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Physics 9 Online
OpenStudy (anonymous):

If A.B=2 and IAI=2, IBI=undroot2 then angle b/w vectors is 45 degree...how?? :/someone explain me please...

OpenStudy (anonymous):

To find the angle we use A.B=ABCOStheta right??

OpenStudy (amistre64):

cos = a.b/|a||b| yes

OpenStudy (amistre64):

to determine the angle itself requires you to take the arccos of the setup

OpenStudy (anonymous):

But angle 45 doesn't come when i solve it :/

OpenStudy (amistre64):

of course it does ...

OpenStudy (amistre64):

\[2\sqrt2~cos=2\] \[\sqrt2~cos=1\] \[cos=\frac 1{\sqrt2}\] \[\theta=cos^{-1}\left(\frac 1{\sqrt2}\right)=\frac{pi}{4}\]

OpenStudy (amistre64):

|dw:1371128093993:dw|

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