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 $$\vec{r}= ((1/2)\vec{j}+ (1/2)\vec{k})+ (\vec{i}+ (1/2)\vec{j}+ (1/2)\vec{k})t$$.

 

[Architeuthis Dux]

Hello Peter,

Finding the line of intersection between two planes involves finding the point where the two planes intersect. To do this, you can set the equations of the planes equal to each other and solve for two variables. In this case, you can set x + 2y - 9z = 7 equal to 2x - 3y + 17z = 0 and solve for x and y. This will give you a point on the line of intersection, which you can then use to find other points on the line by adding multiples of the direction vector you found. I hope this helps. Let me know if you have any other questions.