Step 1: Understanding the Concept:
We are given a subspace W of a 3-dimensional vector space (presumably \(\mathbb{R}^3\)) defined by a single linear constraint. The dimension of a vector space is the number of vectors in its basis. We need to find a basis for W and count the number of vectors.
Step 2: Key Formula or Approach:
1. The dimension of the parent space \(V^3\) is 3.
2. The subspace W is defined by the equation \(a+b+c=0\).
3. We can express one of the variables in terms of the others. For example, \(c = -a-b\).
4. An arbitrary vector \((a,b,c)\) in W can then be written in terms of fewer, independent parameters. The number of these independent parameters will be the dimension of W.
5. Alternatively, the Rank-Nullity theorem can be used. The subspace W is the null space of a linear transformation.
Step 3: Detailed Explanation:
An arbitrary vector in W is of the form \((a,b,c)\) and satisfies the condition \(a+b+c=0\).
From this condition, we can write \(c = -a-b\).
So, any vector in W can be written as:
\[ (a, b, -a-b) \]
We can decompose this vector into a linear combination of vectors involving only \(a\) and only \(b\):
\[ (a, b, -a-b) = (a, 0, -a) + (0, b, -b) \]
\[ = a(1, 0, -1) + b(0, 1, -1) \]
This shows that any vector in W can be expressed as a linear combination of the two vectors \(\vec{v}_1 = (1,0,-1)\) and \(\vec{v}_2 = (0,1,-1)\).
These two vectors span the subspace W.
Now we need to check if they are linearly independent. Suppose:
\[ k_1(1,0,-1) + k_2(0,1,-1) = (0,0,0) \]
\[ (k_1, k_2, -k_1-k_2) = (0,0,0) \]
This gives \(k_1=0\) and \(k_2=0\).
Since the only solution is the trivial one, the vectors \(\vec{v}_1\) and \(\vec{v}_2\) are linearly independent.
Because the set \(\{(1,0,-1), (0,1,-1)\}\) spans W and is linearly independent, it forms a basis for W.
The number of vectors in the basis is 2.
Therefore, the dimension of W is 2.
Step 4: Final Answer:
The dimension of the subspace W is 2, which corresponds to option (A).