How do we know if a third vector is on the plane of the first and second vector? I have to add its components and if it gives me zero,then is it on the plane?
Suppose that the vectors have three components.
Thank you.
 A: Use the fact that a plane is a two dimensional object. That means that the size of the base is two (you can describe any vector in that plane as a linear combination of the vectors in the base). Now create a matrix with the components of those three vectors as rows (or columns). If the determinant is not zero, then the vectors are independent, so they span a three dimensional space, not two. You need to have the determinant equal to $0$
A: Other answers are right, but I personally avoid the use of determinant when I can, so here is an alternative solution.

I have to add its components and if it gived me zero,then is it on the plane?

I don't really get what you mean with that. 
Let's call your vectors $u$,$v$ and $w$. Let's say that $u$ and $v$ are your first vectors spanning a plane, and you're asking if $w$ is in that plane.
By "adding it's components", are your adding all the components of $w$ (i.e $w_1+w_2+w_3$)? That doesn't really make sense in this context.
However, if $w$ is in this plan, it is a linear combinaison of $u$ and $v$. So, it can be written in the form of: $$\alpha u + \beta v = w$$
Hence, you have to solve the system:
$$\begin{cases}
\alpha u_1+\beta v_1=w_1 \\
\alpha u_2+\beta v_2=w_2 \\
\alpha u_3+\beta v_3=w_3 \\
\end{cases}$$
You then have three possibilities:


*

*If you find a uniq solution for $\alpha,\beta$, all vectors are in the same plane. 

*If you find infinite solutions, that means that all vectors are colinear, and $u$ and $v$ did't span a plane in the first place. 

*If you don't find any solution, the vectors are linearly independant, and therefore, $w$ isn't in the plane spanned by $u$ and $v$
A: HINT
Recall that


*

*$\vec a\times \vec b=\vec c \implies \vec c\perp \vec a,\vec b$

*$\vec c= x\vec a + y \vec b \iff \det (\vec a,\vec b,\vec c)=0$
