20 Physics -- Vectors

Proof Parallelogram  Law

Proof Parallelogram  Law

Parallelogram Law of Vector Addition Formula

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 ϕ with the vector P, then the formulas for its magnitude and direction are:

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

Parallelogram Law of Vector Addition Proof

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.

parallelogram law of vector addition proof

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

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