Parametric Equations of Line Passing Through (0,1/2,1) and (2,1,-3)

[ineedhelpnow]

find the parametric equations of the line that passes through the points (0,1/2,1) and (2,1,-3)

$v= \left\langle 2-0, 1-1/2, -3-1 \right\rangle = \left\langle 2, 1/2, -4 \right\rangle$

$P_0 (2,1,-3)$

$x=2+2t$
$y=1+\frac{1}{2}t$
$z=-3-4t$so my question is, why are the points (2,1,3) used and not the other ones

 

[Prove It]

find the parametric equations of the line that passes through the points (0,1/2,1) and (2,1,-3)

$v= \left\langle 2-0, 1-1/2, -3-1 \right\rangle = \left\langle 2, 1/2, -4 \right\rangle$

$P_0 (2,1,-3)$

$x=2+2t$
$y=1+\frac{1}{2}t$
$z=-3-4t$so my question is, why are the points (2,1,3) used and not the other ones

Either point can be used.

 

[ineedhelpnow]

so both sets of equations will be correct? because those are the ones that my book gave as answers.

 

[Prove It]

so both sets of equations will be correct? because those are the ones that my book gave as answers.

Yes there will actually be FOUR possible answers, all equivalent and all correct.

I wonder if you can figure out what they would be...

 

[Evgeny.Makarov]

Yes there will actually be FOUR possible answers, all equivalent and all correct.

If the points are denoted $P_0$ and $P_1$, I see two simplest answers: $P_0+(\overrightarrow{P_0P_1})t$ and $P_1+(\overrightarrow{P_1P_0})t$. There are also infinitely many other answers denoting the same line, which differ both in the starting point and the vector.

 

[ineedhelpnow]

well wouldn't you just have to find other points on the same line to get equivalent equations?

 

[Dethrone]

Yes, it doesn't matter which point you choose, as long as it is on the same line. If it is, all equations are equivalent. It only takes one point on the line, and a directional vector.

It's helpful to look at it from first principles.
http://www.criced.tsukuba.ac.jp/grapes/image/vector2.gif

The directional vector, $OB$, establishes the direction, and adding a point $OA$ to it (by vector addition) brings us to some point on the line. Multiplying the directional vector $OB$ by any constant will bring us to a different point on the line.

 

[Prove It]

It often helps to think about the properties of lines and vectors.

Lines: Have a direction (but not a direction of motion), are of infinite magnitude (so they go on forever in both directions), and are positioned somewhere in 3D space.

Vectors: Have a direction, have a direction of motion, have a magnitude. Do not have a direction.As you can see, vectors and lines are similar, but not identical.

If we know something about a line though (such as two points it goes through), then we CAN find the equation of the line if we can find its direction, make it infinitely long, and position it properly (i.e. to go through those two points).

To get the direction, we need a DIRECTION VECTOR.

So by finding EITHER of the two vectors defined by those two points (the reason I say either vector is because the line will not have a direction of motion, and thus the arrow can point either way).

Then we need to make sure that this vector has infinite magnitude. The trouble is, to scale a vector you would normally multiply it by a number. But infinity is not a number. Thus you multiply the direction vector by a PARAMETER, which you can then change the value of (i.e. to make it as large as needed).

Finally, even if we are able to find the direction vector, all vectors are DEFINED as starting at the origin. The line might not go through the origin though. Thus we need to move it to the right position. To do this, we think "well we know where the direction vector 'starts' (at the origin) and we know a point it has to go through (two actually), so we can translate the direction vector by adding the same number of units to each component as is instructed by the point it has to go through (e.g. if it has to go through the point (1, 5, -3), then it will be moved 1 unit in the positive x direction, 5 units in the positive y direction, and 3 units in the negative z direction). This process is essentially the same as adding two vectors, which is why we loosely say "add one of the points". The reason we can use either point is because by translating the direction vector according to either point, because it is going in the right direction, it will still go through both points!

So there you are, to work out the equation of a line, you need its direction vector, multiply it by a parameter to make it infinitely long, and then position it correctly by "adding one of your known points".

And since there are two possible direction vectors, and two possible points to use, there will be FOUR possible solutions.

 

[ineedhelpnow]

Either point can be used.

so B was used only because the numbers are nicer?