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State parallelogram law of vector addition and derive an expression for its magnitude.

State parallelogram law of vector addition and derive an expression for its magnitude.

Grade:upto college level

2 Answers

Shoaib Malik
20 Points
4 years ago
Parallelogram Law of Vector AdditionStatement of Parallelogram Law If two vectors acting simultaneously at a point can be represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through that point.Derivation of the lawNote: All the letters in bold represent vectors and normal letters represent magnitude only.Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure.Let θ be the angle between P and Q and R be the resultant vector. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.So, we have R = P + QNow, expand A to C and draw BC perpendicular to OC.From triangle OCB, In triangle ABC, Also, Magnitude of resultant:Substituting value of AC and BC in (i), we get which is the magnitude of resultant.Direction of resultant: Let ø be the angle made by resultant R with P. Then,From triangle OBC, which is the direction of resultant
Shoaib Malik
20 Points
4 years ago
If two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors.The proof for the resultant vector in Parallelogram addition is as follows,Consider a parallelogram OABCOABC as shown in the figure,Parallelogram Law Of Vector AdditionLet P&QP&Q be two adjacent sides of parallelogram, and RR be the resultant vector obtained by addition of vectors P&QP&Q,Now, drop a perpendicular from CC on OAOA so that they meet at AA.From right angled triangle ΔOCD,OC2=OD2+DC2ΔOCD,OC2=OD2+DC2[OD=OA+AD][OD=OA+AD]R2=(OA+AD)2+DC2R2=(OA+AD)2+DC2R2=OA2+AD2+2OA.AD+DC2R2=OA2+AD2+2OA.AD+DC2From ΔΔADC, AC2=AD2+DC2AC2=AD2+DC2and also cosθ=ADACcos⁡θ=ADACAnd hence R2=OA2+AC2+2OA.ACcosθR2=OA2+AC2+2OA.ACcos⁡θAnd substituting A and BR2=A2+B2+2A.Bcosθ (R2 here is square of R and A2 is for A square likewise respectively)

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