\theta = \tan^{-1}\left(\displaystyle\frac{4}{3}\right)\).
Write \(\vec{p}\) in component form without the use of a calculator.
In this case, we are told the direction of the vector in a bit of a weird way! Let us go through our usual steps and see how we should deal with this.
Step 1: Draw a labelled diagram of the triangle of decomposed vectors.
Step 2: Use the argument \(\boldsymbol{\theta}\) to find the relevant angle \(\boldsymbol{A}\) in this triangle.
In this case, the argument \(\theta\) is the same angle that is within the triangle, i.e.
\[A = \tan^{-1} \left(\frac{4}{3}\right)\]
\[\,\]
Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{p_x}| = |\vec{p}|\cos A} \) and \(\boldsymbol{|\vec{p_y}| = |\vec{p}|\sin A} \).
It seems that we need to take a different approach for this step as we are not allowed to use a calculator to determine \(\cos A\) and \(\sin A\).
If we could use a calculator, we could of course directly find what the angle is:
\begin{align}A &= \tan^{-1} \left(\frac{4}{3} \right)\\ &\approx 53.13^{\circ}\end{align}
Using this angle, as well as the magnitude, we could then find the components.
\[\begin{align}|\vec{p_x}| &= |\vec{p}| \cos A \\&= 10 \cos (53.13^{\circ}) \\& \approx 10(0.6) \\&= 6\end{align}\]
\[\begin{align}|\vec{p_y}| &= |\vec{p}| \sin A \\&= 10 \sin (53.13^{\circ})\\& \approx 10(0.8)\\&= 8\end{align}\]
But how do we find this without a calculator? Well, we can use this weird way of writing the angle to our advantage!
We can draw the following separate triangle which correctly includes this same angle \(A\).
This is indeed the same \(A \) that we are looking for as \(\tan A = \displaystyle\frac{4}{3}\) for this angle also.
What is the benefit of that? Well, we can now more precisely write down what \(\cos A \) and \(\sin A \) are as fractions rather than as irrational decimals!
To do this, we first need to calculate the hypotenuse \(H\) of this triangle using Pythagoras’ Theorem.
\begin{align}H &= \sqrt{3^2+4^2} \\&= 5\end{align}
Now, we can easily find \(\cos A\) and \(\sin A\) by using their definitions.
\[\begin{align} \cos A &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}}\\& =\frac{3}{5}\end{align}\]
\[\begin{align} \sin A &= \frac{\mbox{opposite}}{\mbox{hypotenuse}}\\& =\frac{4}{5}\end{align}\]
Using these values, we can now directly calculate our components as
\[\begin{align}|\vec{p}_x| &= |\vec{p}| \cos A \\& = 10 \left(\frac{3}{5}\right) \\&= 6\end{align}\]
\[\begin{align}|\vec{p}_y| &= |\vec{p}| \sin A \\&= 10 \left(\frac{4}{5}\right) \\&= 8\end{align}\]
which agrees with what we obtained above when using a calculator.
In general, any time that we are given an angle in this form, we shall use this strategy.
\[\,\]
Step 4: Calculate \(\boldsymbol{p_x}\) and \(\boldsymbol{p_y}\) using \(\boldsymbol{|\vec{p_x}|}\) and \(\boldsymbol{|\vec{p_y}|}\), i.e. be careful with signs!
According to the above diagram, \(\vec{p_x}\) is in the same direction as \(\vec{i}\) and \(\vec{p_y}\) is in the same direction as \(\vec{j}\). Therefore
\begin{align}p_x = 6 && p_y = 8\end{align}
\[\,\]
Step 5: Write \(\boldsymbol{\vec{p} =p_x \vec{i} + p_y \vec{j}}\).
\[\vec{p}=6\vec{i} + 8\vec{j}\]
