Find the parametric equation of a curve which goes through the points (1,1,3), (2,1,4) and (3,6,9)?

Answer 1

call P=(1,1,3), Q=(2,1,4), O=(3.6.9)
vector PQ=(1,0,1)
vector PO=(2,5,6)
find normal vector n= PQ x PO=(-5,-4,5)
i am not sure what you mean by parametric equation. i think that is plane equation or symmetric equation.
if youhave 2 points you can write parametric equation
x=x0+at, y= y0+bt, z=z0+ct
but we have 3 points that should be plane
my answer is
-5(x-1)-4(y-1)+5(z-3)=0
-5x+5-4y+4+5z-15=0
-5x-4y+5z-6=0
or 5x+4y-5z+6=0
good luck

Answer 2

Find the parametric equation of a curve which goes through the points P(1,1,3); Q(2,1,4) and R(3,6,9).

Define the vectors:
u = PQ = <2-1, 1-1, 4-3> = <1, 0, 1>
v = PR = <3-1, 6-1, 9-3> = <2, 5, 6>

u and v are not parallel so the points are not collinear. There are an infinite number of curves that could go thru the three points. There is not enough information to answer the question as stated.

Are you possibly looking for the parametric equation of the plane thru the three points? That can be calculated.

P(1,1,3)
u = PQ = <2-1, 1-1, 4-3> = <1, 0, 1>
v = PR = <3-1, 6-1, 9-3> = <2, 5, 6>

Plane = P + su + tv = <1, 1, 3> + s<1, 0, 1> + t<2, 5, 6>
Plane = <1 + s + 2t, 1 + 5t, 3 + s + 6t>
where s and t are scalars that range over the real numbers

Expressed parametrically we have:

x = 1 + s + 2t
y = 1 + 5t
z = 3 + s + 6t

Answer 3

The "easy and cheating" way to do it, is to make your curve not-continuous, and define it by parts, thus getting the equations
x={-t+(1+t)2 if t<0
{3t+(1-t)6 if t>=0
y={1 if t<0
{6t+(1-t) if t>=0
z={-3t+(1+t)4 if t<0
{4t+(1-t)9 if t>=0
goes through the point A for t=-1, B for t=0 and C for t=1.


Finding a regular curve (i.e. continuous, etc.) is undoubtedly possible, but harder.
You can see the problem as finding a polynomial such that, coordinate by coordinate, their value is, say x_a, y_a, z_a, at a time pre-determined.
There is a general formula for such polynomials, called Lagrange interpolation polynomials.
Note that L_i(X)=product(X-x_j)/product{(X-x_i), i different from j} is zero on every x_j with j different from x, and L_i(x_i)=1. Thus, sum(L_i(x)a_i) is a polynomial with value a_i at point x_i.
Thus you can get a polynomial curve passing through your three points.

Answer 4

You have given incomplete information.

With the given data there can be an infinite number of curves passing through the three points in a space of three dimensions.