Finding the Parametric Equations of a Line (0,4) & Alpha 60°; Beta 30°, 150°

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Homework Help Overview

The discussion revolves around finding the parametric equations of a line that passes through the point (0,4) with a given direction angle of 60°. The problem involves understanding direction angles and their implications in a two-dimensional context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the definition of direction angles and their relationship in two dimensions. Questions arise regarding the meaning of "second direction angles" and the necessity of multiple angles. Various attempts to derive parametric equations based on the given angle and point are discussed.

Discussion Status

Some participants provide vector representations and equations based on the slope derived from the direction angle. There is a recognition of differing interpretations regarding the number of direction angles and their relevance to the problem. No explicit consensus is reached, but several lines of reasoning are presented.

Contextual Notes

Participants note the potential confusion surrounding the term "second direction angles" and the implications of direction angles in two-dimensional geometry. The original poster's reference to a book answer is also acknowledged, indicating a possible discrepancy in understanding.

PiRsq
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I tried to do this but I don't get the answer as in the book...


A line passes through point (0,4). Its first direction angle is 60°, meaning alpha is 60°.

I found that the second direction angles are 30° and 150° that's is beta is 30° and 150°. But what are the parametric equations of the line for each set of direction angles?

Thx any help appreciated
 
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My understanding of "direction angles" for a line is that they are the angles the line makes with each of the coordinate axes. In a two dimensional problem, there are only two direction angles. In this case, since one angle is given as 60 degrees (the angle the line makes with the x-axis) the other angle has to be 30 degrees (in two dimensions, the angles have to add to 90 degrees- direction angles (and direction cosines) are more often used in three or more dimensions). I don't understand why "second direction angles" is plural. I also do not understand what you mean by " the parametric equations of the line for each set of direction angles".
 
I agree with HallsofIvy that a third direction angle makes no sense in this problem. So ignoring that ...

A vector along that line would be
v = i + √(3)j
or <x,y> = <1,&radic;(3)>

So a vector equation for the line in the form
r = r0 + tv (where r0 is the position vector of your point (0,4) and t is the parameter might be
<x,y> = <0,4> + t<1,&radic;(3)> or
<x,y> = <0+t, 4+t&radic;(3)>

and then the parametric equations of the line would be
x = t
y = 4 + t&radic;(3)

Is that the answer given in your book?
 
Last edited:
That was probably too involved. Maybe this is a better answer:

Since you know that &alpha; is 60o, you know that the slope of the line is tan60o = &radic;(3)

So, an equation for the line through point (0,4) with slope &radic;(3) is
y - 4 = &radic;(3) * (x - 0)
y = x&radic;(3) + 4

To parametrise this, let the parameter be t.
Now, since there are no restrictions on x, we can simply let
x = t
and then, since we require that y = x&radic;(3) + 4, substitute the parameter t for x and you get
y = t&radic;(3) + 4

So those are the parametric equations.
 

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