maeveoneill
Junior Member
- Joined
- Sep 24, 2005
- Messages
- 93
Find parametric equations for the line through the point (o,1,2) that is perpendicular to the line x= 1+t, y= 1-t, z= 2t and intersects this line.
Is this correct??
Okay, first directional vector is (1,-1,2).
Now, put in equation:
(x-0)-(y-1)+2(z-2)=0, or
x-y+1+2z-4=0, or
x-y+2z-3=0.
Okay, I substituted x=1+t, y=1-t, z=2t in
x-y+2z-3=0 to get
1+1-(1-t)+2(2t)-3=0 to find t, and i get t=1/2.
SO, the point where a line perpendicular line and point would intersect is
(1+1/2,1-1/2,(2*1/2) = (3/2,1/2,1)
So, using x=a+by, and x=c+dz to find line I plugged in 3/2=a+(b/2), 3/2=c+d, 0=a+b, 0=c+2d, to find a,b,c,d.
SO, I get x=3-3y and x=3-(3z/2) which shows the line i think.
oh, I finally got:
x=t+1
y=-1/3(t-2)
z=-2/3(t-2)
(And the way I got that, I just let x=t+1, and I was able to solve for y and z.)
Is this correct??
Okay, first directional vector is (1,-1,2).
Now, put in equation:
(x-0)-(y-1)+2(z-2)=0, or
x-y+1+2z-4=0, or
x-y+2z-3=0.
Okay, I substituted x=1+t, y=1-t, z=2t in
x-y+2z-3=0 to get
1+1-(1-t)+2(2t)-3=0 to find t, and i get t=1/2.
SO, the point where a line perpendicular line and point would intersect is
(1+1/2,1-1/2,(2*1/2) = (3/2,1/2,1)
So, using x=a+by, and x=c+dz to find line I plugged in 3/2=a+(b/2), 3/2=c+d, 0=a+b, 0=c+2d, to find a,b,c,d.
SO, I get x=3-3y and x=3-(3z/2) which shows the line i think.
oh, I finally got:
x=t+1
y=-1/3(t-2)
z=-2/3(t-2)
(And the way I got that, I just let x=t+1, and I was able to solve for y and z.)