# The Line of Intersection between two Planes

### Problem:Find theline of intersection between the two planes given by the vector equations r1.[3, 4, 0]  = 5 and  r2.[1, 2, 3]  = 6

A diagram of this is shown on the right.  is a normal vector to Plane 1 is a normal vector to Plane 2.

By simple geometrical reasoning;
the line of intersection is perpendicular to both normals.
To find the position vector, r, of any point on the line of intersection;

find a vector, v, to which the line is parallel,

find the position vector, a, of specific apoint on the line,

then;      r  = a  +  tv, is the required result.
To find v;

the cross product of two vectors gives a third vector which is perpendicular to both vectors.
v is perpendicular to both and ,
therefore, v will beparallel to × ,

From the equations to the planes, .[3, 4, 0]  = 5       and, .[1, 2, 3]  = 6 =   [3, 4, 0]       and =  [1, 2, 3]

therefore × = [3, 4, 0] × [1, 2, 3]

= = i     - j    + k

=  [12,   -9,  9]                                             1.

To find a;

at any point on the line the position vector, ,  = let the point be (x, y, z), then;

From the equations to the planes,

(x, y, z) . [3, 4, 0]  = 5    and   (x, y, z) . [1, 2, 3]  = 6

3x  +  4y   =  5   and    x  +  2y  +  3z  =  6

Let the point have an x co-ordinate = 0 then;

y  = and   z  = a is therefore [ 0, , ]                            2.

Substituting result 1. and 2. into   r  = a  +  tv  gives;

r  = [ 0, , ]  +  t [12,   -9,  9]

or, more simply,   r  = [ 0, , ]  +  t [4,   -3,  3]

since the vector     [12,   -9,  9]  is parallel to   [4,   -3,  3]