Scalar product of 2 vectors. The scalar product of two vectors $\overrightarrow u$ and $\overrightarrow v$ represented as $\overrightarrow u\cdot\overrightarrow v$ One now knows that three vectors $\overrightarrow a$,$\overrightarrow b$ and $\overrightarrow c$ such that $\overrightarrow a\cdot\overrightarrow b $ + $\overrightarrow b \cdot \overrightarrow c =0 $ Hence :
A $\overrightarrow c =-\overrightarrow a$
B $\overrightarrow b$ is orthogonal to $\overrightarrow a + \overrightarrow c$
C $\overrightarrow b = \overrightarrow 0$ or $\overrightarrow a + \overrightarrow c =0$
D $\overrightarrow a$,$\overrightarrow b$ and $\overrightarrow c$ must be collinear.
E $\overrightarrow a$ is orthogonal with $\overrightarrow b$ and $\overrightarrow c$ is orthogonal with $\overrightarrow b$

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*My work
If it's A then the answer should have been the $\overrightarrow 0$.
I think the answer is B because if $\overrightarrow b$ is orthogonal to the sum of $\overrightarrow a$ + $\overrightarrow c $ this leads to a scalar value which is represented as a dot with a circle around it on the cartesian system.
It's not C because the answer should be $\overrightarrow 0$ and not 0.
For D I thought for 3 vectors to be collinear they can't be orthogonal to eachother so reaching 0 isn't a possibility.
For E I thought you'll get 2 times the scalar value of 0 which is not what we want.
Could someone guide me to the right answer?
Thanks in advance.
 A: The correct answer is $B$.
The dot product is commutative and distributive. That is $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}$ and $\vec{a}\cdot(\vec{b}+\vec{c})=\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}$.
This is why $\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}=\vec{b}\cdot(\vec{a}+\vec{c})=0$ This leads to the conclusion in option $B$ becasue this IS the definiton of orthogonality of two vectors.
For $A$ you can take $a=\hat{i}$ ,$ b=\hat{j}$ and $c=2\hat{i}$ to disprove the claim. You say that if it was so then you get $0$. That is already given to you and it does not lead to a contradiction of any kind . That is wrong reasoning.
Your reasoning for $C$ is NOT totally correct. Most times $0$ and $\vec{0}$ in Linear Algebra means the same thing. Although as you correctly say that $0$ is the scalar $0$(That is the $0$ element of the field and $\vec{0}$ is the null vector in the vector space) . Ignore the brackets if it does not makes sense to you right now . I assume you are only familiar with high school level vector algebra and calculus(The bare minimum needed for physics).
The complete reasoning for $C$ should be like this. Let $\vec{a}=\hat{i}$, $\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=-\hat{j}$. Then you have $\vec{b}\neq \vec{0}$ and $\vec{a}+\vec{c}=\hat{i}-\hat{j}\neq \vec{0}$.
For $D$ .Let  $\vec{b}=\hat{i}+\hat{j}+\hat{k}$, $\vec{a}=-\hat{i}$ and $\vec{c}=\hat{j}$. Then these are not collinear.
For $E$ your reasoning is totally incorrect. $2\cdot\vec{0}=\vec{0}$. These are serious misconceptions on which I think you should work on. You have to come up with counterexamples which show that the claim is false.
For this again the example for $D$ suffices to show that the claim is not true.
A: Option B is obviously true.
All others are false:A: Take $\vec c=-2\vec a$, for instance.
C: My guess is that $\vec0$ here is a typo and that it should be $0$ here. And there are plenty of examples of non-zero vectors with scalar product equal to $0$ (unless your space is $1$-dimensional).
D: For instance, in $\Bbb R^2$ endowed with the usual scalar product, take $\vec a=(1,0)$, $\vec b=(0,1)$ and $\vec c=(1,0)$.
E: For instance, in $\Bbb R^2$ endowed with the usual scalar product, take $\vec a=(1,0)$, $\vec b=(1,1)$, and $\vec c=(-1,0)$.
