Given two equations of planes, what is the line of intersection?

Help!

i have two equations of two planes that intersect at the origin:

$\displaystyle $ax + by + cz = 0$ $

and

$\displaystyle $dx + ey + fz = 0$ $

as I have three variables, $\displaystyle $x, y, z$$ and two equations, then I will have an infinite number of solutions: either they are both the same plane or they intersect in exactly one line.

I need to prove this!

I am able to show what happens if they are the same plane:

If there is a non-zero $\displaystyle $\lambda$$ such that $\displaystyle $a = \lambda d $$ ; $\displaystyle $b = \lambda e $$ ; $\displaystyle $c = \lambda f $$then the same line, i.e. the points whose co-ordinates satisfy $\displaystyle $dx + ey + fz = 0$$ also satisfy $\displaystyle $\lambda adx + \lambda ey + \lambda fz = 0$$that is: $\displaystyle $ax + by + cz = 0$$Thus they are the same plane.

but I'm completely at a loss to prove they intersect in precisely one line - any help very gratefully apprectiated!

(I've tried solving simultaneously: multiplying the first equation through by d; finding y in terms of z, then x in terms of z... and then substituting back into the original equations, but everything cancels out to produce $\displaystyle $0=0$$ !

Re: Given two equations of planes, what is the line of intersection?

Quote:

Originally Posted by

**Whiskyfied** i have two equations of two planes that intersect at the origin:

$\displaystyle $ax + by + cz = 0$ $

and

$\displaystyle $dx + ey + fz = 0$ $

as I have three variables, $\displaystyle $x, y, z$$ and two equations, then I will have an infinite number of solutions: either they are both the same plane or they intersect in exactly one line.

I need to prove this!

I am able to show what happens if they are the same plane:

If there is a non-zero $\displaystyle $\lambda$$ such that $\displaystyle $a = \lambda d $$ ; $\displaystyle $b = \lambda e $$ ; $\displaystyle $c = \lambda f $$then the same line, i.e. the points whose co-ordinates satisfy $\displaystyle $dx + ey + fz = 0$$ also satisfy $\displaystyle $\lambda adx + \lambda ey + \lambda fz = 0$$that is: $\displaystyle $ax + by + cz = 0$$Thus they are the same plane.

but I'm completely at a loss to prove they intersect in precisely one line - any help very gratefully apprectiated!

This amounts showing that there is a unique line through $\displaystyle (0,0,0)$ with direction $\displaystyle <a,b,c>\times <d,e,f>.$

Re: Given two equations of planes, what is the line of intersection?

Thanks!

i think this means that the equation of the line is calculated by the determinant of

$\displaystyle \begin{vmatrix}

x & y & z\\

a & b & c\\

d & e & f\\

\end{vmatrix}

= $(bf-ce)x + (cd-af)y + (ae-bd)z = 0$$

where not all a, b, c or x, y, z are zero

if I'm wrong, please tell me! (sorry about the breaking space notation - don't know how to get rid of it!)

thanks everso for your help!

Re: Given two equations of planes, what is the line of intersection?

Quote:

Originally Posted by

**Whiskyfied** Thanks!

i think this means that the equation of the line is calculated by the determinant of

$\displaystyle \begin{vmatrix}x & y & z\\a & b & c\\d & e & f\\\end{vmatrix}= $(bf-ce)x + (cd-af)y + (ae-bd)z = 0$$

where not all a, b, c or x, y, z are zero

if I'm wrong, please tell me! (sorry about the breaking space notation - don't know how to get rid of it!)

See that I got rid of you linefeeds.

[TEX]\begin{vmatrix}x & y & z\\a & b & c\\d & e & \\\end{vmatrix}[/TEX]

$\displaystyle \begin{vmatrix}x & y & z\\a & b & c\\d & e & f\\\end{vmatrix}$

NO LINEFEEDS.

Re: Given two equations of planes, what is the line of intersection?

thanks again!

(I of course meant d,e,f instead of x,y,z on the penultimate line of my last message)

Feeling very relieved now to have got that done!