Unizor – Creative Mind through Art of Mathematics: Vectors+ 05 More about N-vectors: UNIZOR.COM – Math+ & Problems

Notes to a video lecture on http://www.unizor.com

Vectors+ 05
More N-dimensional Vectors

Theory

Let’s summarize the characteristics of N-dimensional vectors.
First of all an N-dimensional vector is an ordered set of N real numbers (R₁,R₁,…,R_N) that we will use as a single object R.
Real numbers R_n where n∈[1,N] will be referred as components of R.
Unless specifically noted, we will assume that the word vector in this lecture means N-dimensional vector.

1. Addition
Any two vectors can be added together giving another vector according to a simple rule
R (R₁,R₂,…,R_N) + S (S₁,S₂,…,S_N) =
= T (R₁+S₁,R₂+S₂…,R_N+S_N,)
Addition of vectors is, obviously, commutative. It follows from the commutative property of the addition of real numbers.
R + S = S + R because
R_n + S_n = S_n + R_n
for n∈[1,N]
Also, addition of vectors is associative. It follows from the associative property of the addition of real numbers.
(R + S) + T = R + (S + T) because
(R_n + S_n) + T_n = R_n + (S_n + T_n)
for n∈[1,N]

2. Null-vector
Vector 0 (0,0,…,0) plays the same role for vector addition as real number 0 with other real numbers, namely, adding null-vector to any vector does not change that vector
R + 0 = 0 + R = R
which directly follows from the rules of addition of vectors and properties of real number 0.
For any non-null vector
R (R₁,R₂,…,R_N)
there is an opposite vector with the signs of all components reversed
−R (−R₁,−R₂,…,−R_N)
that, if added to R, results in a null-vector sum
R + (−R) = 0

3. Multiplication by a scalar
Any vector can be multiplied by a scalar, which is defined as all components of this vector to be multiplied by this scalar.
R (R₁,R₂,…,R_N)·f = T (R₁·f,R₂·f…,R_N·f)
Analogously, multiplication of a scalar by a vector is
f·R (R₁,R₂,…,R_N) = T (f·R₁,f·R₂…,f·R_N)
which, obviously, is the same as multiplication of a vector by a constant because of commutative properties of multiplication of real numbers.

4. Scalar (dot) product of two vectors
Any two vectors can form a scalar product giving a scalar according to a simple rule
R (R₁,R₂,…,R_N) · S (S₁,S₂,…,S_N) =
= R₁·S₁+R₂·S₂+…+R_N·S_N
Obviously, it’s commutative because of commutative properties of multiplication of real numbers.

5. Magnitude of a vector
For any vector
R (R₁,R₂,…,R_N)
we define its magnitude (length) ||R|| in a way to be similar to 2- and 3- dimensional cases:
||R|| = √(R·R) = √R₁²+R₂²+…+R_N².

6. Orthogonality
We will call two vectors orthogonal to each other, if their scalar product is zero.

7. Basis
The following set of N vectors
i⁽¹⁾ (1,0,…,0)
i⁽²⁾ (0,1,…,0)
…
i^(N) (0,0,…,1)
consists of mutually orthogonal vectors of unit length.
This directly follows from the definitions of scalar product and magnitude of vectors.
Indeed,
i^(m)·i⁽ⁿ⁾ = 0 for m,n∈[1,N] and m≠n
√(i^(m)·i^(m)) = 1 for each m∈[1,N]
They are called unit vectors.
These unit vectors form orthogonal basis because any vector
R (R₁,R₂,…,R_N)
can be represented as the linear combination of these unit vectors
R = R₁·i⁽¹⁾ + R₂·i⁽²⁾ +…+ R_N·i^(N)

8. Angle
For two-dimensional vectors
U ·V = u·v·cos(φ)
where u is the magnitude of vector U,
v is the magnitude of vector V and
φ is the angle between them in any direction since cos(φ)=cos(−φ).
Accordingly, as explained in the previous lecture, we define an angle between two N-dimensional vectors as
cos(φ) = (U·V)/(||U||·||V||)

Problem A

Prove that multiplication of a vector by a scalar is distributive.
That is,
(a) f·(R + S) = f·R + f·S
(b) (f + g)·R = f·R + g·R

Problem B

Prove that scalar product of two vectors is distributive.
That is,
R·(S + T) = R·S + R·T

Problem C

Two three-dimensional vectors are given
OA (2,3,4)
OB (−2,1,3)
Imagine a triangle in three-dimensional space ΔOAB formed by these two vectors and their difference:
OB−OA = AB
Determine the lengths of three sides of this triangle, cosines of its angles and check the Theorem of Cosine in it.

Solution

According to definition of a difference of two vectors, vector AB is (−2−2,1−3,3−4)=(−4,−2,−1).

Now we can calculate all magnitudes.
OA = √2²+3²+4² = √29
OB = √(−2)²+1²+3² = √14
AB = √(−4)²+(−2)²+(−1)² = √21

Scalar products are
OA·OB =
= 2·(−2)+3·1+4·3 = 11
OA·AB =
= 2·(−4)+3·(−2)+4·(−1) = −18
OB·AB =
= (−2)·(−4)+1·(−2)+3·(−1) = 3

Before calculating the cosines of angle of ΔOAB let’s draw this triangle in the plane it belongs to.
We have calculated its sides, so it should look like this

Angle ∠AOB is an angle between vectors OA and OB.
cos(∠AOB) = (OA·OB)/(OA·OB) =
= 11/√(29·14)

Angle ∠OAB is NOT an angle between vectors OA and AB. It is, actually, an angle between vectors AO (opposite to OA, the vector of the same length but opposite direction) and AB.
The vector AO, as opposite to OA(2,3,4), is a set of real values (−2,−3,−4) and, therefore, its scalar product with AB is
AO·AB =
= (−2)·(−4)+(−3)·(−2)+(−)4·(−1) =
= 18
Therefore,
cos(∠OAB) = (AO·AB)/(AO·OB) =
= 18/√(29·21) = 18/√(29·21)

Angle ∠OBA is vertical (and, therefore, equal) to an angle between vectors OB and AB.
cos(∠OBA) = (OB·AB)/(OB·AB) =
= 3/√(14·21)

Let’s check the Theorem of Cosines.

AB²=OA²+OB²−OA·AB·cos(∠AOB) =
= 29 + 14 − 2·√(29·14)·11/√(29·14) =
= 29 + 14 − 22 = 21 ∴ CORRECT

OA²=OB²+AB²−OB·AB·cos(∠OBA) =
= 14 + 21 − 2·√(14·21)·3/√(14·21) =
= 14 + 21 − 6 = 29 ∴ CORRECT

OB²=OA²+AB²−OA·AB·cos(∠OAB) =
= 29 + 21 − 2·√(29·21)·18/√(29·21) =
= 29 + 21 − 36 = 14 ∴ CORRECT