# FInd the Vector passing thru a point and parallel to two points in a plane

1. May 25, 2012

### kailsen

Hi,
i have three points A',B',C' and want to find a vector passing through B' and parallel to the other two points in the plane containing these 3 points.
So this how i started,

A'=(9,27,-0.6)
B'=(12,27,-6)
C'=(19,25,-8)

I found the vector AB' and AC' first.
A'B'= A'-B'
A'C'=A'-C'

Found cross product of the two resultant vector A'B' and A'C' to get a normal vector.
now using the equation a*[x – xA] + b*[y – yA] + c*[z – zA] = 0,
i found the equation of the plane that contains all these 3 vectors (A',B',C').
Assuming the equation is -48x + 30y -2z = -90.
Now i would like to get a vector (basically x,y,z) that is passing through B' and parallel to the line passing through A' and C'.

I think, finding a line passing through the origin of of the normal vector to the plane will fetch me half the answer, but how to satisfy it to be parallel to the line A'C'

2. May 25, 2012

### HallsofIvy

First, you statement of the problem is strange. A "vector" does not pass through a particular point. It is entirely given by its direction and length. Nor is one vector "parallel" to another though they may be "in the same direction. You may mean "find the equation of the line passing through B' parallel to the line through A' and C'".

But then I have no idea why you would go to the work of finding two distinct vectors and their dot product. The vector from A' to C' is <19- 9, 25- 27, -8- (-0.6)>= <10, -2, -7.4>. The line through B'= (12, 27, -6) in the direction of that vector is given by
x= 10t+ 12, y= -2t+ 27, z= -7.4t- 6.

3. May 25, 2012

### mathman

The line you want can be expressed in parametric form (s is parameter) as B' + s(A'-C').