A line is given by the vector equation r = (2 i - j + 3 k ) + (3 i - 2 j + 6 k ) . Let A be the point on the line corresponding to = 0 . Let B and C be two points on the line which are at a distance of 14 units from A . If the z -coordinate of B is greater than the z -coordinate of C , then the dot product of the position vectors of A and B is equal to :

Step-by-step solution

The position vector of point A (for = 0 ) is r _A = 2 i - j + 3 k , so the coordinates of A are (2, -1, 3) . The direction vector of the line is d = 3 i - 2 j + 6 k . The magnitude of the direction vector is | d | = 3^2 + (-2)^2 + 6^2 = 9 + 4 + 36 = 7 . The direction cosines of the line are ( 3 7 , -2 7 , 6 7 ) . Any point on the line at a distance r from A has coordinates given by (x₁ r l, y₁ r m, z₁ r n) . Given r = 14 , the coordinates of the points B and C are: (2 14 ( 3 7 ), -1 14 ( -2 7 ), 3 14 ( 6 7 ) ) = (2