Determine an orthonormal basis

[subtle quiver]

Determine an orthonormal basis where two of the basevectors are parallel to the plane 6x-5y-5z=6.

My attempt:
The normal vector is (6,-5,-5) and it is perpendicular to the plane.

so for v1 and v2 to be parallel to the plane, the dot product between v1,2 and n needs to be 0.

So if I for example chose (-5,5,-11) as v1, the dot product would be -5*6+5(-5)-11(-5)=0.

This SHOULD mean that v1 is parallel to plane, right? The reason I'm asking is because if I plug in the coordinates (-5,5,-11) into the plane equation it gives me 0, and not 6.

[zenith crane]

Don't worry, that is fine. The left side is the dot product n·r, anyway.
Here's how you can do it generally, by the way.
Pick an arbitraty vector r0, then calculate the cross product of it and n:
r1 = r0⨯n
Since r1 is perpendicular to n, it must belong to a plane. Thus, it satisfies our condition.
Next, let's calculate the cross product of r1 and n.
r2 = r1⨯n
So, r2 is perpendicular to both r1 and n, and this can be taken as the third vector in our basis.
So, we obtain an orthogonal basis: {n, r0⨯n, (r0⨯n)⨯n}. To make it orthonormal, just divide all three vectors by their lengths.

[zenith crane]

Here's an example.
We have n = {6, -5, -5}.
Let's take r0 = {1, 0, 0}. Then:
r1 = r0⨯n = {1, 0, 0}⨯{6, -5, -5} = {0, 5, -5} = 5{0, 1, -1}
Length doesn't matter for now, so let's take r1 = {0, 1, -1}.
r2 = r1⨯n = {0, 1, -1}⨯{6, -5, -5} = {-10, -6, -6} = (-2){5, 3, 3}
Again, we take r2 = {5, 3, 3}.
So, our orthogonal basis is:
n = {6, -5, -5}
r1 = {0, 1, -1}
r2 = {5, 3, 3}
You can verify that it is indeed orthogonal.
So, now we make it orthonormal - we replace vectors by their orts:
e(n) = (1/√(86)){6, -5, -5}
e(r1) = (1/√(2)){0, 1, -1}
e(r2) = (1/√(43)){5, 3, 3}

Of course, as I said, the choice of r0 is arbitraty. Though, at least, take care not to pick r0 parallel to n 😄

[subtle quiver]

Would this be a possible solution then? I will do the exercise again using your way of doing it so I hopefully learn this better 🙂

[zenith crane]

v1 and v2 don't seem to be perpendicular.

[subtle quiver]

Oh, I thought that in order for them to be parallel to the plane they just need to be perpendicular to the normal vector.

Alright, I'll start from scratch doing it your way 😄

[zenith crane]

Well, they are parallel to the plane, but we need the basis to be orthogonal, so they also need to be perpendicular to each other.

[subtle quiver]

Ah yes.. I completely forgot about that part

[zenith crane]

No worries!
That's the reason of my approach: the cross product makes sure that the vectors are all orthogonal to each other, you don't need to check.

[subtle quiver]

How about this? 😬

[zenith crane]

r2 is wrong, should be {-50, -6, -6}. Basically, refer to the example that I gave, since you picked the same r0 as me.

[subtle quiver]

ah yeah, I see what I did wrong now

actually no, I see how u get 50i, but not how you would get -6j and -6k

[zenith crane]

Ah, sorry! I meant -30 and -30.