Find the line of intersection between 2 Planes

[prace]

Hello,

I am trying to find the line of intersection between these two planes:

P_1 = x + 2y -9z = 7
P_2 = 2x - 3y + 17z = 0

I found the direction vector needed for the line of intersection between these two points by taking the cross product of the P_1 normal vector and the P_2 normal vector which gave me \vec{a} = <7,-35,-7>

Now all I need is a point somewhere along the direction of that vector. This is where I am stuck. Any help on finding this point would be awesome. Thanks!

~Peter

 

[HallsofIvy]

Why find the equations of the line that way? Since you have two equations for three unknown variables, just solve for two of them in terms of the third, then use that third variable as parameter.

For example, to find equations of the line of intersection of x+ y+ z= 1 and 2x- y+ z= 0, adding the equations gives 3x+ 2z= 1 so z= 1/2- (3/2)x.
Then y= 1- x- z= 1- x- 1/2+ (3/2)z= 1/2+ (1/2)x. Let x= t. Then the parametric equations are x= t, y= 1/2+ (1/2)t, z= 1/2- (3/2)t. In vector form, \vec{r}= t\vec{i}+ (1/2+ (1/2)t)\vec{j}+ (1/2- (3/2)t)\vec{k} or \vec{r}= ((1/2)\vec{j}+ (1/2)\vec{k})+ (\vec{i}+ (1/2)\vec{j}+ (1/2)\vec{k})t.