# Thread: Line of intersection of planes

1. ## Line of intersection of planes

Prove that the line of intersection of the two planes
$\displaystyle r=\lambda_1i+2\mu_1j+3\nu_1k$ $\displaystyle \lambda_1+\mu_1+\nu_1=1$
$\displaystyle r=2\lambda_2i+\mu_1j+2\nu_1k$ $\displaystyle \lambda_2+\mu_2+\nu_2=1$
can be written in terms of a single parameter t as
$\displaystyle 6r=(3+t)i+4tj+9(1-t)k$

I managed to get the vector equations into this form:
$\displaystyle r.(6i+3j+2k)=6$
and $\displaystyle r.(i+2j+k)=4$
I calculated the direction vector of the line to be (-i-4j+9k)
I don't know how else to continue.
Thanks!

2. Originally Posted by arze
Prove that the line of intersection of the two planes
$\displaystyle r=\lambda_1i+2\mu_1j+3\nu_1k$ $\displaystyle \lambda_1+\mu_1+\nu_1=1$
$\displaystyle r=2\lambda_2i+\mu_1j+2\nu_1k$ $\displaystyle \lambda_2+\mu_2+\nu_2=1$

Too many equality signs...

can be written in terms of a single parameter t as
$\displaystyle 6r=(3+t)i+4tj+9(1-t)k$

I managed to get the vector equations into this form:
$\displaystyle r.(6i+3j+2k)=6$
and $\displaystyle r.(i+2j+k)=4$
I calculated the direction vector of the line to be (-i-4j+9k)
I don't know how else to continue.
Thanks!
How did you get the intersection line's director vector to be such a nice one if in the definition of the two planes you have 6 (!!) parameters??
The director vector of the line is a normal to both plane's normals, and then you only need one single point belonging to both planes to form the parametric euqation of the intersection line. I can't understand what you did above...in fact, I hardly understand your planes' equations.

Tonio

3. I eliminated the $\displaystyle \lambda$ in each, then since the two vectors remaining were parallel to the plane, their cross product I thought would be the direction vector of the plane. And since I had a position vector of a point on each line, found the equation of the plane.
for the first equation
$\displaystyle r=i+\mu_1(-i+2j)+\nu_1(-i+3k)$
I think I did something wrong.