# Angle between radius vector and x,y,z axis

## Homework Statement

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

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

## Homework Equations

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

## 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...

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

What is the answer it states in the book?

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

b)y axis

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

c)z axis

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

tiny-tim
Homework Helper
Find the angle between radius vector of the point M(6,2,9) and:

a)x axis ; b)y axis ; c)z axis
$$\frac{2*\sqrt{10}}{11}$$ for the angle between radius vector and x axis...
b)y axis

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

c)z axis

$$\frac{\sqrt{85}}{11}$$
Hi Physicsissuef!

You've got the question wrong …

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

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

HallsofIvy
Homework Helper
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.

Yes but I didn't get the right answers.

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

tiny-tim
Homework Helper
Draw a diagram … follow what I said …
… it's the angles between the radius vector and the three coordinate planes, not axes.
… and use Pythagoras!

I get

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

tiny-tim
Homework Helper
Hi Physicsissuef!

No, you get:

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

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

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?

HallsofIvy
Homework Helper
Yes but I didn't get the right answers.

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

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.

And what about my last problem?

tiny-tim
Homework Helper
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.
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".

tiny-tim
Homework Helper

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?
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:
Euler angles are one of several ways of specifying the relative orientation of two such coordinate systems. Moreover, different authors may use different sets of angles to describe these orientations, or different names for the same angles. Therefore a discussion employing Euler angles should always be preceded by their definition.
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?

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...

tiny-tim
Homework Helper
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...
(I assume you mean $$\vec{n_1}\cdot\vec{n_2}$$ ?)

hmm … this is your usual problem of putting too much reliance on formulas.

ok, this is to help you see why you have to use what you think is only the formula for vectors even though these are lines …

Using your formulas, what are the angles that the following lines (in a plane) make with the x-axis and with the y-axis:
(1) x - y = 0?
(2) x + y = 0?
(3) y - x = 0?

(and no questions this time … just do it! )

(1),(2),(3) are this all equations of planes?

CompuChip