Given:
[tex]L_1:x-3=\frac{y-2}{2}=4-z[/tex][tex]L_2:(x,y,z)=(1,2,1)+s(1,-3,1)[/tex]
Required:
We need to verify that the lines L1 and L2 do not intersect at any point.
Explanation:
Consider the line.
[tex]L_2=(x,y,z)=(1,2,1)+s(1,-3,1)[/tex]
[tex]L_2=(x,y,z)=(1,2,1)+(s,-3s,s)[/tex]
[tex]L_{2:}:(x,y,z)=(1+s,2-3s,1+s)[/tex]
The point (1+s,2-3s,1+s) lies in line 2.
Consider the equation of line 1.
[tex]x-3=\frac{y-2}{2}[/tex]
Substitute x =1+s in the equation.
[tex]1+s-3=\frac{y-2}{2}[/tex]
[tex]s-2=\frac{y-2}{2}[/tex]
[tex]2s-4=y-2[/tex]
[tex]2s-4+2=y-2+2[/tex][tex]2s-2=y[/tex]
[tex]\frac{y-2}{2}=4-z[/tex]
Substitute y =2s-2 in the equation.
[tex]\frac{2s-2-2}{2}=4-z[/tex]
[tex]2s-4=8-2z[/tex]
[tex]2s-4-8=-2z[/tex]
[tex]2s-12=-2z[/tex][tex]\frac{2s-12}{-2}=z[/tex]
[tex]6-s=z[/tex]
The point (1+s,2s-2,6-s) lies in line 1.
If line 1 and line 2 intersect , then the points should be equal,
Equat the points (1+s,2-3s,1-s) and (1+s,2s-6,6-s).
[tex](1+s,2-3s,1-s)=(1+s,2s-6,6-s)[/tex]
Equate corresponding terms.
[tex]1+s=1+s[/tex][tex]2-3s=2s-6[/tex][tex]1-s=6-s[/tex]
Solve the following equation.
[tex]2-3s=2s-6[/tex]
[tex]2+6=2s+3s[/tex]
[tex]\frac{8}{5}=s[/tex]
Substitute s =8/5 in the third equation.
[tex]1-\frac{8}{5}=6-\frac{8}{5}[/tex]
[tex]\frac{5-8}{5}=\frac{30-8}{5}[/tex][tex]-\frac{3}{5}=\frac{22}{5}[/tex]
This is not true.
So there is no solution for these equations when we equate the points liying on both line 1 and line 2.
For any value of s, the line does not intersect.
Hence the given lines do not intersect at any point.
