Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector `3 hat i+5 hat j-6 hat k`.

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3hati+5hatj-6hatk

Find the vector equation of the plane which is at a distance of 6/sqrt(29) from the origin with perpendicular vector 2i-3j+4k . Convert into Cartesian form. Also find the foot of the perpendicular drawn from the origin to the plane.

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

Find the projection of a vector i+3j+7k on the vector 7i-j+8k .

Consider the vector equation of two planes barr.(2i+j+k)=3,barr.(i-j-k)=4 Find vector equation of the plane through the intersection of the above planes and the point (1,2,-1).

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)

If the projection of the vectors vec a on vec b is |vec a times vec b| and if 3 vec b= hat i+ hat j+ hat k then the angle between vec a and vec b is a) pi//3 b) pi//2 c) pi//4 d) pi//6