may I know what is the scalar product for $\vec{p}=\lambda a + (1-\lambda) b$, $\vec{A}=a$. Should we say $a \cdot p = \lambda a^2 +(1-\lambda) a\times b $ or $a \cdot p=\lambda a^2 +(1-\lambda) a\cdot b$.
Thank you very much for replying to the question, I am quite confused with the concept of the dot product here.
On
I am assuming $a$ and $b$ are vectors so I will use $\vec{a}$ and $\vec{b}$ instead.
$\vec{p}. \vec{A} = \left ( \lambda \vec{a} + (1 - \lambda) \vec{b} \right ). \vec{a} $ $ =\lambda \vec{a}.\vec{a} + (1 - \lambda)\vec{a}.\vec{b} $
In case $\vec{a}$ $\perp$ $\vec{b}$, then this reduces to $\vec{p}. \vec{A} = \lambda \vec{a}. \vec{a}$
With a further assumption that $\vec{a}$ is a unit vector, then $\vec{p}. \vec{A} = \lambda$.
The dot product distributes, so $\bf a\cdot(b+c)=a\cdot b+a\cdot c$. The second expression is correct.
As a sidenote, you seem to have used the symbol $a$ to mean both the vector $\bf a$, as in $a\cdot p$, and its magnitude, as in $a^2$. You may use
\vec afor $\vec a$ or\mathbf afor $\bf a$, to indicate that '$a$' refers to a vector. You can use $|\mathbf a|,|\vec a|$ or just $a$ to denote its magnitude.