The parallelogram law of vector addition is a method that is used to find the sum of two vectors in vector theory. We study two laws for the addition of vectors - the triangle law of vector addition and the parallelogram law of vector addition. The parallelogram law of vector addition is used to add two vectors when the vectors that are to be added form the two adjacent sides of a parallelogram by joining the tails of the two vectors. Then, the sum of the two vectors is given by the diagonal of the parallelogram.

In this article, we will explore the parallelogram law of the addition of vectors, its formula, statement, and proof. We will learn to apply the law with the help of various examples for a better understanding of the concept.

The sum of two vectors can be determined by vector addition and the parallelogram law of vector addition is a law that makes it easier to determine the resultant sum vector. Suppose a fish is going from one side of the river to the other side of the river along with the vector Q and the river water is flowing in a direction along with the vector P as shown in the figure below.

Now, the net velocity of the fish will be the sum of the two velocities - the velocity of the fish and the velocity of the flow of the river which will be a different velocity. As a result, the fish will move along a different vector which will be the sum of these two velocities. Now, to determine the net velocity, we can consider these two vectors as the adjacent sides of a parallelogram and use the parallelogram law of vector addition to determine the resultant sum vector.

Consider two vectors P and Q with an angle θ between them. The sum of vectors P and Q is given by the vector R, the resultant sum vector using the parallelogram law of vector addition. If the resultant vector R makes an angle 蠒 with the vector P, then the formulas for its magnitude and direction are:

Let us first see the statement of the parallelogram law of vector addition:

Statement of Parallelogram Law of Vector Addition: If two vectors can be represented by the two adjacent sides (both in magnitude and direction) of a parallelogram drawn from a point, then their resultant sum vector is represented completely by the diagonal of the parallelogram drawn from the same point.

Now, to prove the formula of the parallelogram law, we consider two vectors P and Q represented by the two adjacent sides OB and OA of the parallelogram OBCA, respectively. The angle between the two vectors is θ. The sum of these two vectors is represented by the diagonal drawn from the same vertex O of the parallelogram, the resultant sum vector R which makes an angle β with the vector P.

Extend the vector P till D such that CD is perpendicular to OD. Since OB is parallel to AC, therefore the angle AOB is equal to the angle CAD as they are corresponding angles, i.e., angle CAD = θ. Now, first, we will derive the formula for the magnitude of the resultant vector R (side OC).

In right-angled triangle OCD, we have

OC2 = OD2 + DC2

⇒ OC2 = (OA + AD)2 + DC2 --- (1)

In the right triangle CAD, we have

cos θ = AD/AC and sin θ = DC/AC

⇒ AD = AC cos θ and DC = AC sin θ

⇒ AD = Q cos θ and DC = Q sin θ --- (2)

Substituting values from (2) in (1), we have

R2 = (P + Q cos θ)2 + (Q sin θ)2

⇒ R2 = P2 + Q2cos2θ + 2PQ cos θ + Q2sin2θ

⇒ R2 = P2 + 2PQ cos θ + Q2(cos2θ + sin2θ)

⇒ R2 = P2 + 2PQ cos θ + Q2 [cos2θ + sin2θ = 1]

⇒ R = √(P2 + 2PQ cos θ + Q2) → Magnitude of the resultant vector R

Next, we will determine the direction of the resultant vector. We have in right traingle ODC,

tan β = DC/OD

⇒ tan β = Q sin θ/(OA + AD) [From (2)]

⇒ tan β = Q sin θ/(P + Q cos θ) [From (2)]

⇒ β = tan-1[(Q sin θ)/(P + Q cos θ)] → Direction of the resultant vector R

Now, we know the formula to determine the magnitude and direction of the sum of the two vectors. Let us consider a few special cases and substutte the values in the formula:

If vectors P and Q are parallel, then we have θ = 0°. Substituting this in the formula of Parallelogram Law of Vector Addition, we have

|R| = √(P2 + 2PQ cos 0 + Q2)

= √(P2 + 2PQ + Q2) [Because cos 0 = 1]

= √(P + Q)2

= P + Q

β = tan-1[(Q sin 0)/(P + Q cos 0)]

= tan-1[(0)/(P + Q cos 0)] [Because sin 0 = 0]

= 0°

If vectors P and Q are acting in opposite directions, then we have θ = 180°. Substituting this in the formula of Parallelogram Law of Vector Addition, we have

|R| = √(P2 + 2PQ cos 180° + Q2)

= √(P2 - 2PQ + Q2) [Because cos 180° = -1]

= √(P - Q)2 or √(Q - P)2

= P - Q or Q - P

β = tan-1[(Q sin 180°)/(P + Q cos 180°)]

= tan-1[(0)/(P + Q cos 0)] [Because sin 180° = 0]

= 0° or 180°

If vectors P and Q are perpendicular to each other, then we have θ = 90°. Substituting this in the formula of Parallelogram Law of Vector Addition, we have

|R| = √(P2 + 2PQ cos 90° + Q2)

= √(P2 + 0 + Q2) [Because cos 90° = 0]

= √(P2 + Q2)

β = tan-1[(Q sin 90°)/(P + Q cos 90°)]

= tan-1[Q/(P + 0)] [Because cos 90° = 0]

= tan-1(Q/P)

Important Notes on Parallelogram Law of Vector Addition

Related Topics on Parallelogram Law of Vector Addition

Example 1: Two forces of magnitudes 4N and 7N act on a body and the angle between them is 45°. Determine the magnitude and direction of the resultant vector with the 4N force using the Parallelogram Law of Vector Addition.

Solution: Suppose vector P has magnitude 4N, vector Q has magnitude 7N and θ = 45°, then we have the formulas:

|R| = √(P2 + Q2 + 2PQ cos θ)

= √(42 + 72 + 2×4×7 cos 45°)

= √(16 + 49 + 56/√2)

= √(65 + 56/√2)

= 12.008 N

β = tan-1[(7 sin 45°)/(4 + 7 cos 45°)]

= tan-1[(7/√2)/(4 + 7/√2)]

= tan-1[(7)/(4√2 + 7)]

Answer: Magnitude is approximately 12 N and the direction is tan-1[(7)/(4√2 + 7)].

Example 2: Two vectors P = (1, 2) and Q = (2, 4) have an angle of 60° between them. Find the magnitude and direction of their sum vector.

Solution: Using the parallelogram law of vector addition formulas, we have

|R| = √(P2 + Q2 + 2PQ cos θ), β = tan-1[(Q sin θ)/(P + Q cos θ)]

For this, first, we need the magnitudes of vectors P and Q.

|P| = √(12 + 22) = √5, |Q| = √(22 + 42) = 2√5

|R| = √(√52 + (2√5)2 + 2PQ cos θ)

= √(5 + 20 + 2×√5×2√5 cos 60°)

= √(25 + 20/2)

= √(35) units

β = tan-1[(2√5 sin 60°)/(√5 + 2√5 cos 60°)]

= tan-1[(2√5 × √3/2)/(√5 + 2√5/2)]

= tan-1[(√15)/(2√5)]

= tan-1[√3/2]

Answer: Magnitude = √35 units and Direction = tan-1[√3/2]

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The parallelogram law of vector addition is used to add two vectors when the vectors that are to be added form the two adjacent sides of a parallelogram by joining the tails of the two vectors. Then, the sum of the two vectors is given by the diagonal of the parallelogram.

The sum of vectors P and Q is given by the vector R, the resultant sum vector. If the resultant vector R makes an angle 蠒 with the vector P, then the formulas for its magnitude and direction are:

The Parallelogram Law of Vector Addition is used when the sum of two vectors is to be determined.

If two vectors can be represented by the two adjacent sides (both in magnitude and direction) of a parallelogram drawn from a point, then their resultant sum vector is represented completely by the diagonal of the parallelogram drawn from the same point.

The resultant vector in the Parallelogram Law of Vector Addition is given by the sum of the two vectors. Its magnitude and direction are given by,