LexyBitch wrote:

In fact, if anyone could give me a quick primer on 3 dimensional vectors in general, I'll give you half of the souls I harvest this year.

You really need to work on your incentive package.

Shadowdarke wrote:

Gughunter wrote:

I think the distance formula is:

sqrt((abs(x1 - x2) ** 2) + (abs(y1 - y2) ** 2) + (abs(z1 - z2) ** 2))

Yes, that's right, but there is no need for the abs() function in there. If the differences come out negative, squaring it still returns a positive.

What other vector questions do you have, Lexy? Adding vectors is easy, you simply add the x components to get the new x, add the y components to get the new y, and add the z components to get the new z. Multiplication is where it get's tricky, but I doubt you'll need it.

Dang it, folks, stop stealing work!

I'm making a library for vectors, angular calculations, distance formulae, etc. Then again, that's only for two dimensions, so I suppose this doesn't apply. =)

Actually, it's releaseable now, if people want it. I just continually add to it... more of a side-project that I use and adapt while working on my games -- need a feature, I add one to the library in the silent hopes that some day I'll release it. =)


Also... scalar multiplication and dot products are nothing to be afraid of! Yes, I've been taking a pre-calculus course this year, and matter-of-factly I'm actually enjoying it. =)

Physics/Geo studies are really fun!

I like the second. I despise the first. =P

I agree with ya on that one...I myself don't like physics...but hey as long as you know what your doing, you won't need to worry :)

What other questions? Well, multiplication would be nice... I figure multiplying a vector by a scalar is pretty straightforward, right? But I could use a vector by a vector...

Also, if I divide a vector by its magnitude, have I reduced its magnitude to 1? I know I could do the math on that one, but I try to avoid doing math to find out something someone else might know.

And figuring the magnitude of a vector... would I just feed it through my distance proc, with 0,0,0 for the second vector?

And figuring the magnitude of a vector... would I just feed it through my distance proc, with 0,0,0 for the second vector?

Yes.

Spuzzum wrote:

The dot product (aka. the scalar product) is what you're looking at when multiplying vectors together. It's pretty hard to explain (a calculus teacher would help immensely =)).

What it results as is

vector_A dot vector_B = Length_of_vector_A * Length_of_vector_B * cos(angle_between_two_vectors)

I think this can be simplified to
A·B = Ax*Bx + Ay*By + Az*Bz
I'll have to dig out my Physics book when I get home to verify.

Note that multiplying two vectors will always yield a scalar quantity (hence the term "scalar product")... that is, you lose the direction.

This is only true of dot products. The cross product yeilds a vector that is perpendicular to the plane defined by the two vectors being multiplied. To confuse things more, cross products are non-communative: A x B = -B x A

I can't even guess the formulas for cross products, since I used them so rarely. I'll have to look them up.

And figuring the magnitude of a vector... would I just feed it through my distance proc, with 0,0,0 for the second vector?

In addition to the info Spuzzum provided, yes the magnitude of a vector can be calculated by the distance formula. One point is the tip of the vector and the other point is the tail, which is usually (0,0,0)

Out of curiousity, why in blazes are you interested in vector multiplication?

LexyBitch wrote:

What's the formula for determining distance between two points established in three-dimensional space? Euclidean distance, rather than Dantomian.

In fact, if anyone could give me a quick primer on 3 dimensional vectors in general, I'll give you half of the souls I harvest this year.

Distance between two points is the same in 3 dimensions as 2, except you add another term to the distance formula:

(This can be derived by pretending the 2-dimensional d2 and dz make up another right triangle, and so d3=sqrt(d2**2+z**2)=sqrt(dx**2+dy**2+dz**2). It works out the same. For any N-dimensional vector, just add as many terms as you have dimensions.)

The magnitude of a vector uses the same formula:

Multiplying by a scalar is as simple as you thought:

v*s=<vx*s,vy*s,vz*s>

Multiplying a vector by a vector isn't possible, because the operation just doesn't exist. However, there are dot products and cross products to consider.

The dot product is the sum of the products of the components of two vectors: I.e., x1*x2+y1*y2+.... This works the same way as the magnitude formula (without the square root), where more dimensions means more terms. Magnitude can also be expressed as mag(v)=sqrt(dot(v,v)) (or v.v as the notation normally goes).

The dot product is equal to mag(u)*mag(v)*cos(UV), where UV is the angle between them. This means that if the dot product equals 0, the two vectors are orthogonal (at a right angle); at >0, they point in the same general direction; <0, away from each other.

A cross product is somewhat different. Given two vectors, u and v, it will give you a vector that's orthogonal to both of them. mag(cross(u,v))=mag(u)*mag(v)*abs(sin(UV)); this means that if u and v are multiples of each other, the result is <0,0,0>. The cross product is usually calculated with matrices, but here's the formula all spelled out:

cross(u,v)=<uy*vz-uz*vy,uz*vx-ux*vz,ux*vy-uy*vx>

As you can see, the order of u and v does matter. cross(<1,0,0>,<0,1,0>)=<0,0,1>, but cross(<0,1,0>,<1,0,0>)=<0,0,-1>.
In two dimensions, there is no cross product; but you can find an orthogonal vector with <-y,x>. In N dimensions, you'd use N-1 vectors, using a matrix determinant to find the cross product. (I doubt you'll have any use for a 4-dimensional cross product, though.)

Other formulas you may find useful:

Distance of a point Q to a plane (normal vector N, centered at P):

(Q-P).N/mag(N) (distance may be negative, indicating Q is "behind" the plane)

Portion of vector u in direction of v (called "projection of u onto v", like u is a sundial and the projection is its shadow on v, the "ground"):

proj(u,v)=v*(u.v/v.v) (u.v and v.v are scalar)

Closest distance from point P to (infinite) line AB:

dist(P,A+proj(P,B-A))

I hope all that helps.

Lummox JR

I can totally say...I feel your pain...lol. I have to do endless algebra questions coming outta my kazoo. I only do em cause its part of my school credit :(