The resultant of and
makes an angle α with
and β with
Let us consider and
along two sides of the parallelogram and the resultant vector
along the diagonal of parallelogram. Now if we take the component of
along
we have B
and component of
perpendicular to
we get A
as illustrated above.
Now consider OAX,
∴ AX =OA
=|A|
Now consider AXR,
,∴ XR=OB
=|B|
In AXR, if we apply the Pythagoras theorem, we get
,∴
, |A|
=|B|
Thus, we have , now if
,
,{
,∴
, implies |A|>|B| for
.
Hence option C satisfies our condition best.