# Applications and Triple Products

Informational note: Vectors can be written in matrix form as well as in the component sum and ordered pair/triple form we've been using.

Now that we've covered the Dot and Cross Products, we can now go over a few applications, some of which will involve the use of triple products.

• Find the Equation of a plane containing the following three points.
P(1,0,0)
Q(0,1,0)
R(0,1,1)

PQ=-i + j
PR=-i + j + k
n = the normal to the plane
= a vector perpendicular to PQ and PR = PQxPR = i + j
Ax + By + Cz = Ax0 + By0 + Cz0
1(x) + 1(y) + 0(z) = 1(1) + 1(0) + 0(0)
x + y = 1

• Two intersecting Planes-- Find the angle between them.

The angle between the planes will be the same as the angle between their normals, which we can find using the dot product formula. The angle A between the planes would be:
cos A = (N1 (dot) N2) / (|N1|*|N2|)

One can also find the vector parallel to the line created by the intersection of the two planes. That vector is a vector perpendicular to both normals, namely (N1xN2)

Example:
3x - 6y - 2z = 15
2x + y - 2z = 5

Find the Parametric equation of the line described by the intersection of these planes.

v = N1xN2 = (3i - 6j - 2k)x(2i + j - 2k)
v=14i + 2j + 15k

Now we need a base point, so we find a solution of the system above.
Take z=0 and the solution is x=3 and y=1. Base Point(3,1,0).
x = 3 +14t
y = -1 + 2t
z = 0 + 15t

• ### Triple Products

1. Scalar Triple Product

You Must do the cross first to make any sense.

2. Vector Triple Product

(AxB)xC or Ax(BxC) [These give different answers, in general].

• #### Uses of Scalar Triple Products

1. The Parallelpipped

2. Projecting Area

Find the Area of the projection of triangle PAB onto the yz plane. To do this we use the formula for the area of a projected parallelogram and half the answer to get triangular area instead.

PAxPB = i + j
Answer = .5 * i (dot) (i + j) = .5
• #### Vector Triple Product

Motivation

Notes by Lawrence C. Weintraub 2-5-95