The positon vector of the centroid of the triangle ABC is `2i+4j+2k`. If the position vector of the vector A is `2i+6j+4k., then the position vector of midpointof BC is (A) `2i+3j+k` (B) `2i+3jk` (C) `2i-3j-k` (D) `-2i-3j-k`

The cross product of the vectors (2i - 3j + 4k) and (I + 4j -5k) is

The unit vector parallel to the resultant of the vectors A = 4i+3j + 6k and B = -i + 3j-8k is

Projection of the vector 2hat(i) + 3hat(j) + 2hat(k) on the vector hat(i) - 2hat(j) + 3hat(k) is :

Find the projection of the vector vec a=2 hat i+3 hat j+2 hat k on the vector vec b=2 hat i+2 hat j+ hat k .

A=4i+4j-4k and B=3i+j +4k , then angle between vectors A and B is

Equation of the plane through three points A, B and C with position vectors -6i+3j+2k, 3i-2j+4k and 5i+7j+3k is equal to

Consider three vectors p=i+j+k, q=2i+4j-k and r=i+j+3k . If p, q and r denotes the position vector of three non-collinear points, then the equation of the plane containing these points is

Let the centre of the parallelepiped formed by vec P A= hat i+2 hat j+2 hat k ; vec P B=4 hat i-3 hat j+ hat k ; vec P C=3 hat i+5 hat j- hat k is given by the position vector (7, 6, 2). Then the position vector of the point P is (3,4,1) (b) (6,8,2) (1, 3, 4) (d) (2, 6, 8)

Let the centre of the parallelepiped formed by vec P A= hat i+2 hat j+2 hat k ; vec P B=4 hat i-3 hat j+ hat k ; vec P C=3 hat i+5 hat j- hat k is given by the position vector (7, 6, 2). Then the position vector of the point P is (3,4,1) (b) (6,8,2) (1, 3, 4) (d) (2, 6, 8)

Find the projection of the vector vec(P) = 2hat(i) - 3hat(j) + 6 hat(k) on the vector vec(Q) = hat(i) + 2hat(j) + 2hat(k)