Finding a unit vector when you have two planar vectors and a normal vector. Find a unit vector in the plane of the vectors $A = i + 2j$ and $B = j + 2k$, perpendicular to the vector $C = 2i + j +2k$.
I'm confused as to what the problem is telling me.
I believe this problem is telling me that $A$ and $B$ span a particular plane and C is normal to that plane. As such $A \times B$ should give me the normal to the plane but $A \times B = 4i - 2j + k \neq 2i +j + 2k = C$.
So how is it possible for these particular $A$ and $B$ to span this plane when this particular $C$ is not normal, even though it is supposed to be?
 A: No, $C$ is not said to be perpendicular to the plane spanned by $A$ and $B$. Just write an arbitrary vector from the span of $A$ and $B$ as $R=\lambda A+\mu B$. Then you do the inner product $(R,C)=0$ and take into account the assumption for unit length of $R$ to find $\lambda,\mu$:
$(R,C)=0\Leftrightarrow (\lambda,2\lambda+\mu,2\mu)\cdot (2,1,2)=2\lambda+2\lambda+\mu+4\mu=0 \Rightarrow 4\lambda+5\mu=0$
also 
$$||R||=1\Leftrightarrow \lambda^2+(2\lambda+\mu)^2+4\mu^2=1$$
 and you get $\lambda,\mu$
Note, that if $R\perp C$ then so is $-R$. Therefore you will get two different solutions for $(\lambda,\mu)$
A: We note that the vectors $$\vec{a}=(1,2,0) \mbox{ and } \vec{b}=(0,1,2)$$
are linearly independent, and thus span a plane. If you are unfamiliar with these words, I hyperlinked the (possibly) problematic phrases to their respective Wikipedia articles. 
Therefore, if we want to find a vector in the plane spanned by $\vec{a}$ and $\vec{b}$, say $\vec{d}$, then 
$$\vec{d} = \lambda \vec{a} + \mu \vec{b}, \mbox{ for some } \lambda,\mu\in\mathbb{R}.$$
Furthermore, we want $\vec{d}$ to be perpendicular to $\vec{c}=(2,1,2)$. That is,
$\vec{d}\cdot \vec{c} = 0. $ So, we see that
$$(\lambda \vec{a} + \mu \vec{b})\cdot \vec{c}=\lambda(\vec{a}\cdot\vec{c})+\mu(\vec{b}\cdot\vec{c})=4\lambda + 5\mu=0.$$
We then simply choose $\lambda$ and $\mu$ such that this is the case. For example, $\lambda=5$ and $\mu=-4$. So, 
$$\vec{d} = 5 \vec{a} -4 \vec{b} = (5,6,-8).$$
Now, we need to normalize the vector; the length of $\vec{d}$ is $\sqrt{25 + 36 + 64}=5\sqrt{5}$. Therefore, the vector
$$\frac{1}{5\sqrt{5}} (5,6,-8)$$
satisfies the requirements.
