- #1

- 174

- 0

## Homework Statement

Find vector and scalar equations that passes through point P(3,7,-1) and is perpindicular to the line of intersection of 2 planes. P1:x-y-2z+3=0 P2: 3x-2y+z+5=0

So initially i started by finding the line between the two planes.

N1 Does Not Equal N2 so they are not paralell (N=Normal)

(1)x-y-2z+3=0 (2)3x-2y+z+5=0

if i multiply equation 1 by 3 i get

(3) 3x-3y-6z+9=0

If i subtract equation 3 from equation 2 i get

-y-7z+4=0

y=4-7z (4)

If i sub (4) into equation 1.

x - (4-7z) -2z + 3 = 0

x -1 +5z=0

x= 1-5z

z=t

therefore.

z=t

y=4-7t

x=1-5t

Equation of line is R=(1,4,0) + t(-5,-7,1)

(Also i noticed, when i did the cross product of the normals of these two planes, you will also find this same direction vector (-5-7,1)

So Now it says Find scalar + Vector equations Given point P(3,7,-1) and PERPINDICULAR to the line of intersection...

The only other case i have dealt with similiar to this, is where you have to find the planes equation when the intersection is PARALELL. In that case you just used the direction vector you find, in the equation of the line, to solve for vector + scalar equations..

***

So what i am lead to beleive with what i learned from that problem, since the plane is perpendicular to the line of intersection, that means that the direction vector of the line is parallel to the normal of the plane***

So this is where i`m stuck, can some one please tell me if this thinking is correct.. If it is correct, here is what i would make my answer...

NormalPlane3 = Direction Vector Of Line = (-5,-7,1)

-5x-7y+z + d = 0

(plugin point P to solve for D)

d= 65

-5x-7y+z+65=0 --SCALAR EQUATION

Also in terms of solving for a vector equation... how can i find a Direction vector i am stuck on this part as ewll.