How do I find the equation of the plane? Given (x-1)/3 = (y+3)/1 = (z)/2 and point (2,3,1)?

Question

Equation of the line:

(x-1)/3 = (y+3)/1 = (z)/2

Given point (2,3,1)

Find the equation of the plane that has this line and point.
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OK so I got to this point:

In vector form, r(t) = (1,-3,0) + t(5,1,2)

Let t = 0 then (1,-3,0) is in the plane
Let t = 1 then (6,-2,2) is in the plane

So I now have 3 points on the plane. How do I use this to find the equation of the plane?

If I assign "P,Q,R" to these and use:

n = (q-p) x (r-p) = (x,y,z) etc.

Does it matter what point is assigned to P,Q or R? This is what I can't get my head around.

Answer 1

Find the equation of the plane containing the line:
t = (x - 1)/3 = (y + 3)/1 = z/2
and the point Q(2, 3, 1).
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First let's recast the equation of the line into something computationally more convenient.

L(t):
x = 1 + 3t
y = -3 + t
z = 2t

L(t) = P + tv
L(t) = <1, -3, 0> + t<3, 1, 2>
where t is a constant ranging over the real numbers

A plane has two directional vectors. One of them is v. Another one is PQ.

PQ = = <2-1, 3+3, 1-0> = <1, 6, 1>

We don't need to find a third point. And no, it doesn't matter what letter designations you give to which point. You just need to be consistent in how the points are used. We needed to get two directional vectors for the plane so we can calculate the normal vector.

The normal vector n, of the plane is orthogonal to both of the directional vectors of the plane. Take the cross product.

n = PQ X v = <1, 6, 1> X <3, 1, 2> = <11, 1, -17>

With the normal vector of the plane and a point in the plane we can write the equation of the plane. We could use either point P or Q. Let's choose Q.

11(x - 2) + 1(y - 3) - 17(z - 1) = 0
11x - 22 + y - 3 - 17z + 17 = 0
11x + y - 17z - 8 = 0