The centre of the circle given by `vec r* (hat i+ 2hat j + 2hat k) = 15 and |vec r-(hat j + 2hat k)| = 4` is

The centre of the circle given by vec r.( hat i+2 hat j+2 hat k)=15a n d| vec r.( hat j+2 hat k)|=4 is a. (0,1,2) b. (1,3,4) c. (-1,3,4) d. none of these

the angle between the planes vec r*(2hat i-hat j+hat k)=6 and vec r*(hat i+hat j+2hat k)=5

The position of a particle is given by vec r = hat i + 2hat j - hat k and its momentum is vec p = 3 hat i + 4 hat j - 2 hat k . The angular momentum is perpendicular to

Find the angle between the planes vec r*(2hat i-hat j+hat k)=6 and vec r*(hat i+hat j+2hat k)=5

Find the shortest distance between the lines vec r = (hat i + 2 hat j + hat k) + lambda (hat i - hat j + hat k) and vec r = ( 2 hat i - hat j - hat k) + mu ( 2 hat i + hat J + 2 hat k)

Find the shortest distance (S.D.) between the lines : vec r = hat i + hat j + lambda (2 hat i- hat j + hat k) and vec r = 2 hat i + hat j- hat k + mu (3 hat i - 5 hat j + 2 hat k) .

Shortest distance between the lines: vec r=(4hat i-hat j)+lambda(hat i+2hat j-3hat k) and vec r=(hat i-hat j+2hat k)+u(2hat i+4hat j-5hat k)

Find the angle between the lines vec r*(hat i+2hat j-hat k)+lambda(hat i-hat j+hat k) and the plane vec r*(2hat i-hat j+hat k)=4

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