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

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To find the parametric equations for the line segment from (-2, 18, 31) to (11, -4, 48), the direction vector is calculated as (13, -22, 17). The parametric equations can be expressed as r(t) = (-2, 18, 31) + (13, -22, 17)t. By setting t=0 at the starting point and t=1 at the endpoint, the equations are derived as x = -2 + 13t, y = 18 - 22t, and z = 31 + 17t. This formulation confirms that when t=0, the point is (-2, 18, 31) and when t=1, it reaches (11, -4, 48).
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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

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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).