![[../../00 Asset Bank/250px-Crossproduct.gif#invert_B]] > [!bug] fucking kill me IB Physics PTSD > An operation for "multiplying" two [[Vector|Vectors]] that creates another [[Vector]] which is [[Orthogonal]] to both inputs, this operation is *exclusive* to $\Rn{3}$. $\huge\begin{align*} \let \vec u, \vec v &\in \R^{3}\\ \mat{v_{x}\\v_y\\v_{z}} &= \mat{u_yv_{z}-u_zv_y\\u_zv_x-u_xv_z\\u_xv_y-u_yv_{x}}\in\R^3\\ \vec u &\perp \vec u \times \vec v \\ \vec v &\perp \vec u \times \vec v \end{align*}$ > [!tip] Rules of Distrobution > All rules of distrobution familiar with standard multiplication apply. > > $\huge \let \vec u, \vec v, \vec w \in \R^3$ > $\huge > \vec u \times (\vec v + \vec w) = \vec u \times \vec v + \vec u \times \vec w > $ > > $\huge \let k \in \R$ > > $\huge$ > > Cross products are **anti communative** > $\huge \vec u \times \vec v = -\vec v \times \vec u$ > > $\huge \vec w \times \paren{\vec u \times \vec v} > \ne \paren{\vec u \times \vec v}\times \vec w > $ # Properties The cross product between two vectors finds a vector [[Orthogonal]] to both vectors. $\huge\begin{align*} \let \vec u, \vec v, \vec w &\in \R^{3}\\ \vec w &= \vec u \times \vec w \\ \vec w &\perp\vec u \\ \vec w &\perp\vec v \\ \end{align*}$ ### Magnitude $\huge \norm{\vec u \times \vec v } = \norms{\vec v}\norms{\vec u} \sin \theta $ ### Relation with 0 Vector All [[Vector|vectors]] crossed with itself will equal the [[Zero Vector]]. $\huge \let \vec v \in \R^{3}$ $\huge \vec 0 \times \vec v $ $\huge \vec v \times \vec v = \vec 0 $ # Ways to solve - Memorize the formula $\huge \vec u \times \vec = \mat{u_yv_{z}-u_zv_y\\u_zv_x-u_xv_z\\u_xv_y-u_yv_{x}}\in\R^3 $ - Ways with basis $\huge\begin{align*} \hat i &= \mat{1\\0\\0} \\ \hat j &= \mat{0\\1\\0} \\ \hat k &= \mat{0\\0\\1} \\ \end{align*} $ $\huge\begin{matrix} \hat i \times \hat i =\vec 0 & \hat i \times \hat j = \hat k & \hat i \times \hat k = -\hat j \\ \hat j \times \hat i =-\hat k & \hat j \times \hat j = \hat 0 & \hat j \times \hat k =-\hat i \\ \hat k \times \hat i =\hat j & \hat k \times \hat j = -\hat i & \hat k \times \hat k =-\vec 0 \\ \end{matrix} $ ### Textbook solution $\huge\begin{align*} \mat{u_x\\u_y\\u_{z}} \times \mat{v_x\\v_y\\v_{z}} &= \left|\begin{matrix} \hat i & \hat j & \hat k \\ u_{x}&u_y&u_z \\ v_x&v_y&v_z \end{matrix}\right| \\ &= \vmat{u_y&u_z\\v_y&v_{z}}\vec z - \vmat{u_x&u_z\\v_x&v_{z}}\vec j + \vmat{u_x&u_y\\v_x&v_{y}}\vec k \end{align*}$ > [!important] $\vec U \vec V$ rays 🤯 (use sunscreen)