I'm reading from Marsden Vector Calculus 6th Edition and this picture is from page 43.
I'm having difficulty understanding how they get to $$ \text{Distance} =|\vec v \cdot \vec n|$$
The way I tried to work it out is using the projection formula:
$$\operatorname{proj}_{\vec n} \vec v= \frac{\vec v \cdot \vec n}{(\frac{A\vec i + B\vec j + C\vec k}{\sqrt{A^2+B^2+C^2}})^2} \cdot \frac{A\vec i + B\vec j + C\vec k}{\sqrt{A^2+B^2+C^2}} $$
$$\operatorname{proj}_{\vec n} \vec v= \frac{\vec v \cdot \vec n}{\frac{A\vec i + B\vec j + C\vec k}{\sqrt{A^2+B^2+C^2}}}$$
$$\operatorname{proj}_{\vec n} \vec v= \frac{\vec v \cdot \vec n}{\vec n}$$
and since $\vec n $ is a unit vector, we know it equals to $1$ so $$\operatorname{proj}_{\vec n} \vec v= \vec v \cdot \vec n$$
Question 1:But I still don't get where they got the absolute values from around the distance formula.
Question 2:Also hoe does the projection of $\vec v$ onto $\vec n$ make sense when projections are defined as dropping a perpendicular from the end of vector $\vec b$ toa line parallel to vector $\vec a$ in case of $proj_{\vec a} \vec b$. However in the case in the derivation we have that the line dropped from $\vec v$ is parallel(not perpendicular) to $\vec n$. Why is that?
To give an idea of what I mean by the last question:
The second picture was taken from: http://tutorial.math.lamar.edu/Classes/CalcII/DotProduct.aspx
Question 1: The dot product can be negative. If the point E is "below" the plane with respect to the given orientation, then without the absolute value, the answer would be negative. The absolute value makes it positive.
Question 2: The projection of a vector $b$ onto a vector $a$ $proj_a b$ is parallel to $a$. The "line dropped" is perpendicular to $a$ and, when subtracted from $b$, yields $proj_a b$ which is then parallel to $a$. In the picture shown, the solid black line is the "line dropped from $v$," and the dotted black line is the projection of $v$ onto the span of $n$.