THE VELOCITY of a particle is

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THE VELOCITY of a particle is

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1327679298439:dw| the component of velocity v parallel to the vector in the form of a is
|dw:1327679486945:dw|
unit 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=|a||b|cos theta so v.x=|v||x|cos (theta) (6i+2j-2k)(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

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Other answers:

nice work ash2326.
hey ash in this question we have to find a component of v parallel to a vector
|dw:1327683618621:dw|yeah , i'll show you how i've done
|dw:1327683675640:dw|
y=|y|* unit vector of x unit vector of x= x/|x|

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