Scalars and Vectors | Complete notes | Important Questions | Class 11 (Physics)

Physical quantities ( measurable quantities ) on the basis of whether they represent magnitude or both magnitude and direction; are broadly classified as scalars and vectors.
What are scalars?
Scalars are quantities which have magnitude but no direction. A car moving on a road at 80 km/h is said to have a speed of 80 km/h. Speed is a quantity which has no direction but only magnitude, like mass or density or temperature . These quantities are thus the examples of scalars.
What are vectors?
Vectors are quantities which have both magnitude and direction. A car moving on a road at 80 km/h in north-east direction is said to have a velocity of 80 km/h NE (i.e., north-east ). Velocity is a quantity which has both magnitude and direction, like displacement or force or acceleration. These quantities are thus the examples of vectors.
Let's point some of the differences between scalars and vectors.

Addition : Scalars obey the ordinary rules of algebra but vectors do not. We should add vectors geometrically. By adding the two vectors, we get a quantity called resultant vector . For example, if we consider two different masses i.e., 4 kg and 3 kg , by adding we obtain the total mass (resultant) to be 7 kg as we don't need to worry while adding the masses (i.e., scalars). However, while adding vectors, it is not that simple (it's simple but not straight-forward as this one!). We need to follow certain rules / laws of addition (will be discussed very soon!).
Representation : Scalars are represented by ordinary letters (for.eg., A). While representing vectors, we use the bold-faced letters ( A ) or letters having arrow over them $\vec{A}$.

Some examples of scalars and vectors in detail

Distance and displacement : Distance is simply the length covered by an object. It doesnot require the specification of direction. But, displacement is a vector quantity: it's magnitude is the distance between the initial and final positions and it's direction from initial position can be represented by plus or minus sign if the motion is along a single axis. Consider that a boy has travelled 5 km on a straight road from the home to the gym, but returned back 2 km after some time. In this case, the total length covered by the object is 5 + 2 = 7 km and this is the distance covered by him. However, the displacement of a boy is +5 + (-2) = +3 km. Here, plus sign indicates the initial direction of his travel and minus sign represnts the opposite direction.

Speed and velocity : Speed is a scalar quantity and velocity is a vector quantity ( already discussed ). We can make a clear distinction between this two by taking an example of a car moving around a circular track at say 80 km/h as in fig. At every point on the track, the speed is the same - it is 80 km/h. At every point, however, the velocity is different. At A, B, or C, for example, the velocity is in the direction of corresponding tangent AP, BQ or CR. So even though they have the same magnitude , the three velocities are all different because they point in different directions.

Some of the frequently encountered vectors are,

Unit vector : Unit vector has a magnitude of unity (i.e., 1). Any unit vector $\^{A}$ is defined as, \[\hat {A} = {\vec{A}\over A}\] Here, A is the magnitude for $\vec{A}$. Unit vectors are generally used to indicate the direction of a vector. In cartesian coordinates (i.e., coordinate system with x, y and z axis ), x- axis is represented by $\color{red}\hat {i}$, y-axis is represented by $\color{blue}\hat {j}$ and z-axis is represented by $\color{green}\hat {k}$ as in fig.

Parallel vectors : Two vectors having the same direction are called parallel vectors. Their magnitude does not necessarily need to be equal. In the figure below, $\vec{A}$ and $\vec{B}$ are parallel vectors.

Equal vectors : Two vectors having same magnitude and direction are called equal vectors. Every equal vectors are parallel but every parallel vectors may not be equal. In the figure below, $\vec{A}$ and $\vec{B}$ are equal vectors.

Vector addition and subtraction for two vectors

Triangle law of vector addition

Triangle law of vector addition states that, " If two sides of a triangle completely represent two vectors both in magnitude and direction taken in same order, then the third side taken in opposite order represents the resultant of the two vectors both in magnitude and direction."
Consider two vectors $\color{red}\vec{A}$ and vector $\color{blue}\vec{B}$ as in figure. Let us represent these two vectors by the the sides OP and PQ respectively of a triangle OPQ. Let θ be the angle between them. From triangle law of vector addition, the third side OQ represents the resultants of this vector $\color{green} \vec{R}$ both in magnitude and direction. $\color{green}\vec{R}$ makes an angle of φ with $\color{red}\vec{A}$. From the vector addition, $\color{green} \vec{R} = \color{red}\vec{A}+\color{blue}\vec{B}$ At first, we find the magnitude of $\color{green} \vec{R}$, For this, let the perpendicular be dropped from Q to meet extension of OP at S. Now, taking the right angled ΔOQS, Using the pythagoras theorem, $(OQ)^2 = (OS)^2 + (QS)^2 $ $(R)^2 = (OP+PS)^2 + (QS)^2$ $(R)^2 = (A+PS)^2 + (QS)^2$ Now, we use the right angled triangle ΔPQS to find the values PS and QS, $sin\theta = {QS \over PQ}$ ∴ $QS = B sin\theta$ and similarly, $PS = B cos\theta$ Using the value of PS and QS to find R, $(R)^2 = (A+B cos\theta)^2 + (B sin\theta)^2$ $(R)^2 = A^2+ABcos\theta+B^2cos^2\theta + B^2 sin^2\theta$ $(R)^2 = A^2 + B^2 + 2ABcos\theta$ ∴ $R = \sqrt{A^2 + B^2 + 2ABcos\theta}$ So, the magnitude of resultant vector is obtained. Now, we find the direction of $\vec{R}$ i.e. $\phi$ by taking the right angled Δ OQS, $tan\phi = {QS \over OS} = {Bsin\theta \over {A+Bcos\theta}}$ ∴ $\phi = tan^{-1} \left( {Bsin\theta \over {A+Bcos\theta}}\right)$ Now, we consider three cases and obtain the magnitude and direction of the resultant vector in each of this cases:
Case I: When two vectors are acting in the same direction (i.e., $\theta = 0^0$).

In this case , $cos\theta = cos 0^0 = 1$ and the magnitude of resultant becomes, $R = \sqrt{A^2 + B^2 + 2AB}$ $R = \sqrt{(A + B)^2}$ $R = A + B$ This is the maximum value of the resultant vector since the maximum value of cos$\theta$ is 1 (i.e., $-1 \leq cos\theta \leq +1$). So , $R_{max} = A + B$ Now, we find the direction of the resultant vector ($\vec{R}$) as, $\phi = tan^{-1}\left({Bsin\theta \over {A+Bcos\theta}}\right)$ $\phi = tan^{-1} (0)$ $\phi = 0^0$ So, the resultant vector will take the same direction of the given two vectors.
Case II: When two vectors are acting in opposite direction (i.e., $\theta = 180^0$).

In this case , $cos\theta = cos 0^0 = -1$ and the magnitude of resultant becomes, $R = \sqrt{A^2 + B^2 - 2AB}$ $R = \sqrt{(A - B)^2}$ $R = A - B$ This is the maximum value of the resultant vector since the maximum value of $cos\theta$ is 1 (i.e., $-1 \leq cos\theta \leq +1$). So , $R_{min} = A - B$ Now, we find the direction of the resultant vector $\vec{R}$ as, $\phi = tan^{-1}\left({Bsin\theta\over {A+Bcos\theta}}\right)$ $\phi = tan^{-1} \left({B sin180^0 \over {A-B}}\right)$ Here, we consider two cases as follows,
a. When A \gt B: $\phi = tan^{-1} \left({B sin180^0 \over {A-B}}\right)$ $\phi = 0^0$ In this case, resultant vector takes the direction of $\vec{A}$.
b. When B>A: $\phi = tan^{-1} \left({B sin180^0 \over {A-B}}\right)$ $\phi = 180^0$ In this case, resultant vector takes the direction of $\vec{B}$ because $\vec{B}$ is in opposite direction (making $180^0$) to $\vec{A}$. Hence, when the two vectors acts in opposite direction, the resultant will take the direction of a greater vector.
Case III: When two vectors are at right angles to each other (i.e., $\theta = 90^0$).

In this case, $cos\theta = cos 90^0 = 1$ and the magnitude of resultant becomes, $R = \sqrt{(A^2 + B^2)}$ This value of resultant lies somewhere in between $R_{max}$ and $R_{min}$. Now, we find the direction of the resultant vector $\vec{R}$ as, $\phi = tan^{-1}\left({Bsin\theta \over {A+Bcos\theta}}\right)$ $\phi = tan^{-1} \left({B \over A}\right)$

Parallelogram law of vector addition

Parallelogram law of vector addition states that, " If two vectors acting simultaneously at a point are represented both in magnitude and direction by two adjacent sides of a parallelogram drawn from the point, then the diagonal of the parallelogram passing through that point represents the resultant of the two vectors both in magnitude and direction."
Consider two vectors $\color{red}\vec{A}$ and vector $\color{blue}\vec{B}$ as in figure. Let us represent these two vectors by the the sides PQ and PT respectively of a parallelogram PQST. Let θ be the angle between them. From parallelogram law of vector addition, the diagonal PR represents the resultants of this vector ($\color{green} \vec{R}$) both in magnitude and direction. $\color{green}\vec{R}$ makes an angle of $\phi$ with $\color{red}\vec{A}$. Now, use the same procedure as in triangle law of vector addition. For complete derivation, watch the video on Physics in Depth.

Addition of more than two vectors / Composition of vectors (Polygon law of vector addition)

The determination of the resultant of the system of more than two vectors is called compostion of vectors . This may be done by the help of polygon law of vector addition . It states that, "if a polygon be drawn, placing the tail-end of each suceeding vector at the head of the arrow-end of the preceding one, as in fig., their resultant $\vec{R}$ is drawn from the tail-end of the first to the head or the arrow-end of the last."

Rectangular components of a vector

Let us consider the vector $\vec{a}$ represented by OP in x-y plane. Suppose it makes an angle $\theta$ with the x-axis. Dropping perpendicular at x-axis and y-axis at points X and Y respectively. Here, $a_x$ represents the x-component of $\vec{a}$ and $a_y$ represents the y-component of $\vec{a}$ as in figure. In right angled triangle ΔOPX, $sin\theta = {PX \over OP}$ here, $ PX = OY = a_y $, so , $sin\theta = \frac{a_y}{a} \implies a_y = a sin\theta$ Similarly, $a_x = a cos\theta$ Squaring and adding, we get, $a = \sqrt{a_x^2 + a_y^2}$ and dividing $a_y$ and $a_x$, we get, $tan\theta = \frac{a_y}{a_x}$
Jump to Vector Multiplication.

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