# Why Do we call a projection of one vector onto another a "component"?

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".

• It is a component using the directions of $b$ and a vector perpendicular to $b$. Dec 23, 2022 at 13:31
• 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 $\vec b$ Dec 23, 2022 at 13:36
• @Peter except a and b are colinear (uniqueness is not fulfilled then) Dec 23, 2022 at 14:50
• @ErnestoIglesias Of course, we do not have even existence in general in this case. Dec 24, 2022 at 10:03

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.
• 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? Dec 23, 2022 at 15:22
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.