Determine the interaction of the line of intersection of the planes [$1$] $x + y - z = 1$ and [$2$] $3x + y + z = 3$ with the line of intersection of the planes [$3$] $2x - y + 2z = 4$ and [$4$] $2x + 2y + z = 1$.
That is the question.
Ok So I need to find out the equation of the line between [$1$] and [$2$] and the equation of a second line between [$3$] and [$4$].
I know how to figure out what kind of interaction two lines have in $3$ space, so please help me solve for the first part of the question; Finding a line equation between two planes.
So my on-line course has provided an explanation on this but I can't make sense of it.
So the way I need to solve this is to use elimination between the two planes, and then I have to make one of the variables a parameter t. and then... I am lost. (please provide an explanation using the method I so vaguely described)
So again, I need two line equations, for the interaction of [$1$] and [$2$], and the planes [$3$] and [$4$].
Please help,
Guest
Here is a start. From your first two plane equations, you can add $x+y-x=1$ to $3x+y+z=3$ to come at $4x+2y=4$ or simplify to $2x+y=2$ Now use a parameter $t$ to set $x=t$ which gives you $y=2-2t$ Substituting these equations into one of the two given plane equations (I picked the second one), you get $z=3-3t-(2-2t)$ which gives $z=1-t$ Now you have a parametric set of the line as an intersection of the first two planes. You can do the same thing for the other two planes. Once you have two lines, you can find if there exists a value of $t$ that "matches" with all $x,y,z$ components of the two lines. You give this a try...