Addition and Subtraction of Vectors
If we have 4 kg of rice in a bag and add 3 kg of rice to it, we will have a total of 4 + 3 = 7 kg of rice in the bag. This is how scalars add, but vectors do not add like this because their direction may be different.Addition of Vectors: We have already added two position vectors to define a displacement vector. We will now learn how this is done. We will illustrate the addition of vectors by taking the example of displacement vectors in a plane.
The addition of vectors in one dimension is straightforward. Suppose we give two displacements PQ and QR one after another to the same object in the same direction. Figure (a) shows a displacement PQ of magnitude, say 4 cm to the â€˜rightâ€™ applied to an object originally at the point P, which brings it to a position Q.
Now a second displacement QR of magnitude, say, 3 cm is applied to it in the same direction, which brings the object to the position R. Obviously the same result could be obtained if a single displacement PR of magnitude 4 + 3 = 7 cm were given to the object in the same direction in which the individual displacements are given. In the language of vectors, the result is written as
PQ + QR = PR
or A + B = C
Addition of vectors in one dimension 
Figure (b) shows how vectors in opposite directions are added. The result is a vector C = PR = 1 cm in the direction of the bigger of the two vectors A and B. This result is also written as
A + B = C
Since the vector C alone produces the same result as the vectors A and B together do, we say that C is the resultant (or sum) of vectors A and B. Thus it is clear that vectors do not add as the scalars do.
We will now consider addition of vectors in two dimensions. We begin with the simplest case of two vectors at right angles to each other.
We will now consider addition of vectors in two dimensions. We begin with the simplest case of two vectors at right angles to each other.
Case Iâ€”Addition of Two Vectors at Right Angles to Each Other
Suppose a man walks a distance of 4 m from P to Q in the east and then walks a distance of 3 m from Q to R in the north. These two perpendicular displacements PQ and QR are respectively represented by two vectors A and B.For convenience, 1 cm can be taken to represent a distance of 1 m. If we join P and R and measure the length of the vector C (or PR), we will find it to be 5 cm which represents a displacement of magnitude 5 m. If instead of walking a distance of 4 m from P to Q and then walking a distance of 3 m from Q to R (a total distance of 4 + 3 = 7 m), the man walks a distance of 5 m from P to R, he would reach the same destination (which is R).
Thus, the displacement of 5 m along PR produces the same effect as the displacement of 4 m along PQ and 3 m along QR together do. We say that the vector PR (or C) is the resultant (or sum) of two vectors PQ (or A) and QR (or B). In the language of vectors, we may write
PQ + QR = PR
or A + B = C (4.1)
Thus the method (or rule) for vector addition is: shift (if necessary) one vector parallel to itself until its tail is at the head of the other vector. The resultant is then given by the vector drawn from the tail of the first vector to the head of the second.
Vector B shifted parallel to itself 
Case IIâ€”Addition of Two Vectors inclined to Each Other:
Consider two vectors A and B represented by PQ and QR respectively and inclined to each other at a certain angle as shown in the given figure. Shift B parallel to itself till the tail O of B touches the head Q of A as shown in figure (b). Then the resultant vector is given by PR which Joins the tail of A to the head of B. Now complete the parallelogram PQRS. Since PR is the resultant of PQ and QR, we havePQ + QR = PR
But PS = QR = B (PS and QR are equal and parallel)
PQ + PS = PR
A + B = C
Now PQ and PS (which represent the two vectors A and B respectively) constitute the two adjacent sides of the parallelogram, meeting at point P and the resultant vector PR is the diagonal of the parallelogram passing through the same point P.
This procedure of finding the resultant of two vectors is known as the parallelogram law of vector addition and may be stated as follows:
If the two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.
Notice that A + B = B + A. The reason is the following: Instead of shifting B, we could shift A parallel to itself. If you do that, you can easily see that the resultant remains the same, no matter which of the two vectors is shifted.
The parallelogram law of vector addition yields the triangle law of vector addition. It follows from figure (b) that in triangle PQR, vector PR is the resultant of vectors PQ and QR. Hence the triangle law of vector addition may be stated as follows: If the two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented in magnitude and direction by the third side of the triangle taken in the opposite direction.
The magnitude and the direction of the resultant vector can be found as follows: Consider two vectors A and B. Let C is the resultant vector of A and B. Let Î² be the angle subtended by the resultant vector C with vector A.
But PS = QR = B (PS and QR are equal and parallel)
PQ + PS = PR
A + B = C
Now PQ and PS (which represent the two vectors A and B respectively) constitute the two adjacent sides of the parallelogram, meeting at point P and the resultant vector PR is the diagonal of the parallelogram passing through the same point P.
This procedure of finding the resultant of two vectors is known as the parallelogram law of vector addition and may be stated as follows:
If the two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.
Notice that A + B = B + A. The reason is the following: Instead of shifting B, we could shift A parallel to itself. If you do that, you can easily see that the resultant remains the same, no matter which of the two vectors is shifted.
The parallelogram law of vector addition yields the triangle law of vector addition. It follows from figure (b) that in triangle PQR, vector PR is the resultant of vectors PQ and QR. Hence the triangle law of vector addition may be stated as follows: If the two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented in magnitude and direction by the third side of the triangle taken in the opposite direction.
The magnitude and the direction of the resultant vector can be found as follows: Consider two vectors A and B. Let C is the resultant vector of A and B. Let Î² be the angle subtended by the resultant vector C with vector A.
To find the magnitude of the resultant, produce PQ and from R draw a perpendicular RT to meet PQ produced at point T. From rightangled triangle PRT, we have
(PR)^{2} = (PT)^{ 2} + (RT)^{ 2} = (PQ + QT)^{ 2} + (RT)^{ 2}
= (PQ)^{ 2} + 2(PQ)(QT) + (QT)^{ 2} + (RT)^{ 2} (i)
In rightangled triangle QRT we have
RT = QR sin Î± and QT = QR cos Î± (ii)
Substituting (ii) in (i), we have
(PR)^{ 2} = (PQ)^{ 2} +2(PQ)(QR) cos Î± + (QR cos Î± )^{2} + (QR sin Î± )^{2}
But (QR cos Î± )^{2} + (QR sin Î± )^{2} = (QR)^{2} (cos^{2 }^{Î± } + sin^{2 }^{Î± } ) = (QR)^{ 2}
âˆ´ (PR)^{ 2} = (PQ)^{ 2} + (QR)^{ 2} + 2(PQ)(QR) cos Î±
or C^{2} = A^{2} + B^{2} + 2AB cos Î±
or C = (A^{2} + B^{2} + 2AB cos Î± )^{1/2}
Thus, if the magnitudes A and B of vectors A and B and the angle Î± between them are known, the magnitude C of the resultant vector C can be calculated by using this relation.
To find the direction of the resultant vector C, we have to determine angle Î² which the resultant C subtends with vector A. From triangle PRT we have
But RT = QR sin Î± and QT = QR cos Î±
Thus, knowing A, B and Î±, tan Î² and hence Î² is determined. The value of angle Î² gives the direction of the resultant vector C.
In the special case, when the two vectors A and B are perpendicular to each other, angle Î± = 90Â°
Hence Cos a = Cos 90 = 0 and Sin Î±=Sin 90 = 1. In this, the magnitude and the direction of the resultant vector C are respectively given by
C = (A^{2} + B^{2})^{ 1/2}
and
(PR)^{2} = (PT)^{ 2} + (RT)^{ 2} = (PQ + QT)^{ 2} + (RT)^{ 2}
= (PQ)^{ 2} + 2(PQ)(QT) + (QT)^{ 2} + (RT)^{ 2} (i)
In rightangled triangle QRT we have
RT = QR sin Î± and QT = QR cos Î± (ii)
Substituting (ii) in (i), we have
(PR)^{ 2} = (PQ)^{ 2} +2(PQ)(QR) cos Î± + (QR cos Î± )^{2} + (QR sin Î± )^{2}
But (QR cos Î± )^{2} + (QR sin Î± )^{2} = (QR)^{2} (cos^{2 }^{Î± } + sin^{2 }^{Î± } ) = (QR)^{ 2}
âˆ´ (PR)^{ 2} = (PQ)^{ 2} + (QR)^{ 2} + 2(PQ)(QR) cos Î±
or C^{2} = A^{2} + B^{2} + 2AB cos Î±
or C = (A^{2} + B^{2} + 2AB cos Î± )^{1/2}
To find the direction of the resultant vector C, we have to determine angle Î² which the resultant C subtends with vector A. From triangle PRT we have
But RT = QR sin Î± and QT = QR cos Î±
In the special case, when the two vectors A and B are perpendicular to each other, angle Î± = 90Â°
Hence Cos a = Cos 90 = 0 and Sin Î±=Sin 90 = 1. In this, the magnitude and the direction of the resultant vector C are respectively given by
C = (A^{2} + B^{2})^{ 1/2}
and
Case IIIâ€”Addition of more than Two Vectors; Polygon Law of Vector Addition:
Suppose we wish to determine the resultant of four vectors A, B, C and D. These vectors may be the four displacements or forces applied to a body. The procedure to find the resultant is that we shift vectors B, C and D parallel to themselves so that the tail of B touches the head of A, the tail of C is at the head of B and the tail of D is at the head of C. The result of this shifting:The resultant of vectors A, B, C and D is given by vector E as
E = A + B + C + D
= PQ + QR + RS + ST
= PR + RS + ST ( PR = PQ + QR)
or E = PS + ST = PT ( PS = PR + RS)
Thus, if a number of vectors are represented in magnitude and direction by the sides of a polygon, taken in order, then the resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order. This is called the polygon law of vector addition.
Note: Equation PQ + QR = PR or A + B = C is a vector equation. It is important to understand that this equation does not say that the sum of the magnitudes of A and B is equal to the magnitude of C. This kind of simple addition holds only for scalars. In fact, a vector equation prescribes an operation.
The equation A + B = C simply says that the operation of moving a body from P to Q and then from Q to R is equivalent to a simple operation of moving it from P to R. It may be mentioned that in writing A + B = C, it is assumed that all vectors lie in the same plane which, in our case, is the plane of the paper.
E = A + B + C + D
= PQ + QR + RS + ST
= PR + RS + ST ( PR = PQ + QR)
or E = PS + ST = PT ( PS = PR + RS)
Note: Equation PQ + QR = PR or A + B = C is a vector equation. It is important to understand that this equation does not say that the sum of the magnitudes of A and B is equal to the magnitude of C. This kind of simple addition holds only for scalars. In fact, a vector equation prescribes an operation.
The equation A + B = C simply says that the operation of moving a body from P to Q and then from Q to R is equivalent to a simple operation of moving it from P to R. It may be mentioned that in writing A + B = C, it is assumed that all vectors lie in the same plane which, in our case, is the plane of the paper.
Subtraction of Vectors
Suppose we wish to subtract a vector B from a vector A. SinceChoose a convenient scale and draw the vectors A and B. If B is to be subtracted from A, draw the vector negative of B, i.e. draw the vector â€“B. Now shift the vector â€“B parallel to itself so that the tail of â€“B is at the head of A. Vector C is the sum of vectors A and â€“B,
C = A + (B) = A â€“ B

Subtraction of vector B from vector A (C = A â€“ B) 