# Analytic method of vector addition

If two vectors display the results from two adjoining side drawn from any point of a parallelogram in the direction of the result, then the result and direction, the parallelogram that is drawn by the diagonal has been drawn from the same point to add the vector Parallel quadratic rules
Assuming vector $\vec{A}$ and $\vec{B}$ are bent at opposite $\theta$ angle, they are displayed in the direction of the parallel quadrilateral OPQS in the direction of OPQS and OP and the OS, then according to the rule of parallelogram, resultant $\vec{R}$ of $\vec{A}$ And $\vec{B}$ will be represented by the diagonal OQ in the result and direction. In order to find the result of the resulting $\vec{R}$, we increase the side OP and pull the vertical QE from the point Q, thus

Thus, in the right angled triangle OEQ

$(OQ)^{2}&space;=&space;(OE)^{2}&space;+&space;(QE)^{2}&space;=&space;(OP+PE)^{2}&space;+&space;(QE)^{2}$

$=&space;(OP)^{2}&space;+&space;(PE)^{2}&space;+&space;2(OP)(PE)&space;+&space;(QE)^{2}$

NOW

$(PE)^{2}&space;+&space;(QE)^{2}&space;=&space;(PQ)^{2}$

SO

$(OQ)^{2}&space;=&space;(OP)^{2}&space;+&space;(PQ)^{2}&space;+&space;2(OP)(PE)$

IN Right-angled triangle

$COS\theta&space;=&space;\frac{PE}{PQ}$
OR

$PE&space;=&space;PQCOS\theta$

So the final equation will be

$(OQ)^{2}&space;=&space;(OP)^{2}&space;+&space;(PQ)^{2}&space;+&space;2(OP)(PQCOS\theta&space;)$   ,  NOW

OP =A , PQ= OS = B , OQ = R ,,    SO

$R^{2}&space;=&space;A^{2}+&space;B^{2}&space;+2AB&space;COS\theta$

$\left&space;[&space;R=\sqrt{A^{2}&space;+B^{2}&space;+2ABCOS\theta&space;}&space;\right&space;]$

To find out the direction of the resulting $\vec{R}$, say that the angle of $\vec{R}$ vector $\vec{A}$ is formed by $\theta$

$Tan&space;\theta&space;=&space;\frac{QE}{OE}&space;,=&space;\frac{QE}{OP+PE}$   NOW  , OP = A , PE = BCOS$\theta$

To find the value of QE, in triangle PEQ

$SIN\theta&space;=&space;\frac{QE}{PQ}$

$\left&space;[&space;Tan\theta&space;=&space;\frac{Bsine\theta&space;}{A+bcos\theta&space;}&space;\right&space;]$

Special cases

(1) when both the vector are in same direction

Again  equation

$R&space;=&space;\sqrt{A^{2}+B^{2}+ABCOS\theta&space;}&space;=&space;\sqrt{A^{2}&space;+B^{2}+2AB}&space;=&space;A+B$   and

$Tan\theta&space;=\frac{b&space;\times&space;0}{a+b}&space;=0&space;,&space;\alpha&space;=0$
Thus the result of result $\vec{A}$ is equal to yoga of both vector $\vec{A}$ and $\vec{B}$ and in $\vec{A}$ and $\vec{B}$ same direction

(2) when the both vector are at right angle to each other

$R=\sqrt{A^{2}+b^{2}+2ABcos\theta&space;}&space;,=\sqrt{A^{2}+B^{2}}$

$Tan\theta&space;=\frac{Bsin90}{A+Bcos90}&space;,&space;=\frac{A}{B}$

(3) When both the vectors are in opposite direction

Again  equation

$R=\sqrt{A^{2}+B^{2}+2ABcos180^{\circ}}&space;=&space;\sqrt{(A-B)^{2}}&space;=A-B,B-A$

$Tan\theta&space;=\frac{Bsin180^{\circ}}{A+Bcos180^{\circ}}&space;=0&space;,nad&space;,\alpha&space;=0,or&space;180^{\circ}$
Thus the result of the resultant vector $\vec{R}$ is equal to the difference of the result of both vectors and in the direction of the large vector
It is clear from the above that the result of the result of both vector is maximized, then both vectors are in the same direction and this is minimal when they are in the direction