Parallelogram LAW of vectors

parallelogram law of vectors 
Two vector quantities can be added using parallelogram law ( velocity vector can be added to velocity vector only). This law is useful to find both magnitude and direction of resultant.
Statement: If two vectors are represented in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, the diagonal passing through that point represents their resultant both in magnitude and direction.

Explanation: \(\overrightarrow P \) and \(\overrightarrow Q \)  are two vectors represented by \(\overrightarrow {AB} \)  and \(\overrightarrow {AD} \). Both vectors act at the common point A and mutually inclined at angle ‘\(\theta\)’ as shown in figure. If the parallelogram ABCD is completed taking AB and AD as adjacent sides, then the diagonal \(\overrightarrow {AC} \) represents their resultant \(\overrightarrow {R} \) both in magnitude and direction.
Magnitude of the resultant: The line of action of  is extended. The perpendicular dropped from ‘C’ meets the extension of AB at E.
    From the figure, it is obvious that \(\overrightarrow {BC} \, = \,\overrightarrow {AD} = \overrightarrow Q \) and 
     Length of AB = magnitude of \( \overrightarrow P\) = P ; 
     Length of BC = magnitude of  \( \overrightarrow Q\)= Q
    Length of AC = magnitude of \( \overrightarrow R\) = R
    From triangle CBE and \(\frac{{BE}}{{BC}}\)=cos \(\theta\),  BE = BC cos\(\theta \to\)  BE = Q cos\(\theta\) ....... (1) 
    From the triangle CBE and  \(\frac{{EC}}{{BC}}\)= sin\(\theta\) and EC = Q sin\(\theta\) .............. (2)
    In right angled triangle ACE
    (AC)² = (AE)² + (EC)²,        but from figure AE = AB + BE
    (AC)² = (AE)² + (EC)²,  
    (AC)² = (AB)² + (BE)² + 2AB . BE + (EC)², from eqns (1) and (2) 
        R² = P² + Q² cos2 \(\theta\) + 2PQ cos\(\theta\) + Q² sin2  \(\theta\)
        R² = P² + Q² + 2PQ cos\(\theta\)
        R \(= \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } \,\,\,\,\,\,\, \to \)  (3) 
Direction of the resultant : Let the resultant makes angle ‘\(\alpha\)’ with \(\overrightarrow P \)
From triangle CAE : \(\tan \alpha = \frac{{EC}}{{AE}} = \frac{{EC}}{{AB + BE}}\)
     \( \Rightarrow \tan \alpha = \frac{{Q\sin \theta }}{{P + Q\cos \theta }}\)
   \(\Rightarrow \alpha = {\tan ^{ - 1}}\left[ {\frac{{Q\sin \theta }}{{P + Q\cos \theta }}} \right]\, \to \)  (4) 
         The expression (3) and (4)gives the magnitude and direction of the resultant of \(\overrightarrow P \) and \(\overrightarrow Q \).
SPECIAL CASES:
1)If   \(\overrightarrow P \) and \(\overrightarrow Q \).  are in same direction ,then \(\theta=0^0\) and cos \(\theta\) = 1
From equation (3) and (4), R = P + Q and \(\alpha\) = 0
Hence the magnitude of resultant is sum of the magnitude of individual vectors. The direction of resultant is same as that of individual vectors.
2)If   \(\overrightarrow P \) and \(\overrightarrow Q \).  are opposite, then  \(\theta\)= 180° and cos \(\theta\)= –1
R = P – Q i.e., R = P – Q or Q – P and  \(\alpha=0^0\)or 180°.
Thus the magnitude of resultant is equal to difference of magnitudes of individual vectors and the direction of resultant is same as that of the vector of larger magnitude.
3) If   \(\overrightarrow P \) and \(\overrightarrow Q \).  are perpendicular, then \(\theta=90^0\) & cos\(\theta\) = 0
R \(= \sqrt {{P^2} + {Q^2}}\) and  \(\alpha\)= Tan^–1 (Q/P)
4) If \(\left| {\overrightarrow P } \right| = \left| {\overrightarrow Q } \right|\), then R = 2P cos \(\theta\)/2 and \(\alpha = \theta /2\)
 If the vectors have equal magnitude, then the resultant will bisect the angle between  them.
5) If \(\left| {\overline P } \right|{\text{ = }}\left| {\overline Q } \right|\) and \(\theta =120^0\), then .\(\left| {\overline R } \right|{\text{ = }}\left| {\overline P } \right|\)
6)Using parallelogram law of vectors  \(\left| {\left( {\overline a {\text{ - }}\overline b } \right)} \right|{\text{ = }}\sqrt {{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}{\text{ - 2abcos \theta }}} \)
7)The magnitude of resultant that can be obtained from the vectors of magnitude P & Q will in be between P+Q & P-Q
8)The resultant makes small angle with the vector of higher magnitude.