When vectors are added, the resultant vector is the sum of the individual components of each vector. For example, for these two vectors: $$\vec{u}=2\boldsymbol{\hat{\imath}}+3\boldsymbol{\hat{\jmath}}$$ $$\vec{v}=4\boldsymbol{\hat{\imath}}+7\boldsymbol{\hat{\jmath}}$$ their sum is: $$\vec{u}+\vec{v}=(2+4)\boldsymbol{\hat{\imath}}+(3+7)\boldsymbol{\hat{\jmath}}=6\boldsymbol{\hat{\imath}}+10\boldsymbol{\hat{\jmath}}$$ In matrix notation, this may written either of two ways: $$\vec{u}+\vec{v}=\begin{bmatrix}2 \\ 3 \end{bmatrix}+\begin{bmatrix}4 \\ 7 \end{bmatrix}=\begin{bmatrix}6 \\ 10 \end{bmatrix}$$ $$\vec{u}+\vec{v}=\begin{bmatrix}2 & 3 \end{bmatrix}+\begin{bmatrix}4 & 7 \end{bmatrix}=\begin{bmatrix}6 & 10 \end{bmatrix}$$
Scalar multiplication of vectors multiplies the magnitude of vectors by a scalar quantity without altering the direction of the vector. For example, the 3-dimensional vector: $$\vec{u}=2\boldsymbol{\hat{\imath}}+4\boldsymbol{\hat{\jmath}}+1\boldsymbol{\hat{k}}$$ A scalar multiplication by a factor of 3 yields: $$3\vec{u}=(3\cdot2)\boldsymbol{\hat{\imath}}+(3\cdot4)\boldsymbol{\hat{\jmath}}+(3\cdot1)\boldsymbol{\hat{k}}=6\boldsymbol{\hat{\imath}}+12\boldsymbol{\hat{\jmath}}+3\boldsymbol{\hat{k}}$$ In matrix notation, this can be written two ways: $$3\vec{u}=3\begin{bmatrix}2 \\ 4 \\ 1 \end{bmatrix}=\begin{bmatrix}6 \\ 12 \\ 3\end{bmatrix}$$ $$3\vec{u}=3\begin{bmatrix}2 & 4 & 1 \end{bmatrix}=\begin{bmatrix}6 & 12 & 3\end{bmatrix}$$ In ordered-set notation, it is written as: $$3\vec{u}=3\langle2, 4, 1\rangle=\langle6, 12, 3\rangle$$
The dot product multiplies together the common components of two vectors to yield a scalar value (not a vector): $$\sum_{i=1}^{n}u_{i}v_{i}$$ Where \(u_{i}\) is the component of vector \(\vec{u}\) in the \(i^{th}\) dimension, and \(v_{i}\) is the component of vector \(\vec{v}\) in the \(i^{th}\) dimension. \(n\) represents the total number of dimensions in vectors \(\vec{u}\) and \(\vec{v}\). A dot product can be performed on any two vectors with identical dimensions. For these 3-dimensional vectors: $$\vec{u}=1\boldsymbol{\hat{\imath}}+4\boldsymbol{\hat{\jmath}}+5\boldsymbol{\hat{k}}$$ $$\vec{v}=6\boldsymbol{\hat{\imath}}+3\boldsymbol{\hat{\jmath}}+2\boldsymbol{\hat{k}}$$ Multiply together the components with the same dimension ... $$\vec{u}\cdot\vec{v}=(1\cdot6)+(4\cdot3)+(5\cdot2)=28$$ Note that orthogonal vectors have a dot product of zero:
$$\vec{u}=-4\boldsymbol{\hat{\imath}}+2\boldsymbol{\hat{\jmath}}$$
$$\vec{v}=1\boldsymbol{\hat{\imath}}+2\boldsymbol{\hat{\jmath}}$$
$$\vec{u}\cdot\vec{v}=(-4\cdot1)+(2\cdot2)=0$$
The direction of the cross-product obeys the right-hand rule, and its magnitude is dependent on the magnitude of the two vectors, as well as the degree of orthogonality between them.
The definition of the cross-product is: $$\vec{u}\times\vec{v}=\| \mathbf{u} \|\| \mathbf{v} \|\sin(\theta)\mathbf{n}$$ Where \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\) are the magnitudes of vectors \(\vec{u}\) and \(\vec{v}\), \(\theta\) is the angle created by the intersection of the two vectors, and \(\mathbf{n}\) is the vector normal to the plane created by the two vectors. Because \(\sin(0°)=0\) and \(\sin(90°)=1\), vectors with high orthogonality will create cross product vectors with higher magnitude than those with low orthogonality.
The cross product of \(\vec{u}\) and \(\vec{v}\): $$\vec{u}=u_{1}\boldsymbol{\hat{\imath}}+u_{2}\boldsymbol{\hat{\jmath}}+u_{3}\boldsymbol{\hat{k}}$$ $$\vec{v}=v_{1}\boldsymbol{\hat{\imath}}+v_{2}\boldsymbol{\hat{\jmath}}+v_{3}\boldsymbol{\hat{k}}$$ can be calculated through the determinant: $$\vec{u}\times\vec{v}= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{vmatrix}$$ This results in: $$\vec{u}\times\vec{v}=(u_{2}v_{3}-u_{3}v_{2})\boldsymbol{\hat{\imath}}+(u_{3}v_{1}-u_{1}v_{3})\boldsymbol{\hat{\jmath}}+(u_{1}v_{2}-u_{2}v_{1})\boldsymbol{\hat{k}}$$ Or in matrix notation: $$\vec{u}\times\vec{v}=\begin{bmatrix} u_{2}v_{3}-u_{3}v_{2} \\ u_{3}v_{1}-u_{1}v_{3} \\ u_{1}v_{2}-u_{2}v_{1} \end{bmatrix}$$