vectors in $3$ dimensions such that $v+(v \times u)=u$ Let $i,j,k$ denote the usual three unit vectors in $\mathbb R^3.$
1) Find all vectors $v \in \mathbb R^3$ such that $v+(v \times i)=j$.
2) Suppose vectors $v$ and $u$ belong to $\mathbb R^3$ and satisfy $v+(v \times u)=u.$ What is the connection between $u$ and $v$?
For part 1) I let $v=xi+yj+zk$ and substituted into expression, I came out with a answer $v=\frac12 j + \frac12 k$, can someone tell me if this is correct.
Also can someone please help me with part 2).
 A: Part $1$ looks good.
For part two, it might help to take the dot product of both sides of the expression with $u$ or $v$ and see what this implies (since $(u \times v)\cdot u = (u \times v) \cdot v = 0$).
In particular, 
$$
u\cdot v + u\cdot(v \times u) = u \cdot u \\
\Rightarrow u\cdot v = \|u\|^2
$$
and
$$
v\cdot v + v\cdot(v \times u) = v \cdot u \\
\Rightarrow \|v\|^2 = u\cdot v
$$
In other words, $$\|u\|^2 = u\cdot v = \|v\|^2$$ What does this tell you geometrically about $u$ and $v$? (Hint: under what circumstances does $u\cdot v = \|u\|\|v\|$?)
A: For the first part,
$$\left(\frac12 j + \frac12 k\right) \times i =
\frac12 (j \times i) + \frac12 (k \times i) =
\frac12 (-k) + \frac12 (j),$$
so your answer seems OK.
For the second part, an equivalent statement is
$(v \times u) = u - v.$
Suppose $u$ and $v$ are vectors such as $u=i+2j$, $v=-2i+k$
that span a plane within $\mathbb R^3$.
What angle does $u - v$ make with that plane? What angle does $v \times u$
make with the same plane?
(The dot product method in the other answer works too!)
