Find two linearly independent vectors that lie within the plane 2x + y +z=4. verify they are linearly independent What confuses me is how to find vectors that fit into a plane. its been a while since ive done linear algebra and im confused about this concept.
I know that being independent of each other means that they are not parallel to each other but i am also uncertain of when looking at an equation where these aspects can be easily picked out.  
 A: Well, the plane is a subset of $\mathbb R^3$, so it is basically a set full of vectors from $\mathbb R^3$, and we say a vector lies in the plane if it is an element of the set. In your particular example, one element of the set (a vector, lying on the plane) is the vector $$[3,-1,-1],$$
i.e. the vector $[x,y,z]$ for which $x=1$ and $y=z=-1$. You know that it lies in the plane because $$2x+y+z = 2\cdot 3 + (-1) + (-1) = 6-2=4$$
A: The plane is defined by
$$\mathcal{P}:2x+y+z=4$$
A point on this plane is $A=(2,0,0)$. Another one is $B=(1,2,0)$. Therefore, a vector on this plane that transforms $A$ into $B$ is $\vec{u}=(-1,2,0)$.
I can now define a second vector $\vec{v}=(v_x,v_y,v_z)$ by taking it to be orthogonal to $\vec{u}$ (two orthogonal vectors form a basis, and are by definition linearly independent) and in the plane, i.e. so that the translation of $A$ by $\vec{v}$ is still a point in the plane. This is given by:
\begin{cases}
\vec{u}\cdot\vec{v}=0\Longleftrightarrow u_xv_x+u_yv_y+u_zv_z=0\\
2(v_x+2)+(v_y+0)+(v_z+0)=4
\end{cases}
The first equation gives:
$$-v_x+2v_y=0\Longleftrightarrow v_x=2v_y$$
Substituting for $v_x$ in the second:
$$5v_y+v_z=0$$
I am free to choose one of the values of $v_y$ and $v_z$. Let $v_y=1$, so that $v_z=-5$. Substituting back into $v_x$, I get $v_x=2$.
Conclusion: $\vec{u}=(-1,2,0)$ and $\vec{v}=(2,1,-5)$ are two linearly independent vectors on $\mathcal{P}$.
A: HINT
Say $\langle a,b,c\rangle $ is a vector in the given plane, this vector has to be perpendicular to the normal vector of given plane $\langle 2,1,1\rangle $. So the dot product between them will be 0 :  $2a+b+c = 0$. You can pick values for $a,b,c$.
