Square roots

Every positive real number has a positive square root. For example,

Often expressions must be evaluated before the square root is taken.

Derivation of law of Cosines

There are two cases where points O,P,Q are not collinear. In both cases drop a line from Q perpendicular to line OP and meeting it at T.

Case I: 0° < θ ≤ 90°    Fig 1
Then cos θ ≥ 0.
(1)    |OT| = |OQ| cos θ
Now in right triangles PTQ and OTQ apply the Pythagorean theorem:
   |QP|² = |TP|² + |QT|²
   |OQ|² = |OT|² + |QT|²
Subtraction of these two equations eliminates |QT|:
(2)    |QP|² − |OQ|² = |TP|² − |OT|²
To eliminate |TP| notice that
   |TP| = |OP| − |OT|
Squaring both sides produces
   |TP|² = |OP|² − 2 |OP| |OT| + |OT|²
Replacing |TP|² in equation (2) by the right side of this equation produces
   |QP|² − |OQ|² = |OP|² − 2 |OP| |OT| + |OT|² − ||OT|²
Simplifying and solving for |QP|² produces
   |QP|²= |OP|² + |OQ|² − 2 |OP| |OT|
Eliminate |OT| by replacing it in this equation by the right side of (1) above
(*)    |QP|²= |OP|² + |OQ|² − 2 |OP| |OQ| cos θ

Case I: 90° < θ < 180°    Fig 2
Then cos θ ≤ 0. Hence (− cos θ) ≥ 0
(3)    |OT| = |OQ| (−cos θ)
Now in right triangles PTQ and OTQ apply the Pythagorean theorem:
   |QP|² = |TP|² + |QT|²
   |OQ|² = |OT|² + |QT|²
Subtraction of these two equations eliminates |QT|:
(4)    |QP|² − |OQ|² = |TP|² − |OT|² To eliminate |TP| notice that
   |TP| = |OP| + |OT|.
Squaring both sides produces
   |TP|² = |OP|² + 2 |OP| |OT| + |OT|²
Replacing |TP|² in equation (4) by the right side of this equation produces
   |QP|² − |OQ|² = |OP|² + 2 |OP| |OT| + |OT|² − |OT|²
Simplifying and solving for |QP|² produces
   |QP|² = |OP|² + |OQ|² + 2 |OP| |OT|
Eliminate |OT| by replacing it by the right side of (#). The result is again equation
(*)    |QP|²= |OP|² + |OQ|² − 2 |OP| |OQ| cos θ

If O,P,Q are collinear then there are three cases to consider:
θ = 0°:   Q is between O and P,
θ = 0°:   P is between O and Q
θ = 180°:   O is between Q and P.
The proofs consist simply of adding and subtracting segments OQ, PQ to get PQ and inserting the squares their lengths into (*). The reader can supply the details.

Derivation of the geometric interpretation of the dot product

A similar proof exists for points O,P,Q is space:
   P(x₁, y₁, z₁)   and   Q(x₂, y₂, z₂)
also lead to
   |p| |q| cos θ   =   p⋅q.

Proof of the following equations

                  ***

Equation  (a)    (p + q)⋅(r + s) = (p + q)⋅r + (p + q)⋅s = p⋅r + q⋅r   +   p⋅s + q⋅s

Equation  (b)    (p + q)⋅(r − s) =

To show |p |² + |q |² − |p − q|² = 2p⋅q    for vector arrays.

Let P(x₁, y₁) and Q(x₂, y₂) be the points located by p and q respectively. Replace the vectors on the left side by their arrays and evaluate the norms:
|p |² + |q |² − |p − q|²   =   x₁² + y₁²   +   x₂² + y₂²   −   ((x₁ − x₂)² + (y₁ − y₂)²)   =  
                                      x₁² + y₁²   +   x₂² + y₂²   −   (x₁² − 2x₁x₂ + x₂²   +   y₁² − 2y₁y₂ + y₂²)   =  
                                      2x₁x₂ + 2y₁y₂   =  
                                      2p⋅q

A similar proof exists for arrays of size 3, where P(x₁, y₁, z₁) and Q(x₂, y₂, z₁) are the points located by p and q respectively.

(c) (Triangle inequality)   The norm of the sum of two vectors is never larger than the sum of the norms of those two vectors.
Notation. |p + q| ≤ |p| + |q| for any position vectors p and q
Equality occurs   if and only if   p and q one or both are zero vectors or they point in the same direction.
The adjacent figure shows the vector sum   OS = p + q   and how to get it, using the triangle method, from OP = p and PS = q. The shortest distance from O to S is along p + q. The distance   |OP| + |PS| cannot be any shorter.
But if point P is on segment OS so that O is not between P and S, then the length of OS = length of OP + length of PS, that is
   |p + q| = |p| + |q|
This happens if and only if position vectors p and q point in the same direction.

(d) The norm of a product of a scalar and a vector is equal to the product of their individual norms.
   |λp| = |λ| |p|
This equality comes directly from the physical definition of multiplying a geometric vector by a real number. The length of the resulting vector is the product of the real number, ignoring the sign, and the length of the original vector.

Proof of the Schwarz inequality for arrays of size 3

It will be shown that the difference of squares

The inequality
|(x₁, y₁, z₁)⋅(x₂, y₂, z₂)|   ≤   |(x₁, y₁, z₁)| |(x₂, y₂, z₂)|
is called in these notes the "super Schwarz inequality: |p⋅q| ≤ |p| |q|. It implies the regular Schwarz inequality.

By removing z's, setting z=0, and modifying appropriately the argument, a proof of   (x₁, y₁)⋅(x₂, y₂)   ≤   |(x₁, y₁)| |(x₂, y₂)|   for arrays of size 2 is obtained.

Proof of the triangle inequality for arrays of size 3

Triangle equality

If |u + v| = |u| + |v| then the one inequality in the chain (#) becomes an equality, namely u⋅u + 2u⋅v + v⋅v = |u|² + 2|u| |v| + |v|² which forces u⋅v = |u| |v| . This means that |u| |v| cos θ = |u| |v|. This equality happens if either or both vectors are zero, or when cos θ = 1. Then angle θ = 0, which means that the vectors point in the same direction.

Sums of squares of vector sums and differences

[2.9b] (Sums of squares of vector sums and differences) The sum of the square of the norm of the vector sum and the square of the norm of the vector difference of any two position vectors is equal to twice the sum of the squares of the norms of e|AC|h vector.
Notation:
(#)                          |p + q|² + |p − q|² = 2(|p|² + |q|²)

The proof of (#) uses the f|AC|t that the square of the norm of any vector is equal to the inner product of that vector with itself: |v|² = v⋅v. Also used is the distributive law for inner products. Then
  |p + q|² + |p − q|² = (p + q)⋅(p + q)       +      (p − q)⋅(p − q)
                             = p⋅p + 2p⋅q + q⋅q  +  p⋅p − 2p⋅q + q⋅q
                             =         |p|² + |q|²        +          |p|² + |q|²
                             =                      2(|p|² + |q|²).

Now to show that [2.9b] is the vector interpretation of [2.9a].

Construct the addition parallelogram OPRQ for finding the sum of position vectors p and q. (See adj|AC|ent figure.). The following equates non-negative scalar values:
(##)                          OP = |p|, QO = |q|, OR = |p + q|, QP = |p − q||, PR = |q| and RQ = |p|
Since OPRQ is a parallelogram with diagonals QP and OR, the verbal part of [2.9a] justifies the equation
(###)                          QP² + OR² = OP² + PR² + RQ² + QO²
Repl|AC|e non-negative lengths QP, OR, OP, PR, RQ, QO by non-negative norms |p − q|, |p + q|, |p|, |q|, |p|, |q| respectively (see (##)) and collect like terms to obtain equation (#).

Two non-zero vectors locating points with array coordinates are perpendicular if and only if the dot product of those arrays is zero

The proof is for arrays of size 2. There is a similar proof for arrays of size 3.
Let position vectors p,q locate points P(x₁, y₁) and Q(x₂, y₂). To prove that p,q are perpendicular if and only if p⋅q = x₁x₂ + x₁x₂ = 0.

The points P,O,Q form a triangle. Then p and q are perpendicular if and only if the triangle is a right triangle with the PQ as the hypotenuse. From plane geometry, a triangle is a right triangle if and only if the square of the length of the longer side = sum of the squares of the lengths of the other two sides.This means that

Then the squares of the lengths of the sides of this triangle are:

Vector product of parallel vectors

Let p be any position vector, and let q be a parallel position vector. Then by definition of being parallel, q = λp for some real number λ. Suppose p = (x,y,z).   Then q = λ(x,y,z) = (λx, λy,λz). So the three 2x2 determinants for p x q become:

Since PQ² = PQ², equating expressions in (3) and (4) produces:
(5)      |p||q| cos θ = x₁x₂ + y₁y₂

If   P(x₁,y₁,z₁),   Q(x₂,y₂,z₂)   are points in space, a similar derivation produces    |p||q| cos θ = x₁x₂ + y₁y₂ + z₁z₂.

Proofs that the inner product of arrays satisfies the axioms for inner products

(a)
p⋅q = array₁⋅array₂ = x₁y₁ + x₂y₂ = sum of real numbers.

(b)
array₁⋅array₁ = (x₁,y₁)⋅(x₁,y₁) = x₁x₁ + y₁y₁ = (x₁)² + (y₁)²
which is never negative, because it is a sum of squares of real numbers. For this discussion consider position vectors u,v,w as locating points (x₁,y₁), (x₂,y₂), (x₃,y₃).

Evaluating left side: u⋅(v + w)
v+w = (x₂ + x₃, y₂ + y₃)
u⋅(v+w) = x₁(x₂ + x₃) + y₁(y₂ + y₃)

Evaluating right side: u⋅v + u⋅w
u⋅v = x₁x₂ + y₁y₂
u⋅w = x₁x₃ + y₁y₃
u⋅v + u⋅w = x₁x₂ + y₁y₂ + x₁x₃ + y₁y₃

Simple algebra shows that
x₁(x₂ + x₃) + y₁(y₂ + y₃) = x₁x₂ + y₁y₂ + x₁x₃ + y₁y₃
Therefore, u⋅(v + w) = u⋅v + u⋅w

Similar equations are true for position vectors in sp|AC|e where u,v,w locate points (x₁,y₁,z₁), (x₂,y₂,z₂), (x₃,y₃,z₃).

Each of two non-zero position vectors is perpendicular to their vector product

A similar argument proves that q is perpendicular to p x q.

Area of a parallelogram

Right hand screw method for direction of vector product

The vector product of two position vectors is perpendicular to both vectors. There are ex|AC|tly two ways that position vector   p x q   can be perpendicular to this plane, and those ways are vectors pointing in opposite directions. The angle   θ   is measured from the first vector   p   mentioned in the vector product   p x q   to the second vector   q   in that same vector product. Always choose the angle that is between 0° and 180°.

Most screws are right-hand screws. This means that if the head of the screw is turned to the right, as indicated by the wavy red arrows in the two center figures, the screw will advance forward into the wood. Now pl|AC|e the screw at the origin O so that the f|AC|e of the head is in the plane of   p,q  . At the same time, the direction of   θ   be the same as the wavy red arrows on the head of the screw. Then   p x q   will point in the direction of the advancing screw when the head is turned (by a screwdriver). This direction is chosen, and not its opposite, so that vector products can be used in vector equations that more simply model some physical phenomena such as induction coils.

3x3 determinants and their evaluations

= x₁y₂z₃ + y₁z₂x₃ + z₁x₂y₃ − z₁y₂x₃ − x₁z₂y₃ − y₁x₂z₃

For example, to evaluate the determinant

Switch the vector product and the inner product in a triple of vectors:   p x q ⋅ r   =   p ⋅ q x r

If   p = (x₁, y₁, z₁),     q = (x₂, y₂, z₂),     r = (x₃, y₃, z₃),   then the expressions E₁ and E₂ are equivalent respectively to
     E'₁:    p x q ⋅ r
     E'₂:    p ⋅ q x r
Since expressions E₁ and E₂ are equal, then these two expressions are equal:   p x q ⋅ r   =   p ⋅ q x r

The expression E₃ is the same as the expression to evaluate a 3x3 determinant given |AB|ove. This means that if the triples of the three position vectors are known then the products   p x q ⋅ r   =   p ⋅ q x r   are equal to the (evaluation of) the determinant

A proof from plane geometry of the following theorem

Drop a perpendicular from Q to line OP meeting that line at M forming right triangles QMP and QMO. Then by the Pythagorean theorem
  |PQ|² = |QM|² + |MP|²,         |QO|² = |QM|² + |OM|²
Eliminate |QM|² from both equations to get
   |PQ|² = |QO|²    +    |MP|²    −    |OM|²
           = |QO|² + (|OP| − |OM|)² − |OM|²
           = |QO|² + |OP|² − 2 (|OP|)(|OM|) + |OM|²    −    |OM|²
SO
(*)  |PQ|² = |QO|² + |OP|² − 2 (|OP|)(|OM|)

Drop a perpendicular from S to line OP meeting that line at N forming right triangles SNO and SNP. Then by the Pythagorean theorem
  |OS|² = |SN|² + |ON|²,         |PS|² = |SN|² + |PN|²
Eliminate |SN|² from both equations to get
   |OS|² = |SP|²    +    |ON|²    +    |PN|² =
               |SP|² + (|OP| +|PN|)² − |PN|² =
              |SP|² + |OP|² + 2 (|OP|)(|PN|) + |PN|²    −    |PN|²
So
(**)  |OS|² = |OP|² + |PS|² + 2 (|OP|)(|PN|)

Therefore, adding equations (*) and (**) above:
   |PQ|² + |OS|² = |QO|² + |OP|² − 2 (|OP|)(|OM|) + |OP|² + |PS|² + 2 (|OP|)(|PN|).
But right triangles QOM and SPN are congruent and therefore |PN| = |OM|. Therefore, the terms 2 (|OP|)(|OM|) and 2 (|OP|)(|PN|) subtract to get zero. Also |OP| = |SQ| because OPSQ is a parallelogram. Hence equation (#).