Why is $(1, 2, 3)$ the normal vector to $x + 2y + 3z = 0$?

Question

Hi — I am confused as to why the normal vector of $x + 2y + 3z = 0$. This is in the context of this problem: Let $T:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ be the orthogonal projection of $\mathbb{R}^{3}$ to the plane $x + 2y + 3z = 0$. Find the kernel of $T$. Find the image of $T$.

I am also confused as to why, when I solve for the kernel of $T$ using the image of $T$, I don't get the correct value $(1, 2, 3)$. Shouldn't I get the correct answer if I solve for the kernel of the image of $T$?

Answer 1

The plane defined by $x+2y+3z=0$ is the colleciton of points that satisfy this equation, which we can rewrite as $$ (1, 2, 3)\cdot (x, y, z) = 0 $$ So the plane is the collection of vectors that are orthogonal to $(1,2,3)$

Answer 2

Let $\vec {n}=(1,2,3) $. $O(0,0,0) $ is a point of the plane

$$P \;: x+2y+3z=0$$

let $M (a,b,c) $ be an other point of $P $.

$$\vec {n}.\vec {OM}=a+2b+3c=0$$

thus $\vec {OM} $ and $\vec {n} $ are perpendicular.

Answer 3

One way to think about a geometric object (e.g., plane, line, sphere) is to ask the question:

"What is invariant numerically about all the points making up the object?"

Let's ask this question about a plane. The question is easier to answer if you visualize a direction vector o pointing perpendicular to the plane (its "orientation"). Now, think of o as representing an alternate x-axis emanating from the origin. What would be the coordinates of a point p on the plane along this new x-axis? Whatever the value is, all the points on the plane project onto this SAME new-x-axis coordinate. So, there is a way to capture the invariance of the points...you need the "normalized normal" o before the invariance is teased out.

What is the coordinate all the points share? It is the signed distance of the plane from the origin. This is L ("location" of the plane).

The vector dot product is the calculation that projects a point onto a new axis direction (specified as a direction vector). This leads to a geometrically meaningful invariance predicate for a plane:

 p • o == L

If you expand the dot product, you'll see the similarity to the algebraic standard equation for a plane:

ax + by + cz = d

o.x*x + o.y*y + o.z*z = L  (vector equation after expanding dot product)

The advantage of the vector geometry form p • o == L is that all its quantities have a spatial meaning that can be sketched (or rendered in CG). o is the 3D direction pointing perpendicular to the plane. L is the signed distance of the plane from the origin (where arrow o defines positive distances). And p is any point p = [ x y z ] in space. For all points p on the plane represented by [ o, L ], the equation will be true. Points where equality fails are not on the plane.

A practical use of this vector equation is that you can easily find the distance of any point p1 from the plane. Again, think of the plane as having its own axis defined by o. All you have to do is project p1 onto this coordinate axis o, and then subtract L from that coordinate (the coordinates of points on the plane):

  signed_proximity(p1) = p1 • o - L

This is very useful in 2D computer graphics for "line hit testing". That problem is to decide when the user is clicking the mouse on a visible line. If we define the pecking forgiveness as pixel_epsilon, then this calculation can make the desired decision:

  click on line(T/F) <-- abs( mousePos • o - L ) <= pixel_epsilon

Applying p • o == L to your example equation, < 1,2,3 > has to be normalized

o = < 1, 2, 3 > / || < 1, 2, 3 > || = < 1/sqrt(14), 2/sqrt(14), 3/sqrt(14) > = < 0.267, 0.534, 0.802 >

L = 0 (the plane passes thru the origin)

Answer 4

Let $\pi$ is our plane.

Hence, easy to check that $O(0,0,0)\in\pi$, $A(1,1,-1)\in\pi$ and $B(2,-1,0)\in\pi$. Hence, $\vec{OA}(1,1,-1)||\pi$ and $\vec{OB}(2,-1,0)||\pi$.

Also, easy to see that these vectors are linearly independent.

Thus, it's enough to prove that $$(1,2,3)(1,1,-1)=0$$ and $$(1,2,3)(2,-1,0)=0,$$ which is obvious.