Ok, I was pretty vague and useless last time, let me elaborate a bit...

So you know the points of A, B, C, therefore you can find the distance from B to C using the distance formula sqrt( (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 ) where x2, x1, y2, ect turn out to just be whatever the variables you choose for the points B and C such as (Bx, By, Bz) and (Cx, Cy, Cz).

Then you use the same method for finding the distance from A to B.

From there you have everything you need, and can simply use the Scalar Projection formula:

comp of BC onto AB = ((A vector) (dot product) (B vector)) / (magnitude of A vector)

stafu

Australia1196 Posts

April 20 2010 20:58 GMT

So how would I solve for H? By the way, thanks for the replies/help

Antimage

Canada1293 Posts

April 20 2010 20:58 GMT

AB is perpendicular to CH. Use either dot product or cross product to solve for the perpendicular vector identity.

Excuse me for being vague, I'm an accounting student and haven't done real math for 3+ years =/

stafu

Australia1196 Posts

April 20 2010 20:59 GMT

Thanks DeathByMonkeys, now to see if I can get my head around this hehe. Life saver.

MER

Bulgaria125 Posts

April 20 2010 21:00 GMT

The point H is the projection of C on the plane. But your plane is not defined because you have only 2 points on it. so you can't solve the problem. There are infinite planes that go through the points A and B.

Tex

United States126 Posts

April 20 2010 21:01 GMT

I see right triangles and am wondering if that would simplify anything at all. Or not since its in 3d.

EtherealDeath

United States8366 Posts

April 20 2010 21:02 GMT

#10

I was about to say what the above said, but then, since 3 points are needed to determine the plane, the plane is in effect undetermined, therefore H may be wherever we wish it to be. Suppose H = C. Then this is distance 0. Of course this is not allowed by the problem. So then we choose H to be arbitrarily close to C, so that given any H = (x,y,z), we can find a closer H to C.

A most trivial case is where A = (0,0,0), B = (1,0,0), C = (0.5,1,1). There can be no explicit solution for H in terms of A,B,C since in this case there is no H closest to C, because we can arbitrarily change the angle between the xy axis and the plane to have the plane approach C.

Could be full of shit though, at first I thought it was a trivial projection problem :/ back to foods.

Chill

Calgary25969 Posts

April 20 2010 21:08 GMT

#11

Just using common sense I don't see any reason C can't be 0.00000000000000000000000001 units away from H. Seems like a retarded question.

DeathByMonkeys

United States742 Posts

April 20 2010 21:13 GMT

#12

On April 21 2010 05:59 stafu wrote:
Thanks DeathByMonkeys, now to see if I can get my head around this hehe. Life saver.

Actually I messed up. You don't need to know the distance between AB and BC, just the vectors.

So disregard the distance formula part.

AB will simply be = (Bx - Ax, By - Ay, Bz - Az)
BC will simply be = (Cx - Bx, Cy - By, Cz - Bz)

Now you can use the projection formula I provided earlier where you project BC onto AB.

So...

comp of BC onto AB = ((AB vector) (dot product) (BC vector)) / (magnitude of AB vector)

This resultant scalar is the distance from B to H.

Oracle

Canada411 Posts

April 20 2010 21:16 GMT

#13

On April 21 2010 06:08 Chill wrote:
Just using common sense I don't see any reason C can't be 0.00000000000000000000000001 units away from H. Seems like a retarded question.

The "Shortest distance to point C" statements implies that H and C are perpendicular. C can be epsilon units away from H but that doesn't mean anything.

]343[

United States10328 Posts

April 20 2010 21:17 GMT

#14

On April 21 2010 06:08 Chill wrote:
Just using common sense I don't see any reason C can't be 0.00000000000000000000000001 units away from H. Seems like a retarded question.

they're asking what the projection of C onto AB is, for general A, B, C

which is ... dot product

specifically, the length of (C-A) projected onto (B-A) is [(C-A) . (B-A)]

so we multiply this by unit vector in direction of (B-A) and then add A

EtherealDeath

United States8366 Posts

April 20 2010 21:20 GMT

#15

On April 21 2010 06:17 ]343[ wrote:

Show nested quote +

That's not what they ask for. If you pick H along the projection of C onto AB, then A,B,C,H will be on the same plane, which they mentioned C is NOT on the same plane as A,B,H.

Severedevil

United States4832 Posts

April 20 2010 21:29 GMT

#16

The unknown point H is located at the shortest distance from Point C.

The shortest distance from point C to what?

If it's the shortest distance from point C to line AB, this is trivial as a lot of people have pointed out. Project C onto AB and you have H. (As has also been pointed out, all four points would be coplanar.)

If it's the shortest distance from point C to the plane containing AB, well... that's a problem, because the plane isn't specified in the problem.

QuickStriker

United States3694 Posts

April 20 2010 21:31 GMT

#17

I can't help ya since I'm not experienced in this but I have to say, I am very impressed of the manpower and help TL always deliver. This is why I love TL, this is what TL truly is: giving and sharing. Another reason why I feel TL is the best community of the entire world!

Chill

Calgary25969 Posts

April 20 2010 21:33 GMT

#18

On April 21 2010 06:17 ]343[ wrote:

Show nested quote +

I guess I don't understand but I'd like to have it explained for my own sake.

ABH is a triangle (or line) in a plane. C is not in that plane but is the closest distance to H possible. To me, the closest distance possible is like right beside it but not in the other plane so it would be 0.00000000000000000_1 units perpendicular from the plane ABC lies in. I feel like there's a line missing or I'm misunderstanding the question entirely.