Vectors Perpendicular to Plane: nhat

[theneedtoknow]

If I have a vector nhat = [nx, ny, nz] which is normal to some plane, how can I write the vectors (I assume there are infinitely many) which are perpendicular to that plane?

 

[Char. Limit]

Well, normal is perpendicular. So the set of vectors perpendicular to that plane would be any non-zero multiple of $\hat{n}$.

 

[theneedtoknow]

Well, normal is perpendicular. So the set of vectors perpendicular to that plane would be any non-zero multiple of $\hat{n}$.

Sorry, not thinking very clearly. I meant to type the set of unit vectors which are PARALLEL to that same plane

 

[Char. Limit]

Ah, well that is different. We need to find the set of all vectors normal to our normal vector. Now, one requirement of normality is that the dot-product be zero, i.e.

$$\vec{T} \cdot \hat{n} = 0$$

I use T here because any vector that's normal to the normal will be tangent to the plane. Now, with that...


hm. This will require a bit more thought. I'll be right back. In the meantime, hopefully someone who has already figured out the answer will stop by!

 

[Char. Limit]

All right, I'm back and I've figured it out. Start with T dot N = 0, which we know to be true. Expand this out to get:

$$T_1 n_x + T_2 n_y + T_3 n_z = 0$$

$$T_1 = - T_2 \frac{n_y}{n_x} - T_3 \frac{n_z}{n_x}$$

And from there, it should be trivial to express T as a linear combination of vectors with coefficients (n_y)/(n_x) and (n_z)/(n_x), respectively. That'll give you your plane. :)

That was fun!

Note: We assume n_x is not zero. If it is, this problem becomes a lot more trivial.

 

[theneedtoknow]

Thank you so much for the help!

 

[HallsofIvy]

Of course, any vector in the plane is a vector parallel to the plane. You don't need a normal vector to find that.

If the plane is given by z= ax+ by+ c and we take x= y= 0, z= c so (0, 0, c) is a point in the plane. And for any numbers, X and Y, (X, Y, aX+ bY+ c) is also a point in the plane. The vector from the first to the second is $X\vec{i}+ Y\vec{j}+ (aX+ bY)\vec{k}$ is a vector in (parallel to) the plane. That can be written as $X(\vec{i}+ a\vec{k})+ Y(\vec{j}+ b\vec{k})$ indicating that $\vec{i}+ a\vec{k}$ and $\vec{j}+ b\vec{k}$ form a basis for the vector space of all vectors parallel to the plane z= ax+ by+ c.