Math Is Fun Forum

OK, here we go.

General background:

In transformation geometry the transform is usually shown before the thing it operates on.

So that forces us to use position vectors (3 rows by 1 column) rather than coordinates.

A 3 by 3 matrix will transform a 3 by 1 vector to produce another 3 by 1 vector.

As M x 0 = 0 only transformations that leave the origin invariant are possible. (rotations around O, reflections if the line goes through O, stretches and enlargements centred on O and shears where the invariant line goes through O)

Multplication reminder:

If M =  a   b   c                      and         V = p
           d   e   f                                           q 
           g   h   i                                           r

then transformed V = V' = ap + bq + cr
                                       dp + eq + fr
                                       gp + hq + ir

Sorry, cannot put large brackets; hope the alignment reaches you as it leaves me.

I looked up the rotations on Wiki at http://en.wikipedia.org/wiki/Rotational_matrix#Three_dimensions

Below the three rotations, it defines the direction of rotation.

As you want a relationship between angles, it doesn't matter how long OR and the rest are so I chose all to be 1 unit.
(OP = OQ = OR =1)

That gives OP =     sin C                               OQ =       0                     OR =   cos A
                            cos C                                           cos B                              0
                               0                                              sin B                             sin A

In my diagram, I'm imagining OX coming towards me, OY to the right and OZ straight up.

To rotate R onto OX                use  M =     cos A        0        sin A
                                                                0            1          0
                                                            -sin A          0        cos A

This gives That gives OP' =     cosAsin C                 OQ' =     sinAsinB                         OR' =   1
                                              cos C                                     cos B                                      0
                                          -sinAsinC                                 cosAsin B                                   0

Next to rotate OQ' onto the XY plane.  Call the angle D.

M = 1            0             0
       0        cosD        -sinD
       0        sinD          cosD

and we only need to work out the Z component of OQ'' and set it equal to zero

sinDcosB + cosDcosAsinB  = 0                         =>      tan D  =  -cosAtanB

After checking the direction of rotation the minus is Ok because the + direction is from Y towards Z, and we want OQ' to drop down onto the plane.

Now to figure out the final rotation around OZ of angle size E.

Apply D to P' to give P'' =            cosAsinC
                                      cosDcosC + sinDsinAsinC
                                      sinDcosC  -  cosDsinAsinC

We only need to calculate ther x component and set it to zero

M  =  cos E         -sin E        0
         sin E          cos E        0
            0               0           1

=>  cosEcosAsinC  - sinEcosDcosc - sinEsinDsinAsinC  = 0

=>  tan E =  cosAsinC / (cosDcosC + sinDsinAsinC)

And that's where I've got to.  As my formula doesn't agree with your values, either I've slipped up somewhere or you're not talking about the same angle as me.  Please have a look and see if (i) it makes sense, (ii) you can spot what's wrong.  By all means come back to me for any clarifications.  Whatever, the method will work; it's just a case of ironing out the bugs.

Bob

Hi bobbym

To find out what the angle was I first used the sine rule.  Then I spent ages trying to establish connections between the sines of angles so that I could get the values without a calculator.  Eventually did it.

Then I thought:  the sine rule is usually proved using angle properties of circles, so there ought to be a circle (circles) that lead to a solution.  Trouble is there's loads of possibilities but eventually I hit on your point as the centre of a very helpful circle and that led to a solution very similar to yours.

But then I thought:  all well defined problems like this can be solved using the sine rule (I think) so maybe they're all solvable if you can find the right circle(s).  That's as far as I've got so far.

Bob

ps.  If you've got a book with more like this, perhaps you would post one (just one please, I've got things to do!)

Hi John

I, too, am retired.  Was a maths teacher for 37 years plus some ICT in recent years.

Values helped a lot.

The negative sign is correct.  To turn as required needs -19.5966 to go towards the plane.  Might seem unimportant but the final formula is complicated and may give the wrong angle if I use the wrong sign at this stage.

What I have is tan E = (cosAsinC) / (cos DcosC + sinDsinAsinC)

             where tanD = -cosAtanB

I'm still not 100% certain on this as it doesn't give 5.4 with your values.

Would you like to see the matrices?  I can include a reminder of matrix multiplication for you.

Explaining it properly might help to validate it or highlight any error.

Bob

My method will work although I wouldn't like to have to explain it as it involves matrices and vectors.

But I have a sign mistake somewhere as I got atan(-tanBcosA) for the angle around the x axis.

I cannot decide why this has happened so, if you can be patient another 24 hours, I try to figure this out, before  giving my final result.

My final angle expression is pretty nasty; I doubt I could get it by high school trig alone.  But it is a function of sine/cos/tan on the three angles as you thought.

Bob

ps.  Do you have a known answer for some values of A, B and C, so I can test it?

Hi Machinist60

Not sure I've fully understood your problem.  Here's how I picture it.

Is this correct?

Please say what you mean by angle A.  eg A = POY or whatever.

whoops.  I've got BBCode on but my image tag hasn't worked.

What no FAQs on how to post images?

But you can follow the link to my site for the diagram

EDIT No you cannot.  See post 24.

"annual rate = 10%  monthly rate = 10%/12"

Don't think so.

monthly multiplier = (1.10)^(1/12)  = m (say)  [ so that 12 monthly multpliers give the yearly multiplier]

then monthly rate = (m-1) x 100 %  [ to convert back to percentage ]

I don't think this is a straight compound interest.  You can use CI when a fixed sum is invested but Dharshi says there's an annual investment.  So new money has to be added each year in addition to the interest.

Before I work out a revised formula let's just check I've got it right:

(i)  each year $1 is invested 

(ii)  for 20 years

(iii) interest rate is compound (ie. add the calculated interest  to the investment for the next year)

(iv) 10% per month !  This seems a bit generous!  Did you mean an equivalent annual rate of 10% but calculated monthly?

The graph of Y^2=X should be a symmetrical parabola (like Y = X^2 but 'on its side'). 

If you use sqrt you'll only get half the graph because the calc will default to the positive root.  To get the whole graph plot two functions:

Y1=sqrt(x) and Y2=negative sqrt(x) [note : use the negative key not the subtract key]