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Advanced vector geometry
Definition For any pair u = (u1, u2, u3), v = (v1, v2, v3) ∈ R3
define the cross product of u, v as follows:
u × v = (u2 v3 − u3 v2, u3 v1 − u1 v3, u1 v2 − u2 v1).
Example
Let u = (2, −1, 2) and v = (4, −1, −4). Calculate u × v. Solution:
From the formula:
u×v = ((−1)(−4)−2(−1), 2(4)−2(−4), 2(−1)−(−1)(4)) = (6, 16, 2).
For vectors u, v, w ∈ R3 and scalar α ∈ R,
1. u × v = −v × u
2. (hence) u × u = 0
3. u × (v + w) = u × v + u × w
4. (u + v) × w = u × w + v × w
5. α(u × v) = (αu) × v = u × (αv)
Scalar triple product
Definition For vectors u, v, w ∈ R3, the scalar triple product is
defined to be u · (v × w).
Theorem Let u, v, w be vectors in R3. Then
u · (v × w) = (u × v) · w.
Theorem Let u, v ∈ R3. Then u × v is orthogonal to both u and v.
Theorem Let u, v ∈ R3 be non-zero vectors. Then
(1) ku × vk = kuk kvk | sin θ| where θ is the angle between u and v;
(2) ku × vk is the area of the parallelogram defined by u and v;
(3) u × v = 0 if and only if u and v are parallel.
Example
Find the area of the parallelogram whose sides are the vectors (1, 3, 0)
and (2, 1, 2). Answer: the area is
√
p
k(1, 3, 0) × (2, 1, 2)k = k(6, −2, −5)k = 36 + 4 + 25 = 65.
Note also that if the angle between them is α, then
s
√
65
1 65
sin α = √ √ =
≈ 0.85.
3 10
10 9
Lines and planes
Let the parametric form of a line be
r = r0 + t1 d1 + t2 d2.
Let n = d1 × d2. Then n is a normal vector to the plane.
Example
Find a Cartesian equation for the plane containing the points (1, 0, 0),
(0, 1, 0) and (0, 0, 1). First of all, we follow the method of Example 2
in Chapter 3 to write:
r = (1, 0, 0) + t1 (−1, 1, 0) + t2 (−1, 0, 1)
for t1, t2 ∈ R. We thus have r0 = (1, 0, 0), d1 = (−1, 1, 0) and
d2 = (−1, 0, 1). Now,
n = d1 × d2 = (−1, 1, 0) × (−1, 0, 1) = (1, 1, 1)
and the Cartesian equation is
0 = n · (r − r0) = (1, 1, 1) · ((x, y, z) − (1, 0, 0)) = x + y + z − 1.
That is,
x + y + z = 1.
The intersection of two planes
Let P1 and P2 be a pair of planes:
P1 : n1 · r = k1
P2 : n2 · r = k2
for normal vectors n1 and n2 and constants k1, k2 ∈ R.
If n1 and n2 are non-parallel, P1 and P2 intersect in a line. Otherwise,
P1 and P2 are parallel (and possibly equivalent).
Example
Do the lines (1, 1, 3) + t (2, 1, 1) and (5, 6, 4) + s (1, 2, 1) lie in a
single plane?
Suppose they do. Then the plane would contain (1, 1, 3) and be
parallel to both (2, 1, 1) and (1, 2, 1), so an acceptable parametric
form would be
(x, y, z) = (1, 1, 3) + t1 (2, 1, 1) + t2 (1, 2, 1).
To check that both lines are contained in this plane, we need to check
whether the point (5, 6, 4) is in the plane. This leads to a system of
equations
(for x)
(for y)
(for z)
5 = 1 + 2 t1 + t2
6 = 1 + t1 + 2 t2
4 = 3 + t1 + t2
Written out in matrix form, as a system of equations for t1 and t2:

2 1 4



 1 2 5 
1 1 1
which has reduced row echelon form (R1 → R1 −R3, R2 → R2 −R3,
R3 → R3 − R1, R3 → R3 − R2)

1 0

3

4 .
0 0 −6

 0 1
This shows that the system is inconsistent, so there is no solution for
t1 and t2. Thus, the two lines do not lie in a common plane.
Example
Do the points (1, 3, 2), (−2, 3, 5), (−1, 0, 1) and (8, 3, −5) lie on
a a common plane (i.e. are they coplanar)?
Find the plane containing the first three points, and then test whether
the fourth point is on it. We’ll take r0 = (1, 3, 2) and let the two direction vectors be (−2, 3, 5) − (1, 3, 2) = (−3, 0, 3) and (−1, 0, 1) −
(1, 3, 2) = (−2, −3, −1). A suitable normal vector is thus
n = (−3, 0, 3) × (−2, −3, −1) = (9, −9, 9).
The plane containing the first three points thus has equation:
(9, −9, 9) · (x, y, z) = n · r = n · r0 = (9, −9, 9) · (1, 3, 2) = 0.
Dividing by 9, this becomes x − y + z = 0. It is easy to see that first
three points satisfy this equation (as we would expect). But also:
8 − 3 + (−5) = 0,
so the fourth point is on the plane. That is, the four points are coplanar.
Example
Find the point of intersection of the line (x, y, z) = (1, 2, 1)+t (2, −3, 1)
and the plane 2 x + y − 3 z = 5.
First, substitute
x = 1 + 2 t, y = 2 − 3 t, z = 1 + t into 2 x + y − 3 z = 5.
Thus,
5 = 2 (1 + 2 t) + (2 − 3 t) − 3 (1 + t) = 1 − 2 t.
This is solved by t = −2, so the point of intersection is (x, y, z) =
(1, 2, 1) + (−2) (2, −3, 1) = (−3, 8, −1).
Check our answer by substitution! 2 x + y − 3 z = 2 (−3) + 8 −
3 (−1) = 5.