vault backup: 2024-12-04 15:43:49

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arc 2024-12-04 15:43:49 -07:00
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@ -23,14 +23,14 @@ A vector with a magnitude of 1 is a **unit vector**.
If $\vec{v}$ is a nonzero vector, the unit vector can be found using the equation $\vec{u} = \dfrac{1}{|\vec{v}|}\vec{v}$ . In other words, to find a unit vector, divide the vector by its magnitude. If $\vec{v}$ is a nonzero vector, the unit vector can be found using the equation $\vec{u} = \dfrac{1}{|\vec{v}|}\vec{v}$ . In other words, to find a unit vector, divide the vector by its magnitude.
# $i$, $j$ Notation # $i$, $j$ Notation
Every 2d vector has a horizontal component and a vertical component. The horizontal unit vector could be written as $i = \langle 1, 0 \rangle$, and the vertical unit vector could be written as $j = \langle 0, 1 \rangle$ Every vector can be made up using a combination of these standard unit vectors. Every 2D vector has a horizontal component and a vertical component. The horizontal unit vector could be written as $i = \langle 1, 0 \rangle$, and the vertical unit vector could be written as $j = \langle 0, 1 \rangle$ Every vector can be made up using a combination of these standard unit vectors.
# Trigonometric Form # Trigonometric Form
Given a vector $\vec{v}$ with a magnitude $|\vec{v}|$ and direction $\theta$: Given a vector $\vec{v}$ with a magnitude $|\vec{v}|$ and direction $\theta$:
The component form is given as: The component form is given as:
$$ \vec{v} = \langle \cos \theta,\ |\vec{v}|\sin\theta \rangle $$ $$ \vec{v} = \langle |\vec{v}||\cos \theta,\ |\vec{v}|\sin\theta \rangle $$
# Standard position # Standard position
- A vector is in standard position if the initial point is at $(0, 0)$. - A vector is in standard position if the initial point is at $(0, 0)$.