एक‌ कण की स्थिति सदिश समय t पर निम्न प्रकार निर्भर करता है।vecr=(2+3t^(3))hati+(-t+t^(2))hatj+(7...

If S=(t^(3))/(3)-2t^(2)+3t+4 , then

Find the Cartesian from the equation of the plane vecr=(s-2t)hati+(3-t)hatj+(2s+t)hatk .

A particle that is moving in an xy plane has a position vector given by vecr=(3.00t^(3)-6.00t)hati+(7.00-8.00t^(4))hatj , where vecr is measued in meters and t is measured in seconds. For t = 3.00 s, in unitvector notation, find (a) vecr , (b) vecv , and ( c) veca . (d) Find the angle between the positive direction of the x axis and a line that is tangent to the path of the particle at t = 3.00 s.

Find the Cartesian equation of the following planes : a. vec(r). (hati + hatj - hatk ) = 2 b. vec(r). (2 hati + 3 hatj - 4 hatk ) = 1 (c ) vec(r). [ (s - 2t) hati + (3 - t ) hatj + (2 s + t ) hatk] = 15

If t+1/t =5, then (2t)/(3t^(2)-5t+3) is equal to ……..

A body is projected up such that its position vector varies with time as vecr={3thati+(4t-5t^(2))hatj} m . Here t is in second.The time when its y -coordinate is zero is

Let C:r(t)=x(t)hati+y(t)hatj+z(t)hatk be a differentiable curve, i.e., lim_(xto0) (r(t+H)-r(h))/(h) exist for all t, therefore r'(t)=x'(t)hati+y'(t)hatj+z'(t)hatk Iff r'(t) , is tangent to the curve C at the point P[x(t),y(t),z(t)] and r'(t) points in the direction of increasing t. Q. The tangent vector to r(t)=2t^(2)hati+(1-t)hatj+(3t^(2)+2)hatk at (2,0,5) is

Find the shortest distance between the following (6 -7) lines whose vector equations are : (i) vec(r) = (1 -t) hati + (t - 2) hatj + (3 - 2t) hatk and vec(r) = (s + 1) hat(i) + (2s - 1) hatj - (2s + 1) hatk (ii) vec(r) = (3 -t) hati + (4 + 2t) hatj + (t - 2) hatk and vec(r) = (1 + s) hati + (3s - 7 ) hatj + (2s -2) hatk. where t and s are scalars.

The velocity of an object is given by vecv = ( 6 t^(3) hati + t^(2) hatj) m//s) . Find the acceleration at t = 2s.