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

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The discussion focuses on finding the parametric equations of a line that passes through the point (0,4) with a direction angle of 60°. The second direction angle is clarified to be 30°, as the two angles must sum to 90° in two dimensions. A vector along the line is identified as v = i + √3j, leading to a vector equation of the line. The parametric equations derived are x = t and y = 4 + t√3, which reflect the slope determined from the angle. The final equations are confirmed to align with the expected results from the book.
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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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