Vectors and Lines

Chapter 4 • Section 4-1

"Direction and Magnitude... the DNA of geometry! 🧬"

📏 Vector Norms (Length)

The length (or magnitude) of a vector $\vec{v} = (v_1, v_2, \dots, v_n)$ is:

$$ \norm{\vec{v}} = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2} $$

This is just the Pythagorean theorem in $n$ dimensions!

A vector with length 1.

To normalize any non-zero vector $\vec{v}$:

$$ \vec{u} = \frac{1}{\norm{\vec{v}}} \vec{v} $$

Distance between points $P$ and $Q$ (vectors $\vec{u}$ and $\vec{v}$):

$$ \text{dist}(\vec{u}, \vec{v}) = \norm{\vec{u} - \vec{v}} $$

Two vectors are parallel if they lie on the same line. Equivalently, one is a scalar multiple of the other.

$$ \vec{v} = k\vec{u} $$

Example: $\vec{u}=(1, 2)$ and $\vec{v}=(2, 4)$ are parallel because $\vec{v} = 2\vec{u}$.

Vector addition $\vec{u} + \vec{v}$ corresponds to the diagonal of the parallelogram formed by $\vec{u}$ and $\vec{v}$.

$\vec{u} - \vec{v}$ is the vector from the tip of $\vec{v}$ to the tip of $\vec{u}$.

Find the midpoint $M$ between $P_1(-1, -4, 3)$ and $P_2(5, 0, -3)$.

The midpoint is the average of the coordinates:

$$ M = \frac{P_1 + P_2}{2} = \left( \frac{-1+5}{2}, \frac{-4+0}{2}, \frac{3+(-3)}{2} \right) = (2, -2, 0) $$

What is the length of vector $\vec{v} = (3, 4)$?

A line through point $P_0$ with direction $\vec{d}$ is given by:

$$ \vec{p} = \vec{p}_0 + t\vec{d} $$

where $t$ is any real number (scalar).

Breaking it down by component:

$$ \begin{cases} x = x_0 + at \\ y = y_0 + bt \\ z = z_0 + ct \end{cases} $$

where $\vec{d} = (a, b, c)$.

Visualize $\vec{p} = \vec{p}_0 + t\vec{d}$. Drag the slider to change $t$.

t = 0

$\vec{p}_0 = (1, 1)$, $\vec{d} = (2, 1)$

Are the vectors $\vec{u} = (2, -4, 1)$ and $\vec{v} = (-6, 12, -3)$ parallel?

Hint: Check if one is a scalar multiple of the other.