Question

I have been told that a line is
x=6-theta y=theta z=2theta... what does this mean? and how do I find a vector for it? Thanks :)

Answer 1

Theta is just a parameter. So it's just a helper variable in a way.

Let's say theta = 0

that means...


x = 6-theta
x = 6-0
x = 6


y = theta
y = 0


z = 2*theta
z = 2*0
z = 0


So we have the ordered triple (6,0,0) when theta = 0

Answer 2

now let's say theta = 1


x = 6-theta
x = 6-1
x = 5


y = theta
y = 1


z = 2*theta
z = 2*1
z = 2


now we have the point (5,1,2)

Answer 3

Now what is the vector from (6,0,0) to (5,1,2)?

Answer 4

so are the points (6,0,0) and (5,12) points on the line? and what is an ordered triple?

Answer 5

it's like an ordered pair, but for 3d space

Answer 6

umm...(-1, 1,2)?

Answer 7

basically coordinates for points in 3d space

Answer 8

oh ok :)

Answer 9

I like to write vectors with angle brackets like this <-1, 1,2>

Answer 10

that vector tells you how to move from point to point

it's like the slope, but this is for 3d space

Answer 11

so the direction vector then?

Answer 12

so if you were to start at (6,0,0), you can move to (5,1,2) by following the vector

Answer 13

to define the general equation for this line in 3space, we simply start with the starting point (6,0,0)

we can think of this as the vector <6,0,0> because we start at the origin (0,0,0) and point to (6,0,0)....essentially the vector <6,0,0>

Answer 14

then we add on multiples of the vector <-1,1,2> by tacking on a variable (usually t or s)

so we have this equation

y = <6,0,0> + t*<-1,1,2>

Answer 15

sorry I guess y is a bad variable choice, how about q

q = <6,0,0> + t*<-1,1,2>

Answer 16

oh thats right...is that the scalar representation of the line?

Answer 17

notice this

when t = 0


q = <6,0,0> + t*<-1,1,2>
q = <6,0,0> + 0*<-1,1,2>
q = <6,0,0> + <0,0,0>
q = <6,0,0>

aka the starting point

Answer 18

when t = 1

q = <6,0,0> + t*<-1,1,2>
q = <6,0,0> + 1*<-1,1,2>
q = <6,0,0> + <-1,1,2>
q = <6 + (-1), 0+1, 0+2>
q = <5,1,2>

which is the next point over (if you use integral values of t)

Answer 19

Thats pretty easy then :)

Answer 20

yeah it's not too bad

Answer 21

Thanks :)

Answer 22

you're welcome

Answer 23

hmmm