pi through points

[sand steppe]

Hi I need help with this question.

"The plane π passes through the points (1, 1, 2), (2, 1, 3) and is parallel to the line l : (x, y, z) = (1 + t, 2 + 2t, −t).
Determine the (orthogonal) projection and mirror image of the point P = (2, 1, 1) in π."

And btw, this is not an ON-base right, because the dot product is 0?

[solid imp]

Suppose the points are r1 and r2 and the line is r = r0 + st.
Then the normal vector of the plane must be perpendicular to (r2 - r1) and s. Thus:
n = (r2 - r1)⨯s
As for the projection:
We have a point P = r3. First, let's pass a line perpendicular to the plane:
r = r3 + nt
As the equation of the plane is n·(r - r1) = 0, we can substitute the equation of the line into it:
n·(r3 + nt - r1) = 0
n·(r3 - r1) + |n|^2 t = 0
t = n·(r1 - r3)/|n|^2
Now we substitute it back into the equation of the line to find the projection. Let's denote the plane by π. Then:
proj(π, r3) = r3 + (n·(r1 - r3)/|n|^2)n
Now, the reflection. As it requires us to move the same distance in the same direction as the projection, but once more, the value of t is doubled:
t = 2n·(r1 - r3)/|n|^2
So:
ref(π, r3) = r3 + 2(n·(r1 - r3)/|n|^2)n

[sand steppe]

thats the projection formula right?

n = (1, 0 , 1)

so the projection of p on pi is (1, 1/2, 1/2) ?

[solid imp]

No, it should be {4/3, 5/3, 5/3}.

[sand steppe]

Hello, I totally missed this answer I still need some assistance if its ok?

r = (2, 1, 1) + (1, 0, 1)t?

so r is 2+t, 1, 1+t ?

[solid imp]

No, n is not {1, 0, 1}.

[sand steppe]

{-1, 0 ,-1?

[solid imp]

Nope. As I said, n = (r2 - r1)⨯s.
In our case r1 = {1, 1, 2}, r2 = {2, 1, 3}, s = {1, 2, -1}. So, what will n be?

[sand steppe]

Oh right the times s, didnt see it.

so s has the coordinates from the te line (1+t, 2+2t, −t).

?

[solid imp]

Yes.

[sand steppe]

{1, 0, 1 } x {1, 2, -1}

[solid imp]

Yes.

[sand steppe]

the scalar product is 0 so thats an ON-base