MHB Another equation of another line

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To find the parametric and symmetric equations for a line through the point (1, -1, 1) and parallel to the line given by the equation x + 2 = (1/2)y = z - 3, one must first determine the direction vector from the original line. The correct direction vector can be derived from the equations, leading to the realization that the components must be separated for clarity. By expressing the x, y, and z components separately, one can derive the parametric equations and then solve for the parameter t. This process allows for the conversion back to the vector equation, facilitating a clearer understanding of the line's properties.
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Please help, how do I do this?
Find the parametric and symmetric equation for a line through (1,-1,1) and parallel to $x+2= \frac{1}{2}y=z-3$

So confused
 
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ineedhelpnow said:
Please help, how do I do this?
Find the parametric and symmetric equation for a line through (1,-1,1) and parallel to $x+2= \frac{1}{2}y=z-3$

So confused

Parallel lines have the same direction vectors.

Can you get a direction vector for $\displaystyle \begin{align*} x + 2 = \frac{1}{2}y = z - 3 \end{align*}$?
 
That's what I'm stuck on.at first I though it would be v=(-2,2,3) but then I came to the conclusion that it was wrong so I am stuck again.
 
ineedhelpnow said:
That's what I'm stuck on.at first I though it would be v=(-2,2,3) but then I came to the conclusion that it was wrong so I am stuck again.

You need to get better at the problem solving strategy of thinking in reverse. How would you go from a vector form of a line to the symmetric form?

Start by writing out the x, y, z components separately to get the parametric equations.

Solve for t in each equation.

Realise since they are all equal to t, they are equal to each other.

Now how do you think you can go in reverse to get back to the vector equation?
 
i figured it out :)