Answer to Question #328178 in Analytic Geometry for Faith

Question

Calculate the equation for the plane containing the lines L1 and L2,where L1 is given by the parametric equation (x,y,z)=(1,0,-1)+t(1,1,1),t€R and L2 is given by the parametric equation (x,y,z)=(2,1,0)+t(1,-1,0),t€R.

Answer 1

let's write equations for L1 and L2 in canonical form.

For L1:

(x,y,z)=(1,0,-1)+t(1,1,1)

x=t+1; y=t; z=t-1

x-1=t; y=t; z+1=t

x-1=y=z+1

For L2:

(x,y,z)=(2,1,0)+t(1,-1,0)

x=t-2; y=-t-1; z=0

x+2=t; -y-1=t;

x+2=-y-1

Line L1 contains point (1,0,-1), line L2 contains point (-2,-1,0). We need to find one point is contained by one of this lines. Lets take L1. From equation x=t+1; y=t; z=t-1 for t=1 could be find x=2, y=1, z=0, so point is (2,1,0).

Now lets find the equation for plane containing points (1,0,-1), (-2,-1,0), (2,1,0).

Using formula

$$\begin{vmatrix} x-x_{1};y-y_{1} ; z-z_{1}\\ x_{2}-x_{1} ; y_{2}-y_{1} ; z_{2}-z_{1}\\ x_{3}-x_{1} ; y_{3}-y_{1} ; z_{3}-z_{1}\\ \end{vmatrix}=0$$

$$\begin{vmatrix} x-1;y-0 ; z+1\\ -2-1 ; -1-0 ; 0+1\\ 2-1 ; 1-0 ; 0+1\\\end{vmatrix}=0$$

$$\begin{vmatrix} x-1;y-0 ; z+1\\ -3 ; -1 ; 1\\ 1 ; 1 ; 1\\\end{vmatrix}=0$$

(x - 1)(-1·1-1·1) - y ((-3)·1-1·1) +( z +1)((-3)·1-(-1)·1) = 0

- 2x + 4y - 2z = 0

x - 2y + z = 0

Answer: the equation for the plane is x - 2y + z = 0.