Is it possible to have a ↓ b = b ↓ a ? Explain why a ↓ c = a ↓ (b ↓ c

Question

Can anyone explain in which cases do the follow occur? Preferably with diagrams as well?

1. Is it possible to have a ↓ b = b ↓ a ?

2. Explain why a ↓ c = a ↓ (b ↓ c )

   a ↓ b means a's projection on b.

Answer 1

1. if $\mathbf a = \mathbf b$, of course the projection of vector $\mathbf a$ onto the line spanned by $\mathbf b = \mathbf a$ is $\mathbf a$, and conversly.

2. If you know the dot product, observe that $$\mathbf a \downarrow \mathbf b = \frac{\langle \mathbf a\cdot \mathbf b\rangle}{\langle \mathbf b\cdot \mathbf b\rangle}\mathbf b$$.

Thus, \begin{align} \mathbf a \downarrow ( \mathbf b \downarrow \mathbf c ) &= \mathbf a \downarrow \left( \frac{\langle \mathbf b\cdot \mathbf c\rangle}{\langle \mathbf c\cdot \mathbf c\rangle}\mathbf c \right) \end{align}

Let $h = \frac{\langle \mathbf b\cdot \mathbf c\rangle}{\langle \mathbf c\cdot \mathbf c\rangle}$. Of course the line spanned by $\mathbf c$ and $h \cdot \mathbf c$ is the same, and that should hint you that the projection should give you the same resul. Indeed we have

$$ \mathbf a \downarrow ( \mathbf b \downarrow \mathbf c ) = \mathbf a \downarrow \left( h \cdot \mathbf c \right) = \frac{\langle \mathbf a\cdot h \cdot \mathbf c\rangle}{\langle h \cdot \mathbf c\cdot h \cdot \mathbf c\rangle} h \cdot \mathbf c = \frac{h^2 \langle \mathbf a\cdot \mathbf c\rangle}{h^2\langle \mathbf c\cdot\mathbf c\rangle}\mathbf c = \frac{\langle \mathbf a\cdot \mathbf c\rangle}{\langle \mathbf c\cdot \mathbf c\rangle}\mathbf c = \mathbf a \downarrow \mathbf c. $$