Determining A Vector Equation — What Am I Doing Wrong?

[moe darklight]

Determining A Vector Equation — What Am I Doing Wrong??

First day and already I'm screwing up?

Problem is the method the proff used in class is not the same as the one used in the solution manual... and the textbook simply doesn't go over this information, which is very odd; it's an out of the blue question about something not covered in the chapter; all it covers is how to determine a vector equation from a line that passes through point x with direction vector y, or with a certain slope, but not how to do this... so basically I have no clue what I'm doing.

Homework Statement

Determine a vector equation:

$$\[<br /> x_2 = 3x_1 + 2<br /> \]$$

Homework Equations

solution given by book: $$\[<br /> x = ( - 1, - 1) + t(1,3)<br /> \]$$

The Attempt at a Solution

$$\[<br /> \begin{array}{l}<br /> x_2 = 3x_1 + 2 \\ <br /> x_2 - 1 = 3x_1 + 1 \\ <br /> \frac{{(x_2 - 1)}}{3} = x_1 + \frac{1}{3} = t \\ <br /> x_1 = - \frac{1}{3} + 1t \\ <br /> x_2 = 1 + 3t \\ <br /> \\ <br /> x = ( - \frac{1}{3},1) + t(1,3) \\ <br /> \end{array}<br /> \]$$

 

[moe darklight]

The way the proff. does it would be by making $$t=x_{1}$$ and working from there... which to me makes sense (as opposed to my attempt at the textbook method above, which makes no sense to me), but this too gives me different numbers from what the textbook says.

The way the textbook answers this type of question is this:

$$\[<br /> \begin{array}{l}<br /> 2x_1 + 3x_2 = 5 \\ <br /> 2x_1 - 2 = - 3x_2 + 3 \\ <br /> (x_1 - 1)/3 = - (x_2 - 1)/2 = t \\ <br /> etc... \\ <br /> \end{array}<br /> \]$$

Nowhere does it explain why it's doing any of this...

 

[Defennder]

Your answer is the same as that given. There are infinitely many points on any given line, so you choosing (-1/3,1) is different from someone else choosing (-1,-1). Both points line on the same line.

 

[moe darklight]

hm... o boy

thanks.

I guess today I've also learned never to underestimate my own stupidity