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

[karush]

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

Spoiler: book example

 

[skeeter]

direction vector, $v = (11,-4,48)-(-2,18,31) = (13,-22,17)$

$r(t) = (-2,18,31) + (13,-22,17)t$

 

[karush]

that it?
I was expecting a steps salad?

MHB replies to some calculus problems

 

[Kansas Boy]

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).