Finding the direction cosines of a vector I am having difficulty solving a problem presented to me by a fellow classmate. I was given a diagram, but I will describe the problem without one, as it is not necessary to fully understand the problem:
Let v and u be an arbitrary vectors in 3 dimensions. If the angle between v and u is $\lambda$ and u makes an angle $\theta$ with the positive x axis ($cos\alpha=\theta$ for u), show that the direction cosines for v can be expressed as
$cos\alpha=cos\lambda cos\theta$
$cos\beta=cos\lambda sin\theta$
$cos\gamma=sin\lambda$
where $cos\alpha$, $cos\beta$, and $cos\gamma$ are the angles v makes with the positive x, y, and z axes respectively.
I approached this problem using the formula for the angle between to vectors, since it gives a easily manageable expression for $cos\lambda$, but I keep arriving at fallacious statements. Any help would be appreciated. Thank you.
 A: The statement is false. Let $\vec{u}=(a,0,0)$ and let $\vec{v}=(0,b,c)$ for any $a,b,c\in\mathbb{R}$, $a>0$. Then
$$\cos(\theta)=\frac{\vec{u}\cdot\hat{i}}{||\vec{u}||\cdot||\hat{i}||}=\frac{a\cdot1+0\cdot0+0\cdot0}{|a|\cdot1}=1$$
(pictorially, we can see that $\theta$, the angle between $\vec{u}$ and the positive $x$-axis, is $0$) 
and
$$\cos(\lambda)=\frac{\vec{u}\cdot\vec{v}}{||\vec{u}||\cdot||\vec{v}||}=\frac{a0+0b+0c}{|a|\cdot\sqrt{b^2+c^2}}=0,$$
(pictorially, we can see that $\lambda$, the angle between $\vec{u}$ and $\vec{v}$, is $\frac{\pi}{2}$). According to the statement this would imply that
$$\cos(\alpha)=\cos(\lambda)\cos(\theta)=0\cdot1=0$$
$$\cos(\beta)=\cos(\lambda)\sin(\theta)=0\cdot0=0$$
$$\cos(\gamma)=\sin(\lambda)=1$$
so that $\vec{v}$ is necessarily along the positive $z$-axis... which is not necessarily true.
So, a specific counterexample to the statement is $\vec{u}=(1,0,0)$ and $\vec{v}=(0,1,0)$.

Intuitively, we can see that no such statement could be true: let's choose our $\vec{u}\in\mathbb{R}^3$ first. This determines the value of $\theta$, and hence also the value of $\cos(\theta)$ and $\sin(\theta)$. If the formulas
$$\cos(\alpha)=\cos(\lambda)\cos(\theta)$$
$$\cos(\beta)=\cos(\lambda)\sin(\theta)$$
$$\cos(\gamma)=\sin(\lambda)$$
were true, then all that would be necessary to determine the direction cosines $\cos(\alpha)$, $\cos(\beta)$, $\cos(\gamma)$ of any $\vec{v}\in\mathbb{R}^3$ would be the angle between $\vec{u}$ and $\vec{v}$ (namely, $\lambda$); but this can't be true, because the set of vectors in $\mathbb{R}^3$ having angle $\lambda$ with $\vec{u}$ forms a cone of vectors, all having different direction cosines in all three directions (so knowing $\lambda$ can't be enough).
