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What is the Statement for the Polygon Law of Vector Addition?
The sum (resultant) of two vectors can be determined by the law of triangle for vector addition. If the number of vectors is three or more, then by applying the law of triangle successively the law of polygon of vectors is derived. By the application of the law of polygon of vectors, addition of any number of vectors is possible.
Suppose the resultant of four vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\) is to be determined [Fig.(a)]. For this, vector \(\overrightarrow{O A}\) equal to \(\vec{a}\) is drawn from any point O [Fig.(b)). Now taking A as ini-tial point, \(\overrightarrow{A B}\) equal to \(\vec{b}\) is drawn. Similarly, \(\overrightarrow{B C}\) equal to \(\vec{c}\) and \(\overrightarrow{C D}\) equal to \(\vec{d}\) are drawn one after another. Then, the vector \(\overrightarrow{O D}\) drawn by joining the initial point O of the first vector and the terminal point D of the last vector represents the resultant of the vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\).
This is called the law of polygon of vectors and with this law the resultant of any number of vectors can be determined.
Statement: If the magnitudes and directions of a number of vectors are represented by the sides of a closed polygon, taken in order, then the last side of the polygon, taken in the opposite order, represents the magnitude and direction of the resultant of the vectors.
The coplanarity of the vectors is not necessary for the validity of polygon law. Vectors can be added by using the polygon law irrespective of their number and sequence.
Proof of the law of polygon of vectors: Let the magnitudes and directions of the vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), \(\vec{d}\) be represented, by the arms of the polygon OABCD, taken in order: \(\overrightarrow{O A}\), \(\overrightarrow{A B}\), \(\overrightarrow{B C}\), \(\overrightarrow{C D}\) [Fig.].
From triangle law,
in triangle OAB, \(\overrightarrow{O B}\) = \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\),
in triangle OBC, \(\overrightarrow{O C}\) = \(\overrightarrow{O B}\) – \(\overrightarrow{B C}\) = \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) + \(\overrightarrow{B C}\) and in triangle OCD,
\(\overrightarrow{O D}\) = \(\overrightarrow{O C}\) + \(\overrightarrow{C D}\) = \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) + \(\overrightarrow{B C}\) + \(\overrightarrow{C D}\)
∴ \(\overrightarrow{O D}\) = \(\vec{R}\) = \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) + \(\overrightarrow{B C}\) + \(\overrightarrow{C D}\) = \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) + \(\vec{d}\)
Hence, \(\overrightarrow{O D}\) is the resultant of the vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\).
The arm \(\overrightarrow{O D}\) is the remaining arm of the polygon taken in the opposite order. This proves the law and also shows that it is nothing but an extension of the law of triangle of vector addition.
Corollary: In the pentagon OABCD,
\(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) + \(\overrightarrow{B C}\) + \(\overrightarrow{C D}\) = \(\overrightarrow{O D}\)
or, \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) + \(\overrightarrow{B C}\) + \(\overrightarrow{C D}\) – \(\overrightarrow{O D}\) = 0
or, \(\overrightarrow{O A}\) + \(\overrightarrow{A B}\) + \(\overrightarrow{B C}\) + \(\overrightarrow{C D}\) + \(\overrightarrow{D O}\) = 0.
In other words, If three or more vectors can be represented by the sides of a closed polygon, taken in order, the resultant of the vectors must be zero.
The physical quantities which have both magnitude and direction and obey all the laws of vector addition such as—
- triangle law of vectors,
- parallelogram law of vectors and
- polygon law of vectors, are called vector quantities.
Numerical Examples
Example 1.
If \(\overrightarrow{\boldsymbol{a}}\) + \(\overrightarrow{\boldsymbol{b}}\) = \(\overrightarrow{\boldsymbol{c}}\) and a + b = c, find the angle between \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\).
Solution:
Let the angle between \(\overrightarrow{\boldsymbol{a}}\) and \(\overrightarrow{\boldsymbol{b}}\) be α.
Here, \(\vec{a}\) + \(\vec{b}\) = \(\vec{c}\)
∴ c2 = a2 + b2 + 2ab cosα ….. (1)
Again, a + b = c or, a2 + b2 + 2ab = c2 ….. (2)
Hence from (1) and (2),
a2 + b2 + 2ab = a2 + b2 + 2ab cosα
or, 2ab = 2ab cosα or, cosα = 1 = cos0°
∴ α = 0°
Example 2.
Can the magnitude of the resultant of two equal vectors be equal to the magnitude of each of the vectors? Explain. [HS 2000]
Solution:
Let the magnitude of each vector and of the resultant be a and the angle between the vectors be α.
Hence, a2 = a2 + a2 + 2a ᐧ a cosα
or a2 = 2a2(1 + cosα) or, 2(1 + cosα) = 1
or cosα = \(\frac{1}{2}\) – 1 = –\(\frac{1}{2}\) = cos 120°
∴ α = 120°
Hence, the magnitude of the resultant of two equal vectors is equal to that of each of the given vectors when they are inclined at an angle of 120° with each other.
Example 3.
2P and P are two vectors inclined to each other at such an angle that if the 1st vector is doubled, the value of the resultant becomes three times. What is the angle between the two vectors?
Solution:
Let the initial resultant be R and the angle between the two vectors 2P and P be α.
∴ R2 = (2P)2 + P2 + 2 ᐧ 2P ᐧ Pcosα
or, R2 = 5P2 + 4P2cosα …. (1)
In the second case, 1st vector = 2P × 2 = 4P and resultant = 3R
∴ (3R)2 = (4P)2 + P2 + 2 ᐧ 4P ᐧ P cosα
or, 9R2 = 17P2 + 8P2 cosα
or, R2 = \(\frac{17}{9} P^2\) + \(\frac{8}{9} P^2\)cosα …. (2)
From (1) and (2),
5P2 + 4P2cosα = \(\frac{17}{9}\)P 2 + \(\frac{8}{9}\)P 2cosα
or (4 – \(\frac{8}{9}\))cosα = \(\frac{17}{9}\) – 5 or, \(\frac{28}{9}\) cosα = –\(\frac{28}{9}\)
or, cosα = -1 = cos 180° ∴ α = 180°
Example 4.
Using vectors, prove that the line joining the mid points of two sides of a triangle, is parallel to the base and half its length.
Solution:
In Fig., from the triangle law of vector addition in ΔABC,
As D and E are the mid-points of sides AB and AC,
\(\overrightarrow{A D}\) = \(\frac{1}{2} \overrightarrow{A B}\) and \(\overrightarrow{A E}\) = \(\frac{1}{2} \overrightarrow{A C}\)
From the triangle law of vector addition in ΔADE
\(\overrightarrow{A D}\) + \(\overrightarrow{D E}\) = \(\overrightarrow{A E}\)
or, \(\frac{1}{2} \overrightarrow{A B}\) + \(\overrightarrow{D E}\) = \(\frac{1}{2} \overrightarrow{A C}\) ….. (2)
From (1) and (2), \(\overrightarrow{D E}\) = \(\frac{1}{2} \overrightarrow{B C}\)
Hence, \(\overrightarrow{D E}\) and \(\overrightarrow{B C}\) are parallel and DE = \(\frac{1}{2}\)BC.
Example 5.
When will the magnitude of the resultant of two equal vectors be
(i) \(\sqrt{2}\) times and
(ii) \(\sqrt{3}\) times the magnitude of each of them?
Solution:
i) Let the value of each vector be a and the angle between the two vectors be α.
∴ In this case, the magnitude of the resultant = \(\sqrt{2}\)a
∴ (\(\sqrt{2}\)a)2 = a2 + a2 + 2a ᐧ acosα
or, 2a2cosα = 0 or, cosα = 0 = cos90°
∴ α = 90°
Hence, the angle between the two vectors will be 90°.
ii) In this case, the magnitude of the resultant = \(\sqrt{3}\)a
∴ (\(\sqrt{3}\)a)2 = a2 + a2 + 2a ᐧ acosα
or, 2a2cosα = a2 or, cosα = \(\frac{1}{2}\) = cos 60°
∴ α = 60°
The angle between the two vectors will be 60°.
Example 6.
The maximum and the minimum values of the resultant of two forces are 15 N and 7 N respectively. If the magnitude of each force is increased by 1 N and these new forces act at an angle 90° to each other, find the magnitude and direction of their resultant.
Solution:
Let the magnitudes of the two forces be P and Q.
As per given condition, P + Q = 15N ……. (1)
and P – Q = 7N ……. (2)
By solving (1) and (2), we get, P = 11 N and Q = 4 N.
When each of the two forces is increased by 1 N in magnitude, the new magnitudes are P = (11 + 1) = 12 N and Q = (4 + 1) = 5 N [Fig.] and since the angle between these new forces is 90°,
new resultant = \(\sqrt{12^2+5^2}\) = 13 N
Let the angle of inclination of the resultant with the force 12N be θ.
Then, tanθ = \(\frac{5}{12}\) or, θ = tan-1\(\frac{5}{12}\)
∴ The magnitude of the resultant is 13 N and it is inclined with the force 12N at an angle tan-1\(\frac{5}{12}\).
Example 7.
The resultant \(\overrightarrow{\boldsymbol{R}}\) of two vectors has magnitude equal to one of the vectors and is at right angle to it. Find the value of the other vector.
Solution:
Let Q be the value of the other vector.
In Fig.
R2 + R2 = Q2 or, Q2 = 2R2
∴ Q = \(\sqrt{2}\)R
Example 8.
The maximum magnitude of the resultant of two vectors, \(\overrightarrow{\boldsymbol{P}}\) and \(\overrightarrow{\boldsymbol{Q}}\) (where P > Q) is x times the minimum magnitude of the resultant. When the angle between P and Q is θ, the magnitude of the resultant is equal to half the sum of the magnitudes of the two vectors. Prove that, cosθ = \(\frac{x^2+2}{2\left(1-x^2\right)}\).
Solution:
P + Q = x(P – Q) (given) or, Q = \(\frac{x-1}{x+1}\) ᐧ P
If R is the resultant of the two vectors when the angle between them is θ, then
R2 = P2 + Q2 + 2PQcosθ …. (1)
Example 9.
Out of two vectors, the larger one is \(\sqrt{2}\) times the smaller one. Show that the resultant cannot make an angle greater than \(\frac{\pi}{4}\) with the larger one.
Solution:
Let the two vectors be \(\sqrt{2}\)Q and Q and the resultant be R. Angle between R and \(\sqrt{2}\)Q is θ and ∠ADB = α [Fig.].
Example 10.
Show that if three forces are represented by the three medians of a triangle, they will be in equllibrium.
Solution:
Let the medians of the ΔABC be AD, BE and CF. Vectors \(\overrightarrow{A D}\), \(\overrightarrow{B E}\) and \(\overrightarrow{C F}\) represent the three forces [Fig.]. We have to show: \(\overrightarrow{A D}\) + \(\overrightarrow{B E}\) + \(\overrightarrow{C F}\) = \(\overrightarrow{0}\).
As per triangle law of vector addition, Fig.
[as sum of 3 vectors, represented by the 3 sides of a triangle taken in order, is zero].