Intersection of two planes

  • Thread starter jwxie
  • Start date
  • #1
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Homework Statement



P1 = 3x+4y=1
P2 = x +y -z=3

Find the intersection.

Homework Equations



x = x0 +at
y = y0 + bt
z = z0 + ct

The Attempt at a Solution


I used a quicker method. I first did P1 x P2 and got its cross product <-4,3,-1>

Then I followed what our teacher had shown us in class - we did substitution.
3x+4y=1
x+y -z =3
let x = 0, then 4y = 1 means y = 1/4 and y -z = 3 will give us -11/4

then i had both direction vector <-4,3,-1> and point (0,3,-11/4) and i could write the equations of the line of intersection using the parametric equations

x = -4t + 0
y = 3t + 1/4
z = -1t -11/4

the book gives
x = 4t +11
y = -3t - 8
z = t

it seems like the author made z = 0 (in the parametric equation z0 = 0)
my vector was <-4,3,-1> and his was <4,-3,1> and they were scalar multiple of -1

i know there are many possible parametric equations describing the same intersection. but is my result valid, even though my x0, y0, and z0 are different?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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Yes they are the same essentially.
 
  • #3
280
0
thank you rock.
 

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