(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

P1 = 3x+4y=1

P2 = x +y -z=3

Find the intersection.

2. Relevant equations

x = x0 +at

y = y0 + bt

z = z0 + ct

3. 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?

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# Homework Help: Intersection of two planes

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