The lineas r = (6 - 6s) a + (4s - 4) b + (4 - 8s) c and r = (2t - 1) a + (4t - 2) b - (2t + 3) c intersect at

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.

Subtract : 3t ^(4) - 4t ^(3) + 2t ^(2) - 6t + 6 from - 4t ^(4) + 8t ^(3) - 4t ^(2) - 2t + 11

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

A particle starts moving rectilinearly at time t = 0 such that its velocity(v) changes with time (t) as per equation – v = (t2 – 2t) m//s for 0 lt t lt 2 s = (–t^(2) + 6t – 8) m//s for 2 le te 4 s (a) Find the interval of time between t = 0 and t = 4 s when particle is retarding. (b) Find the maximum speed of the particle in the interval 0 le t le 4 s .

The angle between the straight line vec r =(2-3t) hat i+(1+2t)hatj+(2+6t) hatk and vec r= (1+4s) hati+(2-s) hatj+ (8s-1) hatk is a) cos^-1 (sqrt(41)/34) b) cos^-1 (21/34) c) cos^-1 (43/63) d) cos^-1 (34/63)

The position of an object moving on a straight line is defined by the relation x=t^(3)-2t^(2)-4t , where x is expressed in meter and t in second. Determine (a) the average velocity during the interval of 1 second to 4 second. (b) the velocity at t = 1 s and t = 4 s, (c) the average acceleration during the interval of 1 second to 4 second. (d) the acceleration at t = 4 s and t = 1 s.

Let f(x)=x^3+3x+2a n dg(x) be the inverse of it. Then the area bounded by g(x) , the x-axis, and the ordinate at x=-2a n dx=6 is (a) 9/4s qdotu n i t s (b) 4/3s qdotu n i t s (c) 5/4s qdotu n i t s (d) 7/3s qdotu n i t s