Thread: Lines in a Plane 13

1. Lines in a Plane 13

Show that the points A(-9,-3,-16) and B(6,2,14) lie on the line that passes through (0,0,2) and has direction numbers (3,1,6). Describe the line segment from A to B using parametric equations with suitable restrictions on the parameter.

I need help with the end of part B)
Describe the line segment from A to B using parametric equations with suitable restrictions on the parameter.
so I found the parametric equations which are:

x=3t, y=t,z=2+6t
But I am having trouble finding their answer of -3<=t<=2 for the restriction.

How did they find this?

I will add part a if it helps:
d=AB=(15,5,30)
d=(3,1,6)

sym eqtn
x/3
y/1
(z-2)/6

sub in point (0,0,2)
0=0=0 therefore point lies on line

Ok, so how did they get the restriction on t? thanks for your help!

2. Originally Posted by skeske1234
But I am having trouble finding their answer of -3<=t<=2 for the restriction.

How did they find this?
It would appear that they have found the equation of the line by using (x,y,z) = (0,0,2) + t(3,1,6)

In this case substituting in t=-3 gives the point A and substituting in t=2 gives the point B. Hence the segment of the line between the points A and B will be defined by using the restriction: -3<=t<=2 .