Answer:
The parametric equations for the given line are x=1-t, y=1+t and z=2t.
Step-by-step explanation:
Given information: P (1,1,0) and Q (0,2,2).
The parametric equation of line are
[tex]x=x_0+at[/tex]
[tex]y=y_0+bt[/tex]
[tex]z=z_0+ct[/tex]
where, [tex](x_0,y_0,z_0)[/tex] is point on line and <a,b,c> is direction vector.
The line passes through the points P (1,1,0) and Q (0,2,2). So, the direction vector is
[tex]\overrightarrow{v}=<x_2-x_1, y_2-y_1, z_2-z_1>[/tex]
[tex]\overrightarrow{v}=<0-1,2-1, 2-0>[/tex]
[tex]\overrightarrow{v}=<-1,1,2>[/tex]
The direction vector is <-1,1,2>. So, a=-1, b=1 and c=2. The parametric equation of line are
[tex]x=1+(-1)t=1-t[/tex]
[tex]y=1+(1)t=1+t[/tex]
[tex]z=0+(2)t=2t[/tex]
Therefore the parametric equations for the given line are x=1-t, y=1+t and z=2t.
