Vectors from 444

We assume vector addition and multiplication of a vector by a scalar (i.e., a real number) are well-understood.

Notation: Let I, J, K be (1, 0, 0), (0, 1, 0), (0, 0, 1).

Center of Mass for equal masses

The center of mass of a set of points (with equal masses at each point) is the mean, or average. Thus for two points, AB, the midpoint is (1/2)(A+B), for a triangle ABC, the center of mass (the centroid) is (1/3)(A + B + C). For a set of four points ABCD, the center of mass is (1/4)(A+B+C+D).

Applications and exercises

• Find the midpoint of (1, 1, 1) and (1, 0, 2).

For an equilateral triangle, all the usual centers (centroid, circumcenter, incenter, orthocenter) are all the same point.

• Compute the circumcenter of triangle IJK
• Compute the centroid of triangle (1, 1, 1), (1, -1, -1), (-1, 1, 1).
• Compute the center of mass of the 4 points, (1, 1, 1), (1, 1, -1), (1, -1, 1), (-1, 1, 1).

Unequal Masses, Barycenters and parametric equation of a line and a plane

Given two points A and B and two masses a, b at those points, the center of mass is the weighted average:

M = (a/(a+b))A + (b/(a+b))B.

If a+b = 1, then M = aA + bB. In the case a+b = 1, ( if we set b = t, then a = 1 - t, so the formula for the center of mass is simpler:

P(t) = (1-t)A + t.B.

The formula P(t) = (1-t)A + t.B can be written in coordinates as, e.g., (x(t), y(t), z(t)). This is the (affine) parametrization of the line AB. The parameter t provides a coordinate system on the line AB.

When both masses are nonnegative, then P is on the segment AB. If one of the masses is negative, then P is on line AB but not on the segment.

For 3 points A, B, C, with masses a, b, c, the center of mass, or barycenter is

M = (1/(a+b+c))(aA+bB+cC).

Again this is simpler if a+b+c = 1, so the point M = aA+bB+cC. Finally, if b = s and c = t, then if a = 1 - s - t the sum = 1.

P(s,t) = (1 - s - t)A + sB + tC

The point P(s,t) = (1 - s - t)A + sB + tC ranges over the whole plane ABC as s and t take on all real values, with P inside triangle ABC if all 3 masses are nonnegative. P(s,t) can be written in coordinates as, e.g., (x(s,t), y(s,t), (s,t)).This is the (affine) parametrization of the plane ABC.

Applications and Exercises

• If A = (1, 2, 3) and B = (2, -1, 6), write the affine parametrization of line AB.
• Find the point on segment AB for which AB/BC = 1/2.
• Use the parametrization of line AB to find the point where line AB intersects the (x,y,0) plane.
  Hint: Write the parametric form (x(t), y(t), z(t)) explicitly, set z = 0 and solve for t. But don't stop there, since t is not the point. To find the point, substitute the value of t in the parametrization to find (x, y, z).
• Given A = (1, 2, 3), B = (2, -1, 6), and C = (1, 0, 0), write the affine parametrization of plane ABC.
• Use this parametrization to find the equation of the line that is the intersection of plane ABC and the (x, 0, 0) axis.
  Hint: Parametrize the plane, set y = 0 and z = 0 and solve for s and t. Then substitute to find the point.

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