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## #1 Re: Help Me ! » Vector Help » 2014-08-27 20:40:28

hi Shelled,

This is a back of the envelop analysis, done whilst on the way to the Royal Albert Hall to hear Tchaikovsky's  6th.

A and C must lie in a plane that is perpendicular to BED:

As B and D have the same z coordinate, so will E, so the z term above will be zero and the plane is vertical.

If you draw a circle, centre E, and radius BE, it will go through A and C.  This shows the view looking along BED.

But A and C can be any diametrically opposite points on this circle.  So there are an infinite number of such pairs!  AE = BE = CE = DE and AC is at 90 to BD.

But, we also know B'.  The line BB' must be perpendicular to the plane ABCD, so another equation is possible.

So, find the line of intersection of the planes.  AC must lie on it.

Then find where that line intersects the sphere

You'll get a quadratic, hence two solutions.  One will be A and the other C.  You can tell from a rough 3D sketch which is which.

Hope that helps,

Bob

## #2 Re: Help Me ! » how to decide the optimal path? » 2014-08-26 19:36:07

hi pari_alf

I'm assuming you want the minimal penalty.  As two are equal you can choose either.

Bob

## #3 Re: Help Me ! » Vector Help » 2014-08-26 19:29:54

hi Shelled,

I've been having a longer think.  This is what you might do:

ABCD must lie in a plane that contains BED, and AC will be perpendicular to BD.  So you can write down the equation of that plane.

A and C must lie on a circle, centre E and radius EB, so you can write down the equation of that circle.

(To get a square, you need AB = BC.  I think you don't need this one.)

So, you could call A (x, y, z) and form equations according to the above.  The given numbers don't make it easy to simplify these.  If I had access (which I don't) to a computer equation solver, then I could probably find those values and hence get  A and C.

Bob

## #5 Re: Help Me ! » Vector Help » 2014-08-26 06:35:44

hi Shelled,

ABCD is a square and you know the diagonal.

So find the midpoint, E, and BE (=ED).

I notice that B and D are in the same z = 11.9312 plane.  Now, A and C may be one above and one below that plane, but it will have to be by equal amounts.

So you can use the dot product to get a constraint on AE.  I will have to try this to see what else is needed so I'll be back on this later.

EDIT:  AC must be of the form (-10.4391K,-24.7594K, z)

and you can get the x and y components by doing AE.EB = 0

ENDEDIT

ANOTHER EDIT:

That didn't work.  I just got 0 = 0

Need more input so I looked for AB=BC=CD=DA.

By trial I got A is approximately (77.81,34,069,15.48)

I think I need to know something about how you are expected to do this.  Is it a long project or just a quick (?) exercise in vectors ?

ENDEDIT

Once you have A and C:

Then AA' etc will be parallel to BB' with z = 0.  That should be enough to get A', C' and D'.

Thereafter it should be plain (plane ?) sailing.

Bob

## #6 Re: Help Me ! » Mandy Jane's Corner » 2014-08-26 06:24:22

Q4. 1.05% is not 0.001

Bob

Bob

## #11 Re: Help Me ! » Proof Check » 2014-08-21 19:17:02

hi Au101,

Looks like integration by parts to me.  I cannot see how else you can deal with the mix of 'x' and trig.  Try it that way and see what happens.

Bob

## #12 Re: Help Me ! » Proof Check » 2014-08-20 19:38:56

hi Au101

That looks good to me.

Well done.

Bob

## #14 Re: Help Me ! » Mandy Jane's Corner » 2014-08-20 06:28:14

I am preparing a large post.  Give me 10 minutes.

Bob

## #16 Re: Introductions » hi » 2014-08-20 04:18:34

hi iluvdogs

Welcome to the forum.  Glad you like the site.

Agnishom wrote:

Obviously, not paying attention.  Surely the correct question to ask is:

What is your favourite breed of dog ?

Bob

## #17 Re: Help Me ! » Difficult partial fractions and/or difficult integral » 2014-08-20 04:15:27

Au101 wrote:

I never really knew how to do that, but thanks bob bundy, that way's much easier.

I just thought, why did they ask for that substitution, unless it made the integral easier.  So in what way ? Thinks .......  Ah ha!

Bob

## #21 Re: Help Me ! » inverse function » 2014-08-19 22:35:36

hi Ramesh ra,

This graph will show you:

The blue curve is y = x^2, and the red curve is y = √x

The line y = x is the mirror line that will reflect one curve into the other.  This always happens with a pair of inverse functions.

eg.  (2,4) is on y = x^2   and (4,2) is on y = √x

Hope that helps,

Bob

## #22 Re: Help Me ! » Mandy Jane's Corner » 2014-08-19 22:25:42

$\frac{?}{?}= 0.?? = ??\%$

## #23 Re: Help Me ! » Difficult partial fractions and/or difficult integral » 2014-08-19 19:06:53

hi Au101,

What's wrong with

Bob

## #25 Re: Help Me ! » The circumference vs. area of a circle. » 2014-08-19 07:10:15

hi

I get BAD = 115.9 and ABC = 64.1   Notice that they add up to 180  (why?).