1. Mar 22, 2012

### Tui

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 :|

2. Mar 22, 2012

### Tui

150 viewers and no one can help? really? :\

3. Mar 22, 2012

### Tui

4. Mar 22, 2012

### rcgldr

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.

Last edited: Mar 22, 2012
5. Mar 22, 2012

### Tui

I understand vectors but I don't know where to even begin with this question.

6. Mar 22, 2012

### rcgldr

Ok, so what is the vector in this case, based on the two points you are given?

Last edited: Mar 22, 2012
7. Mar 22, 2012

### Tui

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

8. Mar 22, 2012

### rcgldr

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

Last edited: Mar 22, 2012
9. Mar 22, 2012

### Tui

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. Mar 22, 2012

### Tui

By the way thanks for the help I really appreciate it

11. Mar 22, 2012

### rcgldr

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. Mar 22, 2012

### Tui

Oh ok cool :)

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

13. Mar 22, 2012

### rcgldr

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.

Last edited: Mar 22, 2012
14. Mar 22, 2012

### Tui

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