# Thread: Vectors : Image Of LIne

1. ## Vectors : Image Of LIne

Hi guys, got a question I need help in:

Find, in the form r.n = d, the equation of the plane containing the points A,B and C whose position vectors are -i + j + k, i + 2j + 2k and 2i + 2j + 4k respectively.
Find, in the form r = p + tq the equation of the image of the line r = -i + j + k + \lambda (i - 5k) reflected in this plane.

I found the equation of the plane to be r = (-1, 1 ,1) + \lambda (1, 0, -5). But I don't know how to find the equation of the reflected line (L'). I know L' would definitely contain A but I will need the direction vector of the reflected line to form the equation, and to get the direction vector, I need at least one other point lying on L'. Any idea how to find the 2nd point?

2. $\displaystyle (-1, 1, 1) + \lambda (1, 0, -5)$ is not the equation of a plane, it's the vector form of a line.
To find the equation of the plane, use your three points to create two vectors that lie in the plane. Take their cross product to find the normal to the plane. The equation of your plane $\displaystyle ax + by + cz = d$ will have the same coefficients as your normal vector $\displaystyle \textrm{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$. Then use one of the points in the plane to find $\displaystyle d$.