# Span of a Vector Space in $\mathbb{R}^3$

Consider the subspaces $$W_1$$ and $$W_2$$ of $$\mathbb{R}^3$$ given by $$W_1= \{(x,y,z) \in \mathbb{R}^3:x+y+z=0 \}$$ and $$W_2=\{(x,y,z) \in \mathbb{R}^3:x-y+z=0 \}$$.

If $$W$$ is a subspace of $$\mathbb{R}^3$$ such that

• $$W \cap W_2= \mathrm{span}\bigl\{(0,1,1)\bigr\}$$

• $$W \cap W_1$$ is orthogonal to $$W \cap W_2$$ with respect to the usual inner product of $$\mathbb{R}^3$$

then which of these are true?

1. $$W = \mathrm{span} \bigl\{ (0,1,-1),(0,1,1) \bigr\}$$

2. $$W = \mathrm{span} \bigl\{ (1,0,-1),(0,1,-1) \bigr\}$$

3. $$W = \mathrm{span} \bigl\{ (1,0,-1),(0,1,1) \bigr\}$$

4. $$W = \mathrm{span} \bigl\{ (1,0,-1),(1,0,1) \bigr\}$$

My Attempt:
$$x+y+z=0 \implies x+y=-z$$ so that free variables are two so $$\mathrm{dim}(W_1)=2$$ and similarly $$x-y+z=0 \implies x+z=y$$ so that $$\mathrm{dim}(W_2)=2$$.

Also $$W \cap W_2 = \mathrm{span}\bigl\{(0,1,1) \bigr\}$$ implies $$(0,1,1)$$ is one element of $$W$$ so options 2,4 discarded.

How to approach this type of problems in general?

• Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
– Dave
Commented Dec 24, 2018 at 23:35
• I edited it but I use the symbol \$ then curly braces removed. Commented Dec 25, 2018 at 0:40
• You have to "escape" the braces, by typing \{, since they are usually used for something else.
– user403337
Commented Dec 25, 2018 at 1:09
• Thanks for the hint Commented Dec 25, 2018 at 5:57

You reasoned correctly and discarded $$2$$ and $$4$$. It must be $$1$$, since $$(1,0,-1)$$ isn't orthogonal to $$(0,1,1)$$.