I have a math task which says: find the intersection point and the plane for these lines: p: (x-2) / 1 = (y-1) / (-1) = (z-3) / 1 and q: x / 1 = (y-1) / 1 = (z-1)/ 1
I dont know how to find the int point in this form, I know I should make them equal and put z=1 but it doesnt work.. And also for the plane equation, should I choose 3 points? 1 point from p, 1 from q and the intersection point? Help please :-)
$p$ is parallel to the vector $\vec p=(1, -1, 1)$ and passes through the point $P(2,1,3)$
its parametric equation is
$\left\{ {\begin{array}{*{20}{l}} {x = 2 + t} \\ {y = 1 - t} \\ {z = 3 + t} \end{array}} \right.$
$q$ is parallel to the vector $\vec q=(1,1,1)$ and passes through $Q(0,1,1)$
its parametric equation is
$\left\{ {\begin{array}{*{20}{l}} {x =u} \\ {y = 1 +u} \\ {z = 1+u} \end{array}} \right.$
The intersection point is given by the system
$\left\{{\begin{array}{*{20}{l}} {2 + t = u} \\ {1 - t = 1 + u} \\ {3 + t = 1 + u} \end{array}}\right.$
$t=-1;\;u=1$
which gives the point of intersection of $p$ and $q$
$A(1,\;2,\;2)$
The plane containing the two lines has a normal vector $\vec n$ which is the cross product
$\vec n=\vec p\times \vec q=(-2,\;0,\;2)$ and passes through $A(1,\;2,\;2)$
$-2(x-1)+0(y-2)+2(z-2)=0$
which simplified gives
$ x - z +1=0$
Hope this can be useful