How Do You Solve a Parametric Vector Equation for a Line?

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To solve a parametric vector equation for a line through points (2,4,1) and (8,-2,4), first, find the direction vector by subtracting the coordinates of the first point from the second, resulting in {6, -6, 3}. The parametric equation can be expressed as {x, y, z} = {2, 4, 1} + t{6, -6, 3}, where t is a parameter. For part B, to find unit direction vectors, normalize the direction vector by dividing by its magnitude, yielding unit vectors of {-2/3, 2/3, -1/3} and {2/3, -2/3, 1/3}. The discussion emphasizes understanding vector equations and the flexibility in choosing the parameter t.
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Please help! Parametic vector equation??

My lecturer is incredibly hard to understand and I have NO idea how to do this assignment. If someone could help me with this first question I think I might be able to do the rest by myself:

A) Find a parametric vector equation for the line through the points (2,4,1) and (8,-2,4)
B) find both unit direction vectors for the same line

Please show working and explain best you can how your doing it. I'm trying really hard in this course but the lecturer just writes on the black board without taking time to make sure anyone understand its :|
 
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150 viewers and no one can help? really? :\
 


Please help, everything is appreciated
 


Do you understand how to describe a vector given two points?

Do you understand what a parametric equation is? You'd have x, y, and z as functions of some 4th variable, such as t.

x(t), y(t), z(t).

You might want to choose t so it's corresponds to a distance of 1.
 
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I understand vectors but I don't know where to even begin with this question.
 


Ok, so what is the vector in this case, based on the two points you are given?
 
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I don't know how to find the vector given 2 points. I know that's probably really stupid of me and it's really obvious :\ I try to pay attentions in the lectures but it's really hard to keep up. So far the only question I've managed to do on the assignment was proving two lines are skewed
 


To get the vector, you just subtract one of the points from the other:

{8,-2,4} - {2,4,1} = {6, -6, 3}

Then an equation for the line is the second point + t times the vector:

{x, y, z} = {2, 4, 1} + t{6, -6, 3}

x = 6 t + 2
y = -6 t + 4
z = 3 t + 1
 
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Alright looking at a similar question on yahoo answers I think I might be on the right track (?)

(2,4,1) - (8,-2,4) = -6i+6j-3k
=> Equation is: (8,-2,4) + t(-6,6,-3)

Am I right or..?
 
  • #10


By the way thanks for the help I really appreciate it
 
  • #11


Tui said:
Alright looking at a similar question on yahoo answers I think I might be on the right track.

(2,4,1) - (8,-2,4) = -6i+6j-3k
=> Equation is: (8,-2,4) + t(-6,6,-3)

Am I right or..?

Yes, you can use either point for the base. The vector can go in either direction. Sometime a problem will state that the vector goes from one point to another, but in this case the problem just mentions two points, so the vector could point either way.
 
  • #12


Oh ok cool :)

What about part b? Is that just asking for the other equation (Using the opposite point)?
 
  • #13


Tui said:
What about part b? Is that just asking for the other equation (Using the opposite point)?
A unit vector has a length of 1. To convert a vector to unit length, divide by the square root of the sum of the squares of the 3 parameters.

vector = {-6, 6, -3}

unit vector = {-6, 6, -3} / sqrt( (-6)2 + (6)2 + (-3)2 )
unit vector = {-6, 6, -3} / sqrt (36 + 36 + 9)
unit vector = {-6, 6, -3} / sqrt (81)
unit vector = {-6, 6, -3} / 9
unit vector = {-6/9, 6/9, -3/9}
unit vector = {-2/3, 2/3, -1/3}

The other direction just flips the signs

other unit vector = {2/3, -2/3, 1/3}

although not asked for, if you wanted to write parametric equations for the line based on unit vector you could have:

{x, y, z} = {8, -2, 4} + t{-2/3, 2/3, -1/3}

or

{x, y, z} = {2, 4, 1} + t{2/3, -2/3, 1/3}

There's no rule that you have to use just t, you could use 9 t or even t3, but normally you use the simplest case, unless the problem specifies how t should be used.
 
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  • #14


Oh wow I remember the lecturer doing that on the board and wondering what it was. Thanks so much for all your help !
 
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