# Homework Help: Angle between radius vector and x,y,z axis

1. May 17, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

Find the angle between radius vector of the point M(6,2,9) and:

a)x axis ; b)y axis ; c)z axis

2. Relevant equations

$$cos(\vec{a},\vec{b})=\frac{\vec{a} \vec{b}}{|\vec{a}||\vec{b}|}$$

3. The attempt at a solution

I tired with

$$cos\alpha=\frac{(6,2,9)(6,0,0)}{\sqrt{121}\sqrt{36}}$$

but it is not like in my textbook results...

2. May 17, 2008

### CompuChip

Why don't you switch to polar coordinates? Then you can just read them off.

3. May 17, 2008

### BryanP

What is the answer it states in the book?

4. May 17, 2008

### Physicsissuef

$$\frac{2*\sqrt{10}}{11}$$ for the angle between radius vector and x axis...

5. May 17, 2008

### BryanP

6. May 17, 2008

### Physicsissuef

b)y axis

$$\frac{3*\sqrt{13}}{11}$$

c)z axis

$$\frac{\sqrt{85}}{11}$$

7. May 17, 2008

### tiny-tim

Hi Physicsissuef!

You've got the question wrong …

… it's the angle betwen the radius vector and the three coordinate planes, not axes.

8. May 17, 2008

### Physicsissuef

I still don't have the right results...

9. May 17, 2008

### HallsofIvy

If <x, y, z> is a unit vector, $cos(\theta_x)= x$, $cos(\theta_y)= y$, and $cos(\theta_z)= z$ give the angles it makes with the axes. In other words, divide <6, 2, 9> by its length.

10. May 17, 2008

### Physicsissuef

Yes but I didn't get the right answers.

$$\frac{6}{11} ; \frac{2}{11} ; \frac{9}{11}$$

11. May 17, 2008

### tiny-tim

Draw a diagram … follow what I said …
… and use Pythagoras!

12. May 17, 2008

### Physicsissuef

I get

$$cos \alpha=\frac{(6,2,9)(6,2,0)}{11*\sqrt{40}}=\frac{40*\sqrt{40}}{11}$$

13. May 17, 2008

### tiny-tim

Hi Physicsissuef!

No, you get:

$$\cos\alpha=\frac{(6,2,9)(6,2,0)}{11*\sqrt{40}}=\frac{40}{11\sqrt{40}}=\frac{\sqrt{40}}{11}=\frac{2\sqrt{10}}{11}\,.$$

14. May 17, 2008

### Physicsissuef

Ok, thanks... Probably there is mistake in my text book. It clearly says that they are axis and not planes. Thanks again...

15. May 17, 2008

### Physicsissuef

Find the angles of the directions of the normal line of the plane 2x-y+2z+9=0

Probably the normal is $$n(2,-1,2)$$, probably the directions are the x,y,z axis.

$$cos\alpha=\frac{(2,-1,2)(2,0,0)}{3*2}=\frac{2}{3}$$

and in my book says it is $cos\alpha=\frac{-2}{3}[/tex] for the second it says that it is [itex]cos\beta=\frac{1}{3}$

and the third $cos\gamma=\frac{-2}{3}$, what is the problem?

16. May 17, 2008

### HallsofIvy

Those are the cosines of the angles the line makes with the coordinate axes.

Now, solve $cos(\alpha)= 6/11$, $cos(\beta)= 2/11$, and $cos(\gamma)= 9/11$.

I have no idea why tiny tim is assuming "angles made with planes". A line or vector does NOT make a single angle with a plane. We can interpret "angle with normal to plane" as an "angle with plane" but that puts you back to "angle with coordinate axes" again.

17. May 17, 2008

### Physicsissuef

And what about my last problem?

18. May 17, 2008

### tiny-tim

Hi HallsofIvy!

hmm … never could think three-dimensionally …

but my method seemed to get the right result!

I've now noticed that the book answers are the sines of the angles with the axes, while 6/11 2/11 and 9/11 are the cosines.

I'm perplexed as to why the book would expect sines as answers … so I still think it's the planes not the axes.

After all, one often sees "a pole makes an angle x with the ground".

19. May 17, 2008

### tiny-tim

You get cos = 2/3, and the books says cos = -2/3.

Yours is the relative angle "the short way round", and the other is "the long way round".

wikipedia, at http://en.wikipedia.org/wiki/Euler_angles says:
So you'll have to check which system the book is using.

I hope that if it comes up in the exam, the question will tell you which system to use. Is that right, guys?

20. May 18, 2008

### Physicsissuef

The book is using:

$$cos\alpha=\frac{\vec{n_1}\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}$$ for vectors
and
$$cos\alpha=\frac{|\vec{n_1}\vec{n_2}|}{|\vec{n_1}||\vec{n_2}|}$$
for lines... Anyways if I use the both formulas I get positive number, I don't know what is the problem...