Finding the span of 3 vectors

Question

Let $V = \mathbb R^3$, a vector space over the reals. Find the span $W$ of $\{(1, 2, 1), (3, −1, −4), (0, 7, 7)\}$ in the form $\{(x, y, z) ∈ V \mid ax + by + cz = 0\}$ for some $a, b, c$. Find a basis for $W$.

I'm just really having trouble here; I know the definition of span, but not how to apply it here. And to write in a particular form and find a basis is confusing.

Answer 1

First check if the vectors are linearly independent. You can do this by putting the matrix $$\left[ \begin{matrix} 1&2&1\\ 3&-1&-4\\ 0&7&7 \end{matrix} \right]$$ into reduced row echelon form. This gives you $$\left[ \begin{matrix} 1&0&-1\\ 0&1&1\\ 0&0&0 \end{matrix} \right]$$ So the three vectors are not linearly independent, and any two vectors will be sufficient to find the span, which is a plane. I will use the vectors $(1,2,1)$ and $(3,-1,-4)$, which are linearly independent and form the basis you require. The cross product of these is $(-7,7,-7)$, giving you the normal vector of your plane. Therefore your span will be $$\{(x, y, z) ∈ V | -7x + 7y + -7z = 0\}$$

Answer 2

Hint:

Write the condition(s) for matrix $\;A=\begin{bmatrix}1&3&0\\2&-1&7\\1&-4&7\end{bmatrix}$ and the augmented matrix $$B=[A|X],\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},$$ to have the same rank.