Answer to Question #211903 in Analytic Geometry for Faith

Question #211903

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.


1
Expert's answer
2021-06-30T06:40:52-0400

1.


x=3+t,y=1+t,z=2tx=83s,y=2+2s,z=7s\begin{matrix} x=3+t,& y=-1+t,& z=2-t \\ x=-8-3s, &y=2+2s, & z=-7s \end{matrix}


3+t=83s1+t=2+2s2t=7s\begin{matrix} 3+t=-8-3s \\ -1+t=2+2s \\ 2-t=-7s \end{matrix}

3+t=83s4=105s1=25s\begin{matrix} 3+t=-8-3s \\ 4=-10-5s \\ 1=2-5s \end{matrix}


3+t=83s5s=145s=1\begin{matrix} 3+t=-8-3s \\ 5s=-14\\ 5s=1 \end{matrix}

No solution. Therefore the given lines do not intersect.


2.


x=1+3t,y=1,z=5+2tx=2+s,y=35s,z=42s\begin{matrix} x=-1+3t,& y=1,& z=-5+2t \\ x=-2+s, &y=3-5s, & z=-4-2s \end{matrix}


1+3t=2+s1=35s5+2t=42s\begin{matrix} -1+3t=-2+s \\ 1=3-5s \\ -5+2t=-4-2s \end{matrix}

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

3+t=83s5s=145s=1\begin{matrix} 3+t=-8-3s \\ 5s=-14\\ 5s=1 \end{matrix}

No solution. Therefore the given lines do not intersect.


3.


5x+y2z=32x3y+5z=7\begin{matrix} -5x+y-2z=3\\ 2x-3y+5z=-7 \end{matrix}


y=5x+2z+313xz=2\begin{matrix} y=5x+2z+3 \\ -13x-z=2 \end{matrix}

y=21x1z=13x2\begin{matrix} y=-21x-1 \\ z=-13x-2 \end{matrix}

We can write our parametrization of the line of intersection as


0,1,2+t1,21,13,tR\langle0,-1,-2\rangle+t\langle1,-21, -13\rangle, t\in \R


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