vector math question. Help pls!!

[hata]

My question is I have two objects. ObjA at posA and ObjB at posB. ObjA has a normal(facing) and want to know how much deg to turn in order to face ObjB.


vectorBtoA = posA - posB
arccos(dotproduct(normalize(ObjA's facing)), normalize(vectorBtoA)))


what I get is a degree but don't have a sign to indicate which way to turn. Any suggestion or better approach?
Thank you very much.

[Abuse]

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Vector ab = objectB.getPosition().minus(objectA.getPosition()); ab.normalize(); Vector normalA = objectA.getDirection(); normalA.normalize(); double theta = Math.arccos(normalA.dot(ab)); // I've got no idea what this operation is called; // it's kinda like a cross product... but gives a scalar, not a vector double cross = ab.x * normalA.y - normalA.x * ab.y; if(cross<0) theta = -theta;

[crystalsquid]

// I've got no idea what this operation is called;
// it's kinda like a cross product... but gives a scalar, not a vector

double cross = ab.x * normalA.y - normalA.x * ab.y;

if(cross<0) theta = -theta;  

Abuse: It's still the cross product
You are generating just the z component of the cross product from the x and y. You can tell whether the y is to the right/left of the x by the sign.

Hata: Do you really need the angle & the sign?
If you just wish to orient the object A to face object B, then the best way is to build the orientation matrix directly, otherwise you will be doing a lot of normalising & trig to get the same result.

For this code, I will assume X is the axis for the 'facing' vector, and Y is up:

(Pseudocode)

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// Build the true x axis: vectorX = posA-posB; vectorX.normalize(); // Now build a true z axis: vectorY.set(0.0f, 1.0f, 0.0f); vectorZ.cross( vectorX, vectorY ); vectorZ.normalise(); // Now build the true Y axis: vectorY.cross( vectorZ, vectorX);

The orientation matrix is actually these three vectors packed into the matrix, either as rows (vectorX, VectorY, VectorZ) or as columns depending on your coordinate scheme.

- Dom

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[hata]

i think Abuse's answer is quite smart and it certainly solves this 2d problem  

thank you Crystalsquid also, i think what u r talking about is orthogonal martix, rite? the reason i asked this question is becos i do want to find out the degree and the sign.

Thank you all.

[crystalsquid]

Ah, ok.

You can also do this in one step, as the size of the cross product result is the sin of the angle:

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Vector ab = objectB.getPosition().minus(objectA.getPosition());   ab.normalize();   Vector normalA = objectA.getDirection();   normalA.normalize();     double cross = ab.x * normalA.y - normalA.x * ab.y; double theta = Math.arcsin(cross);

- Dom

[hata]

that is even better. thanks