Explain what this tells you about parametric and symmetric equations in R^3?

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parametric and symmetric equations in R^3??

Homework Statement



Recall that there are three coordinates planes in 3-space. A line in R3 is parallel to xy-plane, but not to any of the axes. Explain what this tells you about parametric and symmetric equations in R3. Support your answer using examples.


Homework Equations





The Attempt at a Solution





This question had me thinking for a while, but i want to confirm my understanding or if i am even correct.

I think the line is perpendicular to the Z-axis, So does this mean that the directional vector for z component is 0? basically [ax,ay,0].But couldn't it be perpendicular to xy plane, and still have a directional z component?
But besides that, using that information, in terms of parametric equation it tells us that
the z is just going to equal the position vectors z component.

Now here is where i get confused...with the symmetric equation.
Can you guys please explains this to me in simplest terms...i am trying really hard to understand this vectors in 3 space stuff.
 
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To answer your first questions, yes, it's perpendicular to the Z-axis and there can be no value for the Z component other then 0, otherwise it wouldn't be perpendicular to your xy-plane. If it were perpendicular to the xy-plane, then by definition it would have a Z-component. However, the question is talking about being parallel to the xy-plane.
 


Ah alrite i just got that i decided to draw out a graph...yes what you said sums that up...now onto the parametric and symmetric equations
 


can someone please help with the other problem?
 


Pengwuino said:
To answer your first questions, yes, it's perpendicular to the Z-axis and there can be no value for the Z component other then 0, otherwise it wouldn't be perpendicular to your xy-plane. If it were perpendicular to the xy-plane, then by definition it would have a Z-component. However, the question is talking about being parallel to the xy-plane.

well can you answer the second part? am i atleast close?
 


Given x, y, z as parametric equations with parameter t, you can find the "symmetric" equations by solving each equation for t and setting them all equal. In this case, z is a constant: there is no t so you can't solve for t. The symmetric equations are an equation in x and y with the additional equation "z= constant".
 


HallsofIvy said:
Given x, y, z as parametric equations with parameter t, you can find the "symmetric" equations by solving each equation for t and setting them all equal. In this case, z is a constant: there is no t so you can't solve for t. The symmetric equations are an equation in x and y with the additional equation "z= constant".

So Symmetric would look like:
\frac{x-ax}{bx}= \frac{y-ay}{by}= az
(where 'az' is a constant)
 

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