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Math 311: Topics in Applied Math 1
5: Orthogonality
5.1: The Scalar Product in Rn
Summary
Let x, y ∈ Rn .
Solution
Since x and y be linearly independent, neither vector is a
multiple of the other. Hence the angle θ between them is
not a multiple of π. Hence |cos θ | < 1. Therefore,
T x y = kxk kyk cos θ = kxk kyk |cos θ | < kxk kyk < (2) (3)
or xT y < 6.
n
• The scalar product of x and y is xT y =
∑ xk yk .
k=1
√
• The Euclidean length of x is kxk = xT x.
• The distance between x and y is ky − xk.
• The angle θ between two vectors x and y satisfies
xT y = kxk kyk cos θ .
uT v.
• For unit vectors u and v, we have cos θ =
• Cauchy-Schwarz Inequality: xT y ≤ kxk kyk.
• Vectors x and y are orthogonal if xT y = 0. In this
case, we write x ⊥ y.
• For nonzero vectors x & y, the scalar projection α,
vector projection p, and orthogonal projection q
of x onto y are respectively given by
xT y
kyk
T 1
x y
y
p = projy x = α
y =
kyk
yT y
q = orthy x = x − p.
α = compy x =
Let P1 and P2 be points in 3-dimensional real space. We
identify them with their position vectors P1 and P2 in R3 .
−−→
P1 P2 denotes the vector P2 − P1 ∈ R3 .
• Given a nonzero vector n ∈ R3 and a fixed point P0
in 3-dimensional space, the set of points P in space
−→
such that n ⊥ P0 P is a plane. Here
nT (P − P0 ) = 0
or
nT P = nT P0
We say the vector n and the plane are normal to
each other; n is called a normal vector.
224/9
Find an equation of the plane that passes through
the points P1 (2, 3, 1), P2 (5, 4, 3), and P3 (3, 4, 4).
Solution


 
3
1
−−→
−−→
• Let v = P1 P2 =  1  and w = P1 P3 =  1 .
2
3
Back in Calc 3 we’d compute our normal vector
as the cross product n = v × w. For grins, let’s
alternatively directly obtain a vector that is
orthogonal to both v and w.
T
• Let n = a b c . Then nT v = 0 and nT w = 0.
This gives a linear system in the variables a, b, c,
whose augmented system matrix and reduced row
echelon form are
3 1 2 0
1 0 −1/2 0
→
.
1 1 3 0
0 1 7/2 0
T
• Thus a = 12 c and b = − 72 c. So n = 1 −7 2
is a normal vector. Hence an equation of the plane is
 
x
T
y  = nT P1
n
z
x − 7y + 2z = −17.
225/12+
Prove that if x, y, z ∈ Rn , then
(a) xT x ≥ 0 (b) xT y = yT x (c) xT (y + z) = xT y + xT z
Solution
Examples
n
k=1
Also watch the calculator video for more examples.
Let x and y be linearly independent vectors in R2 . Given
kxk = 2 and kyk = 3, what
can we conclude about the
possible values of xT y?
xT y
is a scalar (1 × 1 matrix), we have
T
xT y = a = aT = xT y = yT x
(b) Since a =
224/4
∑ xk2 ≥ 0.
(a) We have xT x =
whence xT y = yT x.
(c) We have xT (y + z) = xT y + xT z since left matrix
multiplication distributes over matrix addition.