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anonymous
 4 years ago
THE VELOCITY of a particle is
anonymous
 4 years ago
THE VELOCITY of a particle is

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327679298439:dw the component of velocity v parallel to the vector in the form of a is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327679486945:dw

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2unit vector of (i+j+k)= \((i+j+k)/\sqrt{3}\) angle between velocity vector and the second vector let the second vector be x we know that dot product of two vectors a and b is a.b=abcos theta so v.x=vxcos (theta) (6i+2j2k)(i+j+k)= \(\sqrt{44}* \sqrt(3) \cos \theta\) 6/\(\sqrt {132}=\cos \theta \) magnitude of velocity vector in the direction of x let the new vector be y y=v cos theta y= \( \sqrt{44} *6/ \sqrt{132}\) y=6/ \(\sqrt 3\) vector y= y x/x y= 6/ \(\sqrt 3\) *(i+j+k)/\(\sqrt 3\) y=2i+2j+2k

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0nice work ash2326.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey ash in this question we have to find a component of v parallel to a vector

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1327683618621:dwyeah , i'll show you how i've done

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2y=y* unit vector of x unit vector of x= x/x
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