Finding the Intersection of Two Planes: Solving for the Equation of a Line

[Tenshou]

Homework Statement

What is the equation of a line of the intersecting planes
$3x_1-2x_2+x_3=5$
$2x_1+3x_2-x_3=-1$

Homework Equations

The Attempt at a Solution

I didn't know where to start but I started at trying to find the cross product of the planes (needless to say it didn't get me and where) I got $-x_1-5x_2+11x_3$ that equation doesn't make any sense x.x I do not know what to do x.x

 

[member 428835]

What is the equation of a line of the intersecting planes
$3x_1-2x_2+x_3=5$
$2x_1+3x_2-x_3=-1$

first use elimination on the planes, thus add $(1)+(2)=5x_1+x_2=4$:
solving gives: $x_2=4-{5x_1}$
take $x_1=t$ thus $x_2=4-{5t}$
now sub $x_1,x_2$ into $(1)$ or $(2)$ to find $x_3$ in terms of $t$
now you may write up the line in parametric form: $(x_1(t),x_2(t),x_3(t))$

 

[tiny-tim]

Hi Tenshou!

… I started at trying to find the cross product of the planes (needless to say it didn't get me and where) I got $-x_1-5x_2+11x_3$ …

no, your 11 should be 13, shouldn't it?

(and i think one of your signs is wrong)

that method should work …

it gives you the direction that is perpendicular to both normals, and therefore it must lie in both planes …

now you have something like x₂ = Ax₁ + B, x₃ = Cx₁ + D, where A and C are known,

so you substitute that into the original equations, and that gives you two equations in two unknowns (B and D) …

but I'm not sure that's any quicker than joshmccraney's method!​

 

[Tenshou]

Thank you for your insightful answers but I think the easiest solution was IMS... Thank you so much, I mean I just did it this morning (and yes my maths were wrong.) Solved the equation of

$A$$x$=$b$

I was looking for the simplest way to solving it and josh, I think your way is the simplest way for solving it(thanks by the way). Although, I still did not get the equation in the book I got $A$ as a 2x3 matrix

$\left[ \begin{array} {r r r r} \ 3 &\ -2 &\ 1 \\ \ 2 &\ 3 &\ -1 \\ \end{array} \right]$

Then after rref(A) [or something close to it] I got

$\left[ \begin{array} {r r r r} \ -7 &\ 0 &\ 1 \\ \ -5 &\ 1 &\ 0 \\ \end{array} \right]$

I calculated by Nullity-Rank that I should have one free column left. I solved and got
$\left[ \begin{array} {r r r r} \ 0 &\ 4 &\ 13 \\ \end{array} \right] = x_{part}^T$

then I just found out the solution to the null space thank you for correcting my maths tiny-tin

$\left[ \begin{array} {r r r r} \ -1 &\ 5 &\ 13 \\ \end{array} \right] = x_{null}^T$

then I finished up by allowing $x_{comp}=\lambda x_{null}+x_{part}; \forall \lambda \epsilon\mathbb{R}$

Although I went though a page and a half of calculation I will remember your method josh!

the answer in the book is $x_1 = -k+1; x_2 = 5k-1; x_3 = 13k; \forall k \epsilon \mathbb{R}$