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I realize the component of one vector $\vec{a}$ in the direction of the other vector $\vec{b}$ is essentially a projection of $\vec{a}$ onto $\vec{b}$ and I understand how it is calculated. What I don't understand is why exactly we call this projection a "component" in this case. If we take vector $\vec{a}\space(5, 6)$ in $2D$ space with origin in $(0, 0)$, what I am used to call the components of such vector are vectors $\vec{b}\space(5, 0)$, and $\vec{c}\space(0, 6)$. Or, in simple words, such vector has $5$ units towards $X$ direction, and $6$ untis towards $Y$ direction.

Now, looking at this picture projection of $\vec{a}$ onto $\vec{b}$ we still call the resulting projection of $\vec{a}$ onto $\vec{b}\space$ a "component". However this component does not have anything to do with components of vector $\vec{a}$ that are along $X$ or $Y$ axis. I would need some intuition on why we call the resulting projection of one vector onto another a component. It's more intuitive for me to just call it a "projection", not a "component".

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    $\begingroup$ It is a component using the directions of $b$ and a vector perpendicular to $b$. $\endgroup$
    – David Mitra
    Dec 23, 2022 at 13:31
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    $\begingroup$ If we write a vector $\vec v$ as the (always uniquely existing) linear combination $m\cdot \vec a+n\cdot \vec b$, then the coefficients $m,n$ are the components of the vector with respect to the vector space created by $\vec a$ and $\vec b$ $\endgroup$
    – Peter
    Dec 23, 2022 at 13:36
  • $\begingroup$ @Peter except a and b are colinear (uniqueness is not fulfilled then) $\endgroup$
    – Ernesto Iglesias
    Dec 23, 2022 at 14:50
  • $\begingroup$ @ErnestoIglesias Of course, we do not have even existence in general in this case. $\endgroup$
    – Peter
    Dec 24, 2022 at 10:03

2 Answers 2

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The word component here for a vector $\mathbf{v}$ means that $\mathbf{v}$ is already known as a (linear) composition of vectors, that is, that in the given context we have that $\mathbf{v}=\mathbf{v}_1+\ldots +\mathbf{v}_n$ for some vectors $\mathbf{v}_1,\ldots ,\mathbf{v}_n$. Then we can say that each $\mathbf{v}_k$ is a component of (the given/known composition of) $\mathbf{v}$.

Also we can say that $\mathbf{v}$ can be decomposed as the sum of a list of vectors $\mathbf{v}_1,\ldots ,\mathbf{v}_n$, this just means that $\mathbf{v}=\mathbf{v}_1+\ldots +\mathbf{v}_n$, then in this context a component (of the given decomposition) is (again) just one the $\mathbf{v}_k$ vectors already mentioned.

Then we can also talk of a decomposition of $\mathbf{v}$ in a list of linearly independent, or orthogonal, vectors. The kind of decomposition used will depend on the context and why we are decomposing $\mathbf{v}$ in the given way.

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  • $\begingroup$ So do I understand correctly, that a given vector $\vec{a}$ has an infinite number of components in a given $2D$ space, and a vector projection of $\vec{a}$ onto $\vec{b}$ is just a way to find one of such $\vec{a}$ components which is happened to be along $\vec{b}$ direction? $\endgroup$
    – just_a_noobie
    Dec 23, 2022 at 15:22
  • $\begingroup$ @just_a_noobie no. Components of a vector is something related to a specific and previously known or stated decomposition of that vector. So it doesn't make sense to say that a vector have infinite components without referencing to a specific decomposition (that have indeed infinite components). This decomposition can be some times stated implicitly, not explicitly. It depends entirely in the context to know about what decomposition is referencing the word component. $\endgroup$
    – Masacroso
    Dec 23, 2022 at 15:53
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Because every vector is defined with respect to a coordinate system. If your vector $b$ is the unit-vector $(1,0)$ your are indeed computing the "component" in the X-axis. If $b$ would be the vector $(0,1)$ you would compute the component in the Y-axis. But no one forces you to use the X- or Y-axes. You might use another basis to generate your vector $v$ in the $2$-D plane. If you use the standard X,Y-axes then you would you as basis $B=\{(0,1), (1,0)\}$. But you could simply project in any other vector and define it as the component on such vector.

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