1. **State the problem**\nWe will do an example of a vector, find its magnitude, find a dot product, and show one way to use a vector to compute a distance. \n2. **Define a vector (components)**\nLet $\vec{v}=\langle 3,-2\rangle$.\nThis means the vector moves $3$ units in the $x$ direction and $-2$ units in the $y$ direction. \n3. **Find the magnitude (length) of the vector**\nFormula for magnitude in 2D:\n$$\lVert\vec{v}\rVert=\sqrt{v_x^2+v_y^2}$$\nSubstitute $v_x=3$ and $v_y=-2$: \n$$\lVert\vec{v}\rVert=\sqrt{3^2+(-2)^2}$$\n$$\lVert\vec{v}\rVert=\sqrt{9+4}$$\n$$\lVert\vec{v}\rVert=\sqrt{13}$$\nSo the length is $\sqrt{13}$. \n4. **Use the vector in a dot product (work with another vector)**\nLet $\vec{w}=\langle 1,4\rangle$.\nDot product formula:\n$$\vec{v}\cdot\vec{w}=v_x w_x+v_y w_y$$\nCompute: \n$$\vec{v}\cdot\vec{w}=3\cdot 1+(-2)\cdot 4$$\n$$\vec{v}\cdot\vec{w}=3-8$$\n$$\vec{v}\cdot\vec{w}=-5$$\nSo $\vec{v}\cdot\vec{w}=-5$. \n5. **Use vectors to find a distance between two points**\nTake points $A=(1,2)$ and $B=(4,6)$.\nThe vector from $A$ to $B$ is:\n$$\overrightarrow{AB}=B-A=(4-1,6-2)$$\n$$\overrightarrow{AB}=\langle 3,4\rangle$$\nDistance equals the magnitude of this displacement vector:\n$$AB=\lVert\overrightarrow{AB}\rVert=\sqrt{3^2+4^2}$$\n$$AB=\sqrt{9+16}$$\n$$AB=\sqrt{25}=5$$\nSo the distance from $A$ to $B$ is $5$. \n6. **Answer (final results)**\nMagnitude of $\vec{v}=\sqrt{13}$.\nDot product $\vec{v}\cdot\vec{w}=-5$.\nDistance $AB=5$.