Find the base of a set in $\mathbb{R}^3$, and $\mathbb{R}^4$ The set of $4$-th dimensional vectors $V = (x_1, x_2, x_3, x_4)$, where the content's of $V$ that satisfy the following equation: 
$$x_1 − 3 x_2 + 2 x_3 + 4 x_4 = 0$$
make up a subspace L of dimension $3$. Find a basis $\vec e_1, \vec e_2, \vec e_3$ for L. Afterwards find a fourth vector $\vec e_4$ so that $\vec e_1, \vec e_2, \vec e_3, \vec e_4$ form a basis for $\mathbb{R}^4$.
I am familiar with the concept of a space, subspace, span, linear independence and bases. All sorts of help is appreciated!
 A: You need to find three non-coplanar vectors that satisfy your equation.  One easy one to find is $(3,1,0,0)$  You can find another by taking its fourth component to be zero, picking any numbers for the first two, and solving for the third.  It cannot be collinear with the first because it has a component in the third axis.  Now you can pick numbers for the first three components and solve for the fourth.  As long as the fourth does not come out zero it will not be coplanar and you have a basis.  To find a fourth vector to span $\Bbb R^4$ you just need one vector that you cannot express linearly in terms of the first three.  Any vector that does not satisfy your equation will do, such as $(1,0,0,0)$
A: Well $V=\lbrace (x_1,x_2,x_3,x_4) \, : \, x_1=3x_2 - 2x_3 - 4x_4 \rbrace = \lbrace  (3x_2 - 2x_3 - 4x_4, x_2, x_3, x_4) \,:\, x_2,x_3,x_4 \in \mathbb{R} \rbrace = \lbrace x_2(3,1,0,0)+x_3(-2,0,1,0)+x_4(-4,0,0,1) \,:\, x_2,x_3,x_4 \in \mathbb{R} \rbrace $ , so the basis for $V$ is the set of vectors $ \lbrace (3,1,0,0), (-2,0,1,0), (-4,0,0,1) \rbrace $, which is easily provable.
To find a basis for $\mathbb{R} ^ 4$ it is sufficient to find a vector that is not a linear combination of the basis for $V$ ( for example $(1,0,0,0)$), you will thus find a linearly independent set of 4 vectors in $\mathbb{R} ^ 4$ , and since $\dim \mathbb{R} ^ 4 =4$, this set is a basis for $\mathbb{R} ^ 4$
