Question 205178: Hi all, can anyone please show me how to; Find the equation of the line that contains points (2,1,-3) and (-1,4,1)
Would really appreciate if someone could show me with steps and notes.
Thanks, -Nick
Answer by Edwin McCravy(20055) (Show Source):
You can put this solution on YOUR website! Hi all, can anyone please show me how to; Find the equation of the line that contains points (2,1,-3) and (-1,4,1)
Would really appreciate if someone could show me with steps and notes.
Thanks,
A line parallel to the vector v = ‹a,b,c› and passing through the
point P( , , ) is represented by the
parametric equations
, ,
or as the symmetric equations:
if none of a,b, or c are 0.
Begin by using the points P(2,1,-3) and Q(-1,4,1)
to find a direction vector for the line passing through
P and Q, given by
__
v = PQ = ‹-1-(2),4-1,1-(-3)› = ‹-3,3,4›
So we substitute in
, ,
with ‹a,b,c› = ‹-3,3,4› and the point P( , , ) = P(2,1,-3)
, ,
That's the parametric equations for the line.
If you want the symmetric equation of the line, we
substitute in
Edwin
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