AOS5 Topic 5: Vector Proofs

Vector proofs refer to the mathematical demonstrations or arguments that establish the properties, relationships, or theorems related to vectors in mathematics. These proofs typically involve logical reasoning and algebraic manipulations to demonstrate the validity of certain statements or propositions about vectors.

Geometric proofs

In this section we use vectors to prove geometric results in two and three dimensions. The following properties of vectors will be useful:

Parallel vectors

For \( k \in \mathbb{R}^+ \), the vector \( ka \) is in the same direction as \( a \) and has magnitude \( k|a| \), and the vector \( -ka \) is in the opposite direction to \( a \) and has magnitude \( k|a| \).

Two non-zero vectors \( a \) and \( b \) are parallel if and only if \( b = ka \) for some \( k \in \mathbb{R} \setminus \{0\} \).

Three distinct points \( A \), \( B \), and \( C \) are collinear if and only if \( \overrightarrow{AC} = k \overrightarrow{AB} \) for some \( k \in \mathbb{R} \setminus \{0\} \).

Scalar product

Two non-zero vectors \( a \) and \( b \) are perpendicular if and only if \( a \cdot b = 0 \).

\( a \cdot a = |a|^2 \)

Linear combinations of independent vectors

Let \( a \) and \( b \) be two linearly independent (i.e. not parallel) vectors. Then \( ma + nb = pa + qb \) implies \( m = p \) and \( n = q \).

Vector proofs in two-dimensional geometry

Diagonals of a rhombus are perpendicular

Angle subtended by a diameter at a point on a circle is a right angle

Median of a Triangle

A median of a triangle is a line segment from a vertex to the midpoint of the opposite side.

Concurrent Lines

Three or more lines are said to be concurrent if they all pass through a single point.

Centroid of a Triangle

The point where the three medians meet is called the centroid of the triangle.

Vector proofs in three-dimensional geometry

Vector Representation

In three-dimensional geometry, vectors are often represented as directed line segments with magnitude and direction, typically expressed as ordered triples (e.g., \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \)).

Geometric Interpretation

Vectors in three dimensions can represent translations, displacements, or directions in space. They can also represent lines, planes, or spatial shapes.

Example 1

Proofs in 2D geometry
Prove that the diagonals of a rhombus are perpendicular.

Proof:
Let \( \vec{a} = \overrightarrow{OA} \) and \( \vec{c} = \overrightarrow{OC} \).
The diagonals of the rhombus are \( \overrightarrow{OB} \) and \( \overrightarrow{AC} \).
\( \overrightarrow{OB} = \overrightarrow{OC} + \overrightarrow{CB} = \overrightarrow{OC} + \overrightarrow{OA} = \vec{c} + \vec{a} \)
\( \overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = -\vec{a} + \vec{c} \)

Consider the scalar product of \( \overrightarrow{OB} \) and \( \overrightarrow{AC} \):
\( \overrightarrow{OB} \cdot \overrightarrow{AC} = (\vec{c} + \vec{a}) \cdot (-\vec{a} + \vec{c}) \)
\( = \vec{c} \cdot \vec{c} - \vec{a} \cdot \vec{a} \)
\( = |\vec{c}|^2 - |\vec{a}|^2 \)
A rhombus has all sides of equal length, and therefore \( |\vec{c}| = |\vec{a}| \). Hence
\( \overrightarrow{OB} \cdot \overrightarrow{AC} = |\vec{c}|^2 - |\vec{a}|^2 = 0 \)
This implies that \( \overrightarrow{AC} \) is perpendicular to \( \overrightarrow{OB} \).

Example 2

Prove that the angle subtended by a diameter at a point on a circle is a right angle.

Proof
Let \( O \) be the centre of the circle and let \( AB \) be a diameter.
Let \( C \) be a point on the circle (other than \( A \) or \( B \)).
We aim to show that \( \angle ACB \) is a right angle.
We have \( |\overrightarrow{OA}| = |\overrightarrow{OB}| = |\overrightarrow{OC}| = r \), where \( r \) is the radius.
Let \( \vec{a} = \overrightarrow{OA} \) and \( \vec{c} = \overrightarrow{OC} \). Then \( \overrightarrow{OB} = -\vec{a} \).
\( \overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = -\vec{a} + \vec{c} \)
\( \overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \vec{a} + \vec{c} \)

Thus \( BC = (\vec{-a} + \vec{c}) \cdot (\vec{a} + \vec{c}) \)
\( = -\vec{a} \cdot \vec{a} + \vec{c} \cdot \vec{c} \)
\( = -|\vec{a}|^2 + |\vec{c}|^2 \)
\( = -r^2 + r^2 \) (since \( |\vec{a}| = |\vec{c}| = r \))
\( = 0 \)
Hence \( \overrightarrow{AC} \perp \overrightarrow{BC} \). We have shown that \( \angle ACB \) is a right angle.

Example 3

Prove that the medians of a triangle are concurrent.

Proof:
Consider triangle OAB. Let \( A' \), \( B' \), and \( X \) be the midpoints of \( OB \), \( OA \), and \( AB \) respectively.
Let \( Y \) be the point of intersection of the medians \( AA' \) and \( BB' \).
Let \( \vec{a} = \overrightarrow{OA} \) and \( \vec{b} = \overrightarrow{OB} \).
p>We start by showing that \( \frac{{AY}}{{YA'}} = \frac{{BY}}{{YB'}} = 2 : 1 \).

We have \( \overrightarrow{AY} = \lambda \overrightarrow{AA'} \) and \( \overrightarrow{BY} = \mu \overrightarrow{BB'} \), for some \( \lambda, \mu \in \mathbb{R} \).
\( \overrightarrow{AA'} = \overrightarrow{AO} + \frac{1}{2} \overrightarrow{OB} = -\vec{a} + \frac{1}{2} \vec{b} \)
Therefore, \( \overrightarrow{AY} = \lambda \left( -\vec{a} + \frac{1}{2} \vec{b} \right) \)
and \( \overrightarrow{BB'} = \overrightarrow{BO} + \frac{1}{2} \overrightarrow{OA} = -\vec{b} + \frac{1}{2} \vec{a} \)
Therefore, \( \overrightarrow{BY} = \mu \left( -\vec{b} + \frac{1}{2} \vec{a} \right) \)

Now, \( \overrightarrow{BY} \) can also be obtained as follows:
\( \overrightarrow{BY} = \overrightarrow{BA} + \overrightarrow{AY} = \overrightarrow{BO} + \overrightarrow{OA} + \overrightarrow{AY} = -\vec{b} + \vec{a} + \lambda \left( -\vec{a} + \frac{1}{2} \vec{b} \right) \)
\( = (1 - \lambda) \vec{a} + \left( \frac{\lambda}{2} - 1 \right) \vec{b} \)
Since \( \vec{a} \) and \( \vec{b} \) are independent vectors, we now have \( \frac{\mu}{2} = 1 - \lambda \) (1) and \( -\mu = \frac{\lambda}{2} - 1 \) (2)
Multiply (1) by 2 and add to (2):
0 = 2 - 2λ + \( \frac{\lambda}{2} - 1 \)
1 = \( \frac{3\lambda}{2} \)
\( \lambda = \frac{2}{3} \)
Therefore, Substitute in (1) to find \( \mu = \frac{2}{3} \). We have shown that \( \frac{{AY}}{{YA'}} = \frac{{BY}}{{YB'}} = 2 : 1 \).
Now, by symmetry, the point of intersection of the medians \( AA' \) and \( OX \) must also divide \( AA_0 \) in the ratio 2 : 1, and therefore must be \( Y \). Hence the three medians meet at \( Y \).

Example 4

Consider a parallelepiped OABCDEFG as shown.
a. Prove that the diagonals OF and CE bisect each other.
b. Let M be the midpoint of CB, and let N be the midpoint of DE. Prove that the midpoint of MN is the point where the diagonals OF and CE intersect.

Solution
a. We will show that the midpoint of OF and the midpoint of CE coincide.
Let \( \vec{a} = \overrightarrow{OA} \), \( \vec{c} = \overrightarrow{OC} \), and \( \vec{d} = \overrightarrow{OD} \).
We will denote the midpoint of OF as X.
Then \( \overrightarrow{OX} = \frac{1}{2} \overrightarrow{OF} = \frac{1}{2} (\overrightarrow{OA} + \overrightarrow{AB} + \overrightarrow{BF}) = \frac{1}{2} (\vec{a} + \vec{c} + \vec{d}) \).
Let Y be the midpoint of CE. Then \( \overrightarrow{OY} = \frac{1}{2} \overrightarrow{OC} + \overrightarrow{OE} = \frac{1}{2} \overrightarrow{OC} + \overrightarrow{OA} + \overrightarrow{AE} = \frac{1}{2} (\vec{c} + \vec{a} + \vec{d}) = \overrightarrow{OX} \).
Therefore, X = Y, and so the diagonals OF and CE bisect each other.

b. We have \( \overrightarrow{OM} = \overrightarrow{OC} + \frac{1}{2} \overrightarrow{CB} = \vec{c} + \frac{1}{2} \vec{a} \) and \( \overrightarrow{ON} = \overrightarrow{OD} + \frac{1}{2} \overrightarrow{DE} = \vec{d} + \frac{1}{2} \vec{a} \).
Let Z be the midpoint of MN. Then \( \overrightarrow{OZ} = \frac{1}{2} \overrightarrow{OM} + \overrightarrow{ON} = \frac{1}{2} (\vec{c} + \frac{1}{2} \vec{a}) + (\vec{d} + \frac{1}{2} \vec{a}) = \frac{1}{2} (\vec{a} + \vec{c} + \vec{d}) \).
Therefore Z = X, where X is the point of intersection of OF and CE found in part a. Hence X is the midpoint.

Exercise 1

Exercise 2

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