Vectors 1.1.3

Vector Operations and Properties

Midpoint of a Line Segment

The midpoint of a line segment between points A and B is given by:

$\frac{1}{2} (\vec{OA} + \vec{OB})$

Example:

If $\vec{OA} = (1, 0, 4)$ and $\vec{OB} = (5, 6, 12)$, the midpoint is calculated as follows:

$\frac{1}{2} (\vec{OA} + \vec{OB}) = \frac{1}{2} ((1, 0, 4) + (5, 6, 12)) = \frac{1}{2} (6, 6, 16) = (3, 3, 8)$

Length of a Vector

2D Vector Length

For a vector $\,\vec{a} = (a<em>1, a</em>2)$ in 2D, the length (or magnitude) is given by:

$|\vec{a}| = \sqrt{a<em>1^2 + a</em>2^2}$

This formula is derived from the Pythagorean theorem.

3D Vector Length

For a vector $\,\vec{a} = (a<em>1, a</em>2, a_3)$ in 3D, the length is given by:

$|\vec{a}| = \sqrt{a<em>1^2 + a</em>2^2 + a_3^2}$

This is an extension of the Pythagorean theorem in three dimensions.

Derivation using Pythagoras' Theorem:

1. First application: Find the distance $\,d$ in the x-y plane:

   $d = \sqrt{a<em>1^2 + a</em>2^2}$

2. Second application: Find the length of the vector $\,\vec{a}$ using $\,d$ and $a_3$:

   $|\vec{a}| = \sqrt{d^2 + a<em>3^2} = \sqrt{(\sqrt{a</em>1^2 + a<em>2^2})^2 + a</em>3^2} = \sqrt{a<em>1^2 + a</em>2^2 + a_3^2}$

Example:

Find the length of the vector $\,\vec{v} = (2, -3, 6)$:

$|\vec{v}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

Worked Example:

Find the length of the vector $\,\vec{v} = (-1, 4, 7)$:

$|\vec{v}| = \sqrt{(-1)^2 + 4^2 + 7^2} = \sqrt{1 + 16 + 49} = \sqrt{66}$

Triangle Inequality

For any vectors $\,\vec{a}$ and $\,\vec{b}$, the triangle inequality states:

$|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$

This principle implies that the shortest distance between two points is a straight line.

The equality holds (i.e., $\,|\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}|$) only when $\,\vec{a}$ and $\,\vec{b}$ are parallel.

Worked Example:

Verify the triangle inequality for $\,\vec{a} = (1, 2, 2)$ and $\,\vec{b} = (6, 3, 2)$:

• Left-hand side (LHS):

  $|\vec{a} + \vec{b}| = |(1+6, 2+3, 2+2)| = |(7, 5, 4)| = \sqrt{7^2 + 5^2 + 4^2} = \sqrt{49 + 25 + 16} = \sqrt{90} \approx 9.5$

• Right-hand side (RHS):

  $|\vec{a}| + |\vec{b}| = |(1, 2, 2)| + |(6, 3, 2)| = \sqrt{1^2 + 2^2 + 2^2} + \sqrt{6^2 + 3^2 + 2^2} = \sqrt{9} + \sqrt{49} = 3 + 7 = 10$

Since 9.5 < 10, the triangle inequality is verified.

Unit Vector

A unit vector is a vector with a length (or magnitude) of 1.

It represents a