Question

Plane 1 is given by P(1,3,-1) Q(1,-2,0) R(3,-1,1) and Plane 2 is given by the equation 3x +y -4z = 7.

What is the parametric equation of the line of intersection of Plane 1 and Plane 2?

Answer 1

define plane 1 in the form of plane 2 is my first thought

Answer 2

(1,3,-1) (1,-2,0) (3,-1,1)
-(1,3,-1) -(1,3,-1) -(1,3,-1)
--------------------------
(0,0,0) (0,-5,1) (2,-4,2)

Answer 3

cross the 2 vectors from that to get a normal

( i , j, k)
(0,-5,1)
(2,-4,2)

-6i+2j+10k is what i get for that

Answer 4

and if we calibrate it with any point on there: lets use (1,3,-1)

-6(x-1)+2(y-3)+10(z+1) = 0

-6x+6 +2y-6 +10z+10 = 0

-6x +2y +10z+10 = 0 should be the equation of the plane

Answer 5

3x +y -4z - 7 = 0
-6x +2y +10z+10 = 0

when these equal we got a line right?

Answer 6

you have to parametrize the lines within the plane (PQ and PR) and then take the cross product of the two. This gives the n value to the plane. Then you take this n value and cross it with the n value of the 2nd plane (3, 1, -4)

Answer 7

thanks! But I think we need to cross the PQ and PR first to find the equation of the plane, then cross plane 1 and plane 2 to find the line of intersection..

Answer 8

thats what i said...

Answer 9

yep, I was trying to tell it to amistre64. but how will I solve for the plane perpendicula to the two planes and that also passes through the line of intersection?

Answer 10

lol ... thats what I did; well, i got up to the finding the equation of the plane part :) brain froze up after that tho :/