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.
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
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)
You Must do the cross first to make any sense.
(AxB)xC or Ax(BxC) [These give different answers, in general].
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.
Motivation
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
Uses of Scalar Triple Products
PAxPB = i + j
Answer = .5 * i (dot) (i + j) = .5
Vector Triple Product