Find the shortest distance between the lines whose vector equations are` -> r=(1-t) hat i+(t-2) hat j+(3-2t) hat k`and ` -> r=(s+1) hat i+(2s-1) hat j-(2s+1) hat k`

The length of the shortest distance between the two lines r=(-3i+6j)+s(-4i+3j+2k) and r=(-2i+7k)+t(-4i+j+k) is

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 distance between the line vec(r) = (2hat(i) + 2hat(j) - hat(k)) + lambda(2hat(i) + hat(j) - 2hat(k)) and the plane vec(r).(hat(i) + 2hat(j) + 2hat(k)) = 10 is equal to

Find the dot product of vec A= A_x hat i + A_y hat j + A_z hat k and vec B= B_x hat i+ B_y hat j + B_z hat k

Find a vector of magnitude 6 and perpendicular to both vec(a) = 2hat(i) + 2hat(j) + hat(k) and vec(b) = hat(i) - 2hat(j) + 2hat(k)

If the positions vectors of three consecutive vertices of a parallelogram are hat i+ hat j+ hat k, hat i+3 hat j+5 hat k and 7 hat i+9 hat j+11 hat k then the coordinates of the fourth vertex are a)(2,1,3) b)(6,7,8) c)(4,1,3) d)(7,7,7)

A plane is at a distance of 5 units from the origin and perpendicular to the vector 2hat(i) + hat(j) + 2hat(k) . The equation of the plane is a) vec(r ).(2hat(i) + hat(j) -2hat(k))=15 b) vec(r ).(2hat(i)+ hat(j)-hat(k))=15 c) vec(r ).(2hat(i) + hat(j)+ 2hat(k))=15 d) vec(r ).(hat(i) + hat(j) + 2hat(k))=15

The point of intersection of the straight lines r = (3hat(i) - 4hat(j) + 5hat(k)) + lambda(-hat(i) - 2hat(j) + 2hat(k)) and (3-x)/(-1) = (y+4)/(2) = (z-5)/(7) is