Vectors
Topics covered in this section are:
* Further topics would be added based on need.1. Introduction
In our everyday-life, we encounter three quantities - Scalars, Vectors and Higher rank Tensors. To say, Scalar and Vectors are zero and 1st rank tensors, respectively. But, we won't be discussing higher rank tensors here.
- Scalar:
- Vectors:
- Equal vectors and Opposite/ Negative Vectors:
Definition: Scalars are the quantities which have only magnitudes but no direction.
For example, distance, speed, charge, volume, mass, numbers etc.
Now, we can perform additon, subtraction, multiplication and divison on the scalars in the same way we perform these operations on numbers (algebriac operations). Let us consider two scalar values, say \(5C\) and \(-1C\) of charges, where \(C\) is Coulomb.
If they are put together, then we say that \((5-1)C= 4C\), of charge is at that place.
Similarly, other operations are performed.
Definition: Vectors are any physical quantity which have direction as well as magnitude.
For example, displacement, velocity, force, angular speed, etc.
Let's take the example of force. When we push an object with a force, we can feel how hard we are pushing it. This is in a sense a measure of magnitude of the force. Further, we can push the object from different directions as shown in Figure 1.1, this is the direction of the force.
Vectors are denoted by a boldface letter such as \(\textbf{A}\) or with an arrow on head of the letter i.e. as \(\vec{A}\). We would take up the arrow on head notation in rest of the discussion. In diagrams, we represent vectors using arrows, such that its magnitude (denoted as \(|\vec{A}|\) or simply \(A\)) is proportional to the length of arrow and direction indicated by the head of the arrow.
This can be seen in the Figure 1.1.
Two vectors \(\vec{A}\) and \(\vec{B}\) are said to be equal iff the have they same magnitude as well as direction, as shown in Figure 1.2.
And, one vector is said to be opposite of other vector, say \(\vec{C}\) or the negative of that vector iff they have the same magnitude but point in opposite directions and is denoted as \(-\vec{C}\). This is also shown in Figure 1.2.
But, the arithmetic operations on the vectors doesn't operate in the normal algebriac way. We now move to discuss these operations in detail.
2. Addition of Vectors
Additon of two vectors is performed using two laws:
- Triangle Law:
- Parallelogram Law:
The resultant (or sum of) the vectors \(\vec{A}\) and \(\vec{B}\), \(\vec{R}\) is the vector from the the tail of \(\vec{A}\) to the head of \(\vec{B}\).
This is known as Triangle law of vector addition as the three vectors involved are forming a triangle for the rule to apply.
Example:
If we apply force \(\vec{A}\) and \(\vec{B}\) on a plate as shown in Figure 2.2, then the plate would have a net motion in the direction of \(\vec{R}\) as shown. This is the insight over the triangle law. One may notice that the vectors are pararllelly translated and then made into triangle of Figure 2.1. This shows that the vectors, irrespective of their position in space, represent the exactly same thing as long as their magnitude and direction are the same.
Again, consider the vectors \(\vec{A}\) and \(\vec{B}\) arranged (by parallel translation) such that the tail of \(\vec{A}\) and \(\vec{B}\) coincides. Now, the parallelogram law can be stated as follows:
The resultant or the sum of two vectors, \(\vec{A}\) and \(\vec{B}\) is the diagonal of the parallelogram formed by \(\vec{A}\) and \(\vec{B}\) as adjacent sides, directed from the point of coincidence of tail of the two vectors to the opposite vertex of the parallelogram i.e. the vector \(\vec{R}\) in the Figure 2.3.
This is the same as the Triangle law as we can parallelly translate the vector \(\vec{B}\) to the right side of the paralleogram and form the triangle as was in the Figure 2.1. But, the two of them have their own significance as we use Parallelogram law to find the magnitude of resultant vector and triangle law in expressing one vector as difference of two, to say one of many uses.
Example :
Consider the Figure 2.4 as shown. It is just a reimagination of the Triangle Law Example using the Paralellogram Law. Some may find this one more intuitive as the forces \(\vec{A}\) and \(\vec{B}\) are shown so as to form a parallelogram, imagining the forces to be pull forces instead of push forces as was in the Figure 2.2. These eamples illustrates the variety of methods, each having their own beauty, to analyse a given situation.
We now move towards quantifying the above analysis.
- Magnitude and direction of Resultant vector:
- Difference of two vectors:
Consider two vectors \(\vec{a}\) and \(\vec{b}\) (with magnitudes \(a\) and \(b\) respectively), with an angle \(\theta\) between the two vectors, when they are joined tail to tail, as shown in the Figure 2.5.
It can be noted that whenever we say angle between two vectors, we talk about this angle only (i.e. when they are joined tail to tail.)
Using parallelogram law, we can draw the parallelogram \(PSRT\) and assign the diagonal \(PQ\) as the resultant vector \(\vec{R}\). Also, \(SR = PT = b\). Further, using simple trigonometry, we get the side lengths of the right angled triangle, \(\triangle RQS\) as : $$ \begin{align} SQ &= b\cos{\theta} \\ RQ &= b\sin{\theta} \end{align} $$
Now, from the right angled triangle \(\triangle PQR\) and using pythagoras theorem,
$$ \begin{align} |R|^2 &= (a+b\cos{\theta})^2 + (b\sin{\theta})^2 \\ \implies R^2 &= a^2+ b^2 \cos^2{\theta} + 2ab\cos{\theta} + b^2\sin^2{\theta} \\ \implies R &= \sqrt{a^2+b^2 + 2ab\cos{\theta}}\hspace{1cm} (2.1) \end{align} $$ In the last step we used the identity, \(\sin^2{\theta} + \cos^2{\theta} = 1\). Equation 2.1 gives the magnitude of the resultant of two vectors described above.Now, for the direction of the resultant vector, say it is at an angle \(\phi\) with the vector \(\vec{a}\) as shown in Figure 2.5. Then, from \(\triangle PQR\), we get: $$ \begin{align} \tan{\phi} &= \frac{QR}{PQ} = \frac{b\sin{\theta}}{a+b\cos{\theta}} \\ \implies \phi &= \tan^{-1}\left(\frac{b\sin{\theta}}{a+b\cos{\theta}}\right) \hspace{1cm} (2.2) \end{align} $$
Special Case\(\left(|\vec{a}|=|\vec{b}|\right)\): When the two vectors \(\vec{a}\) and \(\vec{b}\) have the same length, say \(a\) and angle between them be \(\theta\). Then, using equation 2.1, the resultant would have magnitude as: $$ \begin{align} R &= \sqrt{a^2 + a^2 + 2a^2 \cos{\theta}}\\ \implies R &= \sqrt{2a^2(1+\cos{\theta})}\\ \implies R &= \sqrt{2a^2\times 2\cos^2{\left(\frac{\theta}{2}\right)}}\\ \implies R &= 2a\cos{\left(\frac{\theta}{2}\right)} \hspace{1cm} (2.3) \end{align} $$ In this case, the resultant vector would be along the angle bisector of between the two vectors i.e. \(\phi=\frac{\theta}{2}\).
The difference of two vectors is the sum of the first vector and the negative of the vector being subtracted. Consider the same vectors \(\vec{a}\) and \(\vec{b}\) again as shown in Figure 2.6.
We have to find the difference \(\vec{a}-\vec{b}\). So, we rewrite it as \((\vec{a}+(-\vec{b}))\). And clearly, this represents the diagonal of the parallelogram formed by \(\vec{a}\) and \(-\vec{b}\) as its resultant, as shown in figure.
Since, here the angle between \(\vec{a}\) and \(\vec{b}\) is \(\theta\), so the angle between \(\vec{a}\) and \(-\vec{b}\) (the vectors which originally have to be "added") would be \((\pi-\theta)\). Now, using the equation 2.1 for the magnitude of \(\vec{R}\), we get:
The magnitude of the resultant vector after subtracting the two vectors:
$$ \begin{align} |\vec{R}| = |\vec{a}-\vec{b}| &= \sqrt{a^2 + b^2 + 2ab \cos(\pi-\theta)} \\ R &= \sqrt{a^2 + b^2 - 2ab\cos\theta}\\ (\because \cos(\pi-\theta) &= -\cos{\theta}) \end{align} $$If you have noticed, we have taken the magnitude of the negative vector, \(-\vec{b}\) to be the same \(b\). This is because the length of the vector remains the same even after negating it. The angle \(\phi\) that gives the direction of \(\vec{R}\) is calculated in the same way.