MHB -12.5.2 Find Parametric eq for line segment from (-2,18,31) to (11,-4,48)

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Line Parametric
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
Find Parametric eq for line segment from (-2,18,31) to (11,-4,48)
ok not sure how to start on this the book example is in the spoiler

Screenshot 2021-12-28 6.59.44 PM.png
 
Physics news on Phys.org
direction vector, $v = (11,-4,48)-(-2,18,31) = (13,-22,17)$

$r(t) = (-2,18,31) + (13,-22,17)t$
 
Parametric equations for a straight line are of the form
x= at+ b
y= ct+ d
z= et+ f

We can take t to be any numbers we t to be whatever we like at the given points. I think it simplest to take t to be 0 and 1.

If t= 0 at (-2,18,31) then a(0)+ b= -2 so b= -2, c(0)+ d= 18 so d= 18, and e(0)+ f= 31 so f= 31.

If t= 1 at (11,-4,48) then a(1)+ b= a- 2= 11 so a= 13, c(1)+ d=c+ 18= -4 so c= -22, and e(1)+ f= e+ 31= 48 so e= 17.

x= -2t+ 13
y= -22t+ 18
z= 17t+ 31.

Check; if t= 0, (x, y, z)= (13, 18, 31). If t= 1, (x, y, z)= (-2+ 13, -22+ 18, 17+ 31)= (11, -4, 48).
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...
Back
Top