1. ## Vectors; Angles made with postive Ox, Oy, Oz

"Find the angle $\displaystyle \theta$, where $\displaystyle 0 \leq \theta \leq 180 ^{\circ}$, made by the position vector $\displaystyle \mathbf{r}$ with the positive directions of the axes $\displaystyle Ox, Oy, Oz$ in the following cases: (a) $\displaystyle \mathbf{r} = (1, 0, 0)$, (b) $\displaystyle \mathbf{r} = (0, 1, 1)$, (c) $\displaystyle \mathbf{r} = (0, 0, -1)$, (d) $\displaystyle \mathbf{r} = (1, 1, 1)$, (e) $\displaystyle \mathbf{r} = (1, 1, -1)$."

I would have thought that the answers would be:

(a) $\displaystyle \theta _x = 0, \, \theta _ y= \theta _ z = 90^{\circ}$.

This agrees with the answer sheet.

(b) I'm not sure of how best to visualise the required answer for $\displaystyle \theta _x$ but I suppose that it equals $\displaystyle 90^{\circ}$. And $\displaystyle \tan{ \theta _ y}= \tan{ \theta _z} = 1$, so $\displaystyle \theta _y = \theta _ z = 45^{\circ}$.

I think this must be wrong, as the answer sheet gives $\displaystyle \theta _x = \theta_y = 45^{\circ}$ and $\displaystyle \theta_z = 90^{\circ}$.

(c) I would have thought that $\displaystyle \theta_x = \theta _ y = 90^{\circ}$ and $\displaystyle \theta _z = 180^{\circ}$. However, I'd really like to make $\displaystyle \theta_y = \theta_y = -90^{\circ}$, to communicate that the angle is measured anticlockwise from the respective positive axes. Though the desired range of angles does not allow for this, nor does it allow for anything greater than $\displaystyle 180^{\circ}$, so $\displaystyle 270^{\circ}$ would be disallowed too.

(d) For visualising $\displaystyle \theta _x$ say, should one imagine hovering orthogonally above the $\displaystyle x$ and $\displaystyle y$ axes, in the region of the positive $\displaystyle z$ axis and visualising the angle between the vector and the axis from this "2 dimensional" viewpoint? In this case, I think that $\displaystyle \theta _ x = \theta _y = \theta _z = 45^{\circ}$.

The answer sheet gives $\displaystyle \theta_x = \theta_y = \theta_z = \arctan{\left(\frac{1}{\sqrt{3}} \right)} \approx 54.7^{\circ}$.

(e) I'm really not sure about this one (not that I've been particularly sure about any of the previous!), but I'd suggest that $\displaystyle \theta _x = \theta _y = 45^{\circ}$ and $\displaystyle \theta _z = 135^{\circ}$.

The answer sheet gives $\displaystyle \theta_x = \theta_y \approx 54.7^{\circ}$ and $\displaystyle \theta_z \approx 125.3^{\circ}$.

Could somebody help me to visualise the required angles? Also, why is it not that negative angles could be given (akin to the system used to describe the arguments of complex numbers), or angles greater than $\displaystyle 180^{\circ}$? Is the system above amiguous, i.e. that the three angles do not imply a single specific half-line in space?

2. Originally Posted by Harry1W
Could somebody help me to visualise the required angles? Also, why is it not that negative angles could be given (akin to the system used to describe the arguments of complex numbers), or angles greater than $\displaystyle 180^{\circ}$? Is the system above amiguous, i.e. that the three angles do not imply a single specific half-line in space?
I don't know how to help unless you tell us what you have to work with.
Do you know about dot products?
Do you know that the angle between vectors $\displaystyle u~\&~v$ is $\displaystyle \arccos \left( {\frac{{u \cdot v}} {{\left\| u \right\|\left\| v \right\|}}} \right)?$

3. Sorry; I'm afraid that I have been victim of the tl;dr scorn in the past, so I am now a bit too careful!

I do remember the dot and cross products, but I'm working in a chapter of a textbook that precedes their discussion, so I was under the impression that it could be worked out via other methods.

But I suppose that one could use the formula you mentioned to find the angle between the given vector and the basis vectors, to give a set of solutions?