i am posting this question from 5 days …..is there anybody who can he

Jitender Singh

Last Activity: 10 Years ago

Ans:
Let the resultant of vectors A & B to be C
$\overrightarrow{C} = \overrightarrow{A}+\overrightarrow{B}$
Angle b/w A & C
$cos\alpha = \frac{\overrightarrow{A}.(\overrightarrow{A}+\overrightarrow{B})}{|\overrightarrow{A}||\overrightarrow{A}+\overrightarrow{B}|}$
$cos\alpha = \frac{\overrightarrow{A}.\overrightarrow{A}+\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{A}+\overrightarrow{B}|}$
$cos\alpha = \frac{|\overrightarrow{A}|^{2}+\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{A}+\overrightarrow{B}|}$
Angle b/w B & C
$cos\beta = \frac{\overrightarrow{B}.(\overrightarrow{A}+\overrightarrow{B})}{|\overrightarrow{B}||\overrightarrow{A}+\overrightarrow{B}|}$
$cos\beta = \frac{\overrightarrow{B}.\overrightarrow{A}+\overrightarrow{B}.\overrightarrow{B}}{|\overrightarrow{B}||\overrightarrow{A}+\overrightarrow{B}|}$
$cos\beta = \frac{\overrightarrow{B}.\overrightarrow{A}+|\overrightarrow{B}|^{2}}{|\overrightarrow{B}||\overrightarrow{A}+\overrightarrow{B}|}$
$\alpha <\beta$
$cos\alpha > cos\beta$
$\frac{|\overrightarrow{A}|^{2}+\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{A}+\overrightarrow{B}|} > \frac{|\overrightarrow{B}|^{2}+\overrightarrow{B}.\overrightarrow{A}}{|\overrightarrow{B}||\overrightarrow{A}+\overrightarrow{B}|}$
$|\overrightarrow{A}|^{2}|\overrightarrow{B}| + |\overrightarrow{B}|\overrightarrow{A}.\overrightarrow{B} = |\overrightarrow{B}|^{2}|\overrightarrow{A}| + |\overrightarrow{A}|\overrightarrow{B}.\overrightarrow{A}$
$|\overrightarrow{A}|^{2}|\overrightarrow{B}| - |\overrightarrow{B}|^{2}|\overrightarrow{A}| = (|\overrightarrow{A}|-|\overrightarrow{B}|)\overrightarrow{A}.\overrightarrow{B}$
$|\overrightarrow{A}||\overrightarrow{B}| (|\overrightarrow{A}|-|\overrightarrow{B}|) = (|\overrightarrow{A}|-|\overrightarrow{B}|)\overrightarrow{A}.\overrightarrow{B}$

$\Rightarrow |\overrightarrow{A}|-|\overrightarrow{B}| > 0$
$\Rightarrow |\overrightarrow{A}|>|\overrightarrow{B}|$
Or you can always say that the resultant will fall toward the vector which have more magnitude.