Answer to Question #211903 in Analytic Geometry for Faith

Question

1.find the point of intersection between the lines:<3,-1,2>+t<1,1,-1> and <-8,2,0>+t<-3,2,-7>.

2.show that the lines x+1=3t,y=1,z+5=2t for tER and z+2=s,y-3=-5s,z+4=-2s for tER intersect, and find the point of intersection.

3.find the point of intersection between the planes: -5x+y-2z=3 and 2x-3y+5z=-7.

Answer 1

1.

$$\begin{matrix} x=3+t,& y=-1+t,& z=2-t \\ x=-8-3s, &y=2+2s, & z=-7s \end{matrix}$$

$$\begin{matrix} 3+t=-8-3s \\ -1+t=2+2s \\ 2-t=-7s \end{matrix}$$

$$\begin{matrix} 3+t=-8-3s \\ 4=-10-5s \\ 1=2-5s \end{matrix}$$

$$\begin{matrix} 3+t=-8-3s \\ 5s=-14\\ 5s=1 \end{matrix}$$

No solution. Therefore the given lines do not intersect.

2.

$$\begin{matrix} x=-1+3t,& y=1,& z=-5+2t \\ x=-2+s, &y=3-5s, & z=-4-2s \end{matrix}$$

$$\begin{matrix} -1+3t=-2+s \\ 1=3-5s \\ -5+2t=-4-2s \end{matrix}$$

$$\begin{matrix} s=0.4 \\ t=-0.2\\ t=0.1 \end{matrix}$$

$$\begin{matrix} 3+t=-8-3s \\ 5s=-14\\ 5s=1 \end{matrix}$$

No solution. Therefore the given lines do not intersect.

3.

$$\begin{matrix} -5x+y-2z=3\\ 2x-3y+5z=-7 \end{matrix}$$

$$\begin{matrix} y=5x+2z+3 \\ -13x-z=2 \end{matrix}$$

$$\begin{matrix} y=-21x-1 \\ z=-13x-2 \end{matrix}$$

We can write our parametrization of the line of intersection as

$$\langle0,-1,-2\rangle+t\langle1,-21, -13\rangle, t\in \R$$