Projecting a vector onto a plane.

[IgF]

Hi, I've been studying up on analytical geometry and trig as I want to start a course in videogame dev. The 3d math primer I am following isn't too descriptive in this particular area so if anyone can help me with this question I can apply it to the rest of section. Thanks for any help or direction.

Project the following vector onto the plane.

Vector: < 18, 52, 42 >

Plane: y = 9x + 13y + 7z + 29

 

[Rasalhague]

First, I'd simplify "y = 9x + 13y + 7z + 29" by collecting all the y terms, subtracting y from both sides, like this: "0 = 9x + 12y +7z + 29". To find the components of a normal vector, n - that is, a vector at right angles to the plane - just read off the coefficients of x, y and z. So n = < 9, 12, 7 >, unless the y on the left of your equation for the plane was a typo! If the equation is actually 0 = 9x +13y + 7z + 29, then n = < 9, 13, 7 >.

This works because, if we let r be any point in the plane, and r₀ some particular, fixed point in the plane, then for every point r, the vector r - r₀ is parallel to the plane. Being parallel to the plane means it's at right angles to the normal vector. We can express this condition with the dot product:

$$\textbf{n} \cdot (\textbf{r}-\textbf{r}_0)=0.$$

In other words, if we let n = < a, b, c>, then

$$a(x-x_0) + b(y-y_0) + c(z - z_0)=0,$$

$$ax + by + cz = ax_0 + ay_0 + bz_0.$$

But the right side of that last equation is a constant: it's the same number for every point r in the plane. So one consistent choice of a, b and c is to take them to be the coefficients of x, y and z in your equation, letting ax₀ + ay₀ + az₀ = -29.

Now, a vector in the direction of the projection is

$$\textbf{a}=\textbf{n} \times (\textbf{v} \times \textbf{n})$$

where the times symbol stands for the cross product, also called the vector product. If this isn't obvious, it might help to draw a diagram. The projection of v onto a is

$$\frac{\textbf{a}\cdot \textbf{v}}{\textbf{a}\cdot \textbf{a}} \, \textbf{a}.$$

Here the denominator cancels out the length of a. It's equivalent to first making a into a unit vector.

 

[eehsun]

You could project your vector onto the normal vector of the plane, and substract the vector you've found in this way from the original vector.

 

[Rasalhague]

Good idea, eehsun. That's simpler than my suggestion.

 

[karawan]

vector 3D

If i given a terminal and initial points of a vector in x,y,z positive plane . how to find three other vectors in the same plane having the same magnitude and the same direction and the same slope ?

 

[Rasalhague]

I'm not sure I understand your question, karawan. R³ is the set of all possible lists of three real numbers: (9,5,-18), (-0.5,421,0), and so on. These lists are the vectors of R³, also called points. A plane is a subset of R³ such that the difference between every pair of vectors - that is, every pair of points - in this subset is orthogonal to a particular non-zero vector. This particular vector is what we've been calling the normal vector in the earlier posts of the thread. For two vectors to be orthogonal (synonyms: normal, at right angles) means for their dot product to be zero.

I think you might be asking: suppose we're given a plane in R³, and a pair of points (=a pair of vectors, two vectors) whose components are all positive, in this plane (NOTE: not every plane contains such vectors), how can we find three other pairs of vectors such that the magnitude of the difference between each pair of vectors is the same (the difference between two vectors, u and v, is the vector, v - u), and the dot product of their difference with any vector orthogonal to the plane is zero? That is, if the vectors are u and v, and the normal vectors n, then n.(v-u) = 0 = n.(u-v).

Perhaps you want the three other pairs of points to also have all positive components.

A clue to finding vectors of the same magnitude: do you know how to find a unit vector in a given direction?

 

[karawan]

thank you for reply. that's help.

 

[IgF]

Thanks everyone I get it now, I did it using eehsun's method. But again thanks so much!

 

[Grandpsykick]

Just more on this, because I got lost as well:

For a vector $\textbf{v}$ being projected on a plane $\mathcal{P}$ with plane unit normal $\textbf{n}$. The part of $\textbf{v}$ on $\mathcal{P}$ can be shown to be:

$\textbf{u} = \textbf{v} - \textbf{n} (\textbf{v} \cdot \textbf{n})$​

This is equivalent to the triple cross product (and BAC CAB rule):

$\textbf{u} = \textbf{n} \times (\textbf{v} \times \textbf{n}) = \textbf{v} (\textbf{n} \cdot \textbf{n}) - \textbf{n} (\textbf{v} \cdot \textbf{n})$​