Vector equation for a line segment in 3 dimensions

[Feodalherren]

Homework Statement

Find the vector equation for a line segment from (2,-1,4) to (4,6,1).

Homework Equations

Arithmetic :p

The Attempt at a Solution

What I did was that I simply solved for the distance between the two points and used it as my vector, I then said

<4-2t, 6-7t, 1+3t> when 0≤t≤1

The book didn't do anything like this at all. They threw in some weird formula that seems like it complicated everything for no reason at all. Is my way of doing it wrong?

 

[MathWarrior]

What is the formula the book threw in ?
Your solution appears to be in the vector form of a line.

Namely:
$$<x_0, y_0, z_0>+t < a, b, c>$$

 

[Feodalherren]

Yeah, well isn't that exactly what they were looking for?
They thew in some (1-t)<vector> + <vector2>. Whatever. I'm not interested in their way as I found a much more intuitive way of doing it on sladder. I was just curious if my answer is OK. It gives the exact same result as their does but it's in a different format.

 

[Ray Vickson]

Yeah, well isn't that exactly what they were looking for?
They thew in some (1-t)<vector> + <vector2>. Whatever. I'm not interested in their way as I found a much more intuitive way of doing it on sladder. I was just curious if my answer is OK. It gives the exact same result as their does but it's in a different format.

The segment from $\vec{a}$ to $\vec{b}$ has the form
$$\vec{v} = \vec{v}(t) \equiv (1-t) \vec{a} + t \vec{b}, \; 0 \leq t \leq 1.$$
When $t = 0$ we have $\vec{v} =\vec{a}$ and when $t = 1$ we have $\vec{v} = \vec{b}$.

Note that just saying $\vec{v} = \vec{a} + t \vec{b}$ won't work (why not?).

 

[SammyS]

Homework Statement

Find the vector equation for a line segment from (2,-1,4) to (4,6,1).
...

The Attempt at a Solution

What I did was that I simply solved for the distance between the two points and used it as my vector, I then said

<4-2t, 6-7t, 1+3t> when 0≤t≤1

The book didn't do anything like this at all. They threw in some weird formula that seems like it complicated everything for no reason at all. Is my way of doing it wrong?

For one thing your solution goes from (4,6,1) to (2,-1,4) as t goes from 0 to 1 .

... if that makes any difference.

 

[Feodalherren]

For one thing your solution goes from (4,6,1) to (2,-1,4) as t goes from 0 to 1 .

... if that makes any difference.

Okay, does it matter? It's still the same line, just drawn in a different direction.

 

[HallsofIvy]

Okay, does it matter? It's still the same line, just drawn in a different direction.

Tha'ts up to you. You said "Find the vector equation for a line segment from (2,-1,4) to (4,6,1)." Was the "from", "to" important?

 

[Feodalherren]

I guess it matters then. Didn't really think about that. Thanks.

 

[SammyS]

I guess it matters then. Didn't really think about that. Thanks.

Even if direction is important, your initial solution is good if t goes from 1 to 0 .

 

[Feodalherren]

Easy fix! I like it, thanks! :)