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if vector a cross vec b=2i+j and a.b=5 then the angle between vec a and vec b is ................... a.tan t =5^1/2 b.tan t=1/(5^1/2)
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|axb| = |ab sin (t) | |a.b| = |ab cos (t) |
take the ratio
|axb | ----- = ? |a.b|
tan t=2i+j/5 ..................... how i get the angle
you need to take |axb| which is the length of axb
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axb = 2i+j |axb| = ?
k, now i became able of solving MAY ALLAH GIVE YOU REWARD OF IT
good to knw :)
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