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Suppose we have the vectors $\vec{AB} ,\vec{AC} ,\vec{AD} $ like in the image , and all of their coordinates are relatively to the two preprinted axes , say I want $\vec{AB} \cdot \vec{AD}$ , is it enough to consider the dot product between their components like they are now , or I should make a traslation of the vector so that they starts at the origin ? (I have never done linear algebra so I hope to have been clear enough)

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    $\begingroup$ I think that if you want $\vec {AB}.\vec {AD}$, then you must simply consider the components as they are now. (you want $\vec {AB}.\vec {AD}$) $\endgroup$ – Devansh Kamra Apr 21 at 10:46
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The components of the vector $\vec{AB}$ as they are now and the components after translating to the origin are the same, so it doesn't matter. Note that the components of $\vec{AB}$ are obtained by subtracting the coordinates of $A$ from those of $B$, this difference doesn't change when $A$ and $B$ are translated together.

In your example we have approximately \begin{align*} \vec{AB} &\approx \begin{pmatrix} (-4.5)-(-1.2) \\ 2.6-5.6 \end{pmatrix} = \begin{pmatrix} -3.3 \\ -3 \end{pmatrix}, \\ \vec{AC} &\approx \begin{pmatrix} (1.5)-(-1.2) \\ 3.1-5.6 \end{pmatrix} = \begin{pmatrix} 2.7 \\ -2.5 \end{pmatrix}, \end{align*} so that $$ \vec{AB}\cdot\vec{AC} \approx (-3.3)\times2.7 + (-3)\times(-2.5) = 1.41. $$

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  • $\begingroup$ right right I was just confused about the fact that I had to subtracte the coordinates of $A$ from those of $B$. Thank you! $\endgroup$ – Tortar Apr 21 at 10:51

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