A position vector is the vector of an arrow drawn to from an origin to a point. Look at the example below - a Cartesian plane in which we see points AA A and BB B The position vector of AA A connects the origin, marked OO O , to AA A - it can be written as a⃗ a→ \vec a or OA→OA→ \vec{OA} . Similarly, the position vector of \(B\) connects the origin to BB B - it can be written as b⃗ b→ \vec b or OB→OB→ \vec{OB} It is important that the letters are written in this order. The vector BO→BO→ \vec{BO} would connect B to the origin and therefore represents −b⃗ −b→ -\vec b . If you are using Cartesian components to describe a position vector, the Cartesian coordinates of a point can be used to describe the position vector of that point. For example, AA A has the coordinates (1,4)(1,4) (1,4) . This means its position vector can be written as a⃗ =a→= \vec a=1ı⃗ +4ȷ⃗ 1ı→+4ȷ→ 1 \vec{\imath} + 4 \vec{\jmath} or a⃗ =[14]a→=[14] \vec a= \begin{bmatrix}1\\4 \\\end{bmatrix} . Similarly, BB B has the coordinates (−3,−1)(−3,−1) (-3,-1) . This means its position vector can be written as b⃗ =b→= \vec b=−3ı⃗ +−1ȷ⃗ −3ı→+−1ȷ→ -3 \vec{\imath} + -1 \vec{\jmath} or b⃗ =[−3−1]b→=[−3−1] \vec b= \begin{bmatrix}-3\\-1 \\\end{bmatrix} .
Position vectors can be used to describe the vector between two points. The position vector \( \vec{AB}\) goes from \(A\) to \(B\) Going from \(A\) to \(B\) is the same as going from \(A\) to \(O\) and then from \(O\) to \(B\) Therefore, \( \vec{AB}= \vec b - \vec a\) This means that the component form of \( \vec{AB}\) can be found by subtracting the components of \(A\) from \(B\), i.e. \[ \vec{AB}= \vec b - \vec a\] \[= \begin{bmatrix}-3\\-1 \\\end{bmatrix} -\begin{bmatrix}1\\4 \\\end{bmatrix}=\begin{bmatrix}-4\\-5 \\\end{bmatrix}\] or \[\Big(-3 \vec{\imath} + -1 \vec{\jmath}\Big) - \Big(1 \vec{\imath} + 4 \vec{\jmath}\Big) = (-3-1) \vec{\imath} + (-1-4) \vec{\jmath}=-4 \vec{\imath}-5 \vec{\jmath}\] The magnitude of this vector can be used to calculate the distance between the two points. \[ |\vec {AB} | = \sqrt{(-4)^2+(-5)^2}= \sqrt{16+25}=\sqrt{41}\] Note: \[ \vec{AB}= \vec b - \vec a\] \[ \vec{BA}= \vec a - \vec b\] \[ \therefore \vec{BA}=-\vec{AB}\]
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