Question

State and prove parallelogram law of vector addition. Discuss some special cases.

Answer 1

It states that if two vectors can be represented completely (i.e., both in magnitude and direction) by the two adjacent sides of a parallelogram drawn from a point then their resultant is represented completely by its diagonal drawn from the same point. 

Proof: Let\(\vec P\) and \(\vec Q\) be the two vectors represented completely by the adjacent sides OA and OB of the parallelogram OACB such that,

\(\vec {OA}\)= \(\vec P\) , \(\vec {OB}\) = \(\vec Q\)

or OA = P, OB = Q

\(\theta\) =angle between them ∠AOB

If R be their resultant, then it will be represented completely by the diagonal OC through point O such that \(\vec {OC}\) = \(\vec R\) 

Magnitude of \(\vec R\) : Draw CD ⊥ to OA produced 

∴ ∠DAC = ∠AOB

= θ 

Now in right angled triangle ODC

 OC² = OD² + DC²

= (OA + AD)² + DC² 

= OA² + AD² + 2OA.AD + DC² 

= OA² + (AD² + DC² ) + 2OA.OD …(i) 

Also in right angled ADC,

 AC² = AD² + DC² …(ii) 

Also, \(\frac {AD}{AC}\) = cos θ

or AD = AC cos θ …(iii) 

and \(\frac{DC}{AC}\) = sin θ

or DC = AC sin θ …(iv)

∴ From eqn. (i), (ii), (iii) we get 

OC² = OA² + AC² + 2OA.AC cos θ 

or OC =\(\sqrt{OA^2 + AC^2 + 2OA. AC cos θ}\)…(v) 

As OC = R, OA = P, AC = OB = Q …(vi) 

∴ From equation (v) and (vi), we get 

R = \(\sqrt{P^2 + Q^2 + 2PQ cos θ}\) …(vii) 

eq. (vii) gives the magnitude of\(\vec R\) .

Direction of \(\vec R\). Let β the angle made by R with \(\vec P\)

∴ In right angle triangle ODC, 

tan β = \(\frac{DC}{OD}\) = \(\frac{DC}{OA+AD}\)

= \(\frac{AC sin θ}{OA+AC cosθ}\) [by using (iii) and (iv)]

tan β =\(\frac{ Q sin θ}{P+Q cos θ}\) …(viii) 

Special cases: (a) When two vectors are acting same direction: 

Then θ = 0º  

∴ R = \(\sqrt{(P+Q)^2}\) = P + Q 

and tan θ = \(\frac{Q.0}{P+Q}\) = 0 or β = 0º  

Thus, the magnitude of the resultant vector is equal to the sum of the magnitudes of the two vectors acting in the same direction and their resultant acts in the direction of P and Q. 

(b) When two vectors act in the opposite directions : 

Then θ = 180º

∴ cos θ = -1 and sin θ = 0 

∴ R =\(\sqrt{P^2+Q^2+2PQ(-1)}\) 

= \(\sqrt{P^2+Q^2+2PQ}\) 

= \(\sqrt {(P-Q)^2 } or \sqrt{(Q-P)^2} \)

= (P – Q) or (Q – P) 

and tan β = \(\frac{Q\times0}{P+Q(-1)}\) = 0

= tan 0º  or tan 180º  

∴ β = 0º  or 180º  

Thus, the magnitude of the resultant of two vectors acting in the opposite direction to the difference of the magnitude of two vectors and it acts in the direction of bigger vector.

(c) If two vectors act perpendicular to each other : 

θ = 90º  , i.e., if \(\vec P\) ⊥ \(\vec Q\) , then cos 90º  = 0 

and sin 90º  = 1

∴ R = \(\sqrt {P^2+Q^2}\)

and tan β = \(\frac{Q}{P}\)