finding normal unit vector

[Henk42]

Problem

See attached drawing.

I need to find the normal unit vector of the plane defined by the points Q1, Q2 and Q3
I have 3 known points P1 (=Q1), P2 and P3.
n2 and n3 are normal to the plane defined by Q1, Q2 and Q3
The lengths of n2 and n3 are known.

There should be 2 solutions as the Q-plane can be at the other side of the P- triangle as well

How to solve this?

 

[Henk42]

Find nu (unit vector n)
N2 = |n2|
N3 = |n3|

I started by noting that:
n2 = N2.nu
n3 = N3.nu

n2 is perpendicular to q1-q2 --> dot product = 0
q12 = p12 + n2
(p12 + n2).n2 = 0
p12.n2 + n2.n2 = 0

N2(p12.nu) + N2^2 = 0
N3(p13.nu) + N3^2 = 0
|nu| = 1

3 equations with 3 unknowns (nux, nuy and nuz)
However this becomes very hairy to solve......

 

[pka]

Problem: I need to find the normal unit vector of the plane defined by the points Q1, Q2 and Q3
How to solve this?

The three points [imath]Q_1,~Q_2,~\&~Q_3[/imath] must be non-collinear otherwise they would not determine a plane.
Define [imath]\overrightarrow v = \overrightarrow {{Q_1}{Q_2}}~\&~\overrightarrow w = \overrightarrow {{Q_1}{Q_3}} [/imath]
Now let [imath]\overrightarrow n = \overrightarrow v \times \overrightarrow w [/imath] is the normal to the plane determined by [imath]Q_1,~Q_2,~\&~Q_3[/imath].
So [imath]\overrightarrow u = \dfrac{{\overrightarrow n }}{{\left\| {\overrightarrow n } \right\|}}[/imath] is a unit normal to the plane.
[imath][/imath][imath][/imath]

 

[Dr.Peterson]

Problem

See attached drawing.

I need to find the normal unit vector of the plane defined by the points Q1, Q2 and Q3
I have 3 known points P1 (=Q1), P2 and P3.
n2 and n3 are normal to the plane defined by Q1, Q2 and Q3
The lengths of n2 and n3 are known.

There should be 2 solutions as the Q-plane can be at the other side of the P- triangle as well

How to solve this?

As I understand it, you are given only P₁, P₂, P₃, and the lengths |n₂| and |n₃|. Thinking geometrically, we'd like to construct points Q₂ and Q₃ so that n₂ and n₃ are parallel and angles P₂Q₂Q₃ and P₃Q₃Q₂ are right angles.

​

So P₂Q₂Q₃P₃ is a right angled trapezoid with three known lengths.

I would imagine making a sphere around P₃ with radius |n₃| - |n₂|, and constructing a tangent P₂R₃ to the sphere. (This assumes |n₂| < |n₃|.)

(I don't have time to fully work this out.)

 

[Henk42]

The three points [imath]Q_1,~Q_2,~\&~Q_3[/imath] must be non-collinear otherwise they would not determine a plane.
Define [imath]\overrightarrow v = \overrightarrow {{Q_1}{Q_2}}~\&~\overrightarrow w = \overrightarrow {{Q_1}{Q_3}} [/imath]
Now let [imath]\overrightarrow n = \overrightarrow v \times \overrightarrow w [/imath] is the normal to the plane determined by [imath]Q_1,~Q_2,~\&~Q_3[/imath].
So [imath]\overrightarrow u = \dfrac{{\overrightarrow n }}{{\left\| {\overrightarrow n } \right\|}}[/imath] is a unit normal to the plane.
[imath][/imath][ima

The three points [imath]Q_1,~Q_2,~\&~Q_3[/imath] must be non-collinear otherwise they would not determine a plane.
Define [imath]\overrightarrow v = \overrightarrow {{Q_1}{Q_2}}~\&~\overrightarrow w = \overrightarrow {{Q_1}{Q_3}} [/imath]
Now let [imath]\overrightarrow n = \overrightarrow v \times \overrightarrow w [/imath] is the normal to the plane determined by [imath]Q_1,~Q_2,~\&~Q_3[/imath].
So [imath]\overrightarrow u = \dfrac{{\overrightarrow n }}{{\left\| {\overrightarrow n } \right\|}}[/imath] is a unit normal to the plane.
[imath][/imath][imath][/imath]

Unfortunately i dont know Q1,2,3. Only P 1,2,3