किसी कण का स्थिति vecr=3t^(2)hati+4t^(2)hatj+7hatk द्वारा प्रदर्शित है प्रथम 10 सेकण्ड में तय की...

The position vector of a particle is given by vecr=(3t^(2)hati+4t^(2)hatj+7hatk)m at a given time t .The net displacement of the particle after 10 s is

The position of a particle is given by vecr = 3t hati + 2t^(2) hatj + 5hatk , where t is in seconds and the coefficients have the proper units for vecr to be in metres. The direction of velocity of the particle at t = 1 s is

The position of a particle is given by vecr = 2t^(2) hati + 3t hatj + 4hatk where t is in second and the coefficients have proper units for vecr to be in metre. The veca(t) of the particle at t = 1 s is

The shortest distance between the lines vecr = (2hati - hatj) + lambda(2hati + hatj - 3hatk) vecr = (hati - hatj + 2hatk) + lambda(2hati + hatj - 5hatk)

Here are four descriptions of the position (in meters) of a puck as it moves in an xy plane: (1) x=-3t^(2)+4t-2andy=6t^(2)-4t (2) x=-3t^(3)-4t andy=-5t^(2)+6 (3) vecr=2t^(2)hati-(4t+3)hatj (4) vecr=(4t^(3)-2t)hati+hatj Are the x and y acceleration components constant? Is acceleration veca constant?

The angle between the planes vecr. (2 hati - 3 hatj + hatk) =1 and vecr. (hati - hatj) =4 is

Find the angle between the line vecr = (hati +2hatj -hatk ) +lambda (hati - hatj +hatk) and vecr cdot (2hati - hatj +hatk) = 4.

Find the points of intersection of the line vecr = 2hati - hatj + 2hatk + lambda(3hati + 4hatj + 2hatk) and the plane vecr.(hati - hatj + hatk) = 5