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OpenStudy (anonymous):
of which?
OpenStudy (anonymous):
a vector?
OpenStudy (anonymous):
i think it is of two vectors.. im not sure
OpenStudy (anonymous):
if a,b are two vectors and angle between them is x then the magnitude's square is a^2+b^2+2ab cos x
OpenStudy (anonymous):
ohh ok thanks
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OpenStudy (anonymous):
or if you're just given one vector let's say {a,b,c} then to find the magnitude of the vector then magnitude = \[\sqrt{(a^2 + b^2 +c^2)}\]
OpenStudy (anonymous):
this is possible only when the vectors are mutually perpendicular
OpenStudy (anonymous):
thank you
myininaya (myininaya):
remember we were finding maginture of like -3
and it was 3
then the magintude of a is |a|
|a|=sqrt(a^2)
what about we went from (x1,y1) and (x2,y2)
it was sqrt{(x1-x2)^2+(y1-y2)^2)
OpenStudy (anonymous):
for any inclined angle between them u have to add a cos x term. when the angle is 90 then the cos term vanishes nd u get a result like sqrt(a^2+b^2)
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OpenStudy (anonymous):
myininaya....u gave the distance between two vectors not the magnitude of the resultant...