0

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 :-)

ellie
  • 1
  • You didn't say how you made "them" equal--exactly what did you set equal to what? Information like that should be included in the body of the question. By the way, setting $z=1$ would help if the intersection point has a $z$ coordinate of $1,$ but in this case the intersection point has a different $z$ coordinate. – David K Aug 18 '17 at 21:08
  • Make a plane from two lines, and then find the point of intersection of this plane with the 3rd line. – John Alexiou Aug 18 '17 at 21:12

1 Answers1

0

$p$ 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

Raffaele
  • 26,371