Writing vector and parametric equations for a line that....

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To write vector and parametric equations for the line through points P(–3, 5, 2) and Q(2, 7, 1), the direction vector PQ is calculated as (5, 2, -1). The vector equation can be expressed as (x, y, z) = (-3, 5, 2) + t(5, 2, -1), leading to the parametric equations x = -3 + 5t, y = 5 + 2t, and z = 2 - t. It is confirmed that either point P or Q can be used to derive the equations without affecting the outcome. The calculations were verified by substituting t = 0 and t = 1 to ensure the equations pass through both points.
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Homework Statement


[/B]
Write vector and parametric equations for the line that goes through the points P(–3, 5, 2) and Q(2, 7, 1).

Homework Equations

The Attempt at a Solution



First I find the direction vector for PQ.

PQ=Q-P = (2,7,1)-(-3,5,2)
=[2-(-3),7-5,1-2]
=5,2,-1

PQ= (5,2,-1)

Now I have a direction vector and two points. I think I can write the vector equation now but I am not sure which point to use, P or Q, and if it matters.

If I were to use P(-3,5,2) the vector equation would look like the following:

(x,y,z)=(-3,5,2)+t(5,2,-1)

Then the parametric equations would be

x=-3+5t
y=5+2t
z=2-t
 
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You can check it yourself. Try ##t=0## and ##t=1## to see if it goes through your two points.
 
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LCKurtz said:
You can check it yourself. Try ##t=0## and ##t=1## to see if it goes through your two points.

Sorry if this is a dumb question but this is my first math course in... a while. Is this what you meant?

I set t=0 and t=1 for the parametric equations and got the coordinates

t=0, (-3,5,2)
t=1, (2,7,1)
 
Yes, that's what LCKurtz meant.

Specter said:
I am not sure which point to use, P or Q, and if it matters.
Doesn't matter -- you can use either one.
 
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Mark44 said:
Yes, that's what LCKurtz meant.

Doesn't matter -- you can use either one.
Thank you.
 
Specter said:

Homework Statement


[/B]
Write vector and parametric equations for the line that goes through the points P(–3, 5, 2) and Q(2, 7, 1).

Homework Equations

The Attempt at a Solution



First I find the direction vector for PQ.

PQ=Q-P = (2,7,1)-(-3,5,2)
=[2-(-3),7-5,1-2]
=5,2,-1

PQ= (5,2,-1)

Now I have a direction vector and two points. I think I can write the vector equation now but I am not sure which point to use, P or Q, and if it matters.

If I were to use P(-3,5,2) the vector equation would look like the following:

(x,y,z)=(-3,5,2)+t(5,2,-1)

Then the parametric equations would be

x=-3+5t
y=5+2t
z=2-t
Well, I wrote this reply and failed to actually post it. :oops:
Doesn't really say anything much different than what the others said.
...

Your results look fine.

It does not matter which of the points you start with. In fact you can find some other point on this line and use that.

Using s as the parameter, start at point Q: (2, 7, 1) .
 
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