Johnny5

You don't even need a proof, all you need to do is come up with a single counterexample. Can't wait to see it.

I already had one in mind Tom, or Thomas.

(x-2)(x-2)=x^2-2x-2x+4=x^2-4x+4

Start with:

x^2 - 4x +4 = 0

The degree is 2. By the meaning of two, there are two roots.

The only root is 2. So there is only one root.

Thus contradicting the hypothesis that there are two roots.

QED

Did you see this coming?

Regards

PS: by the way i didn't use all that logical power to find a counterexample.

well how about you take a few math classes' date=' then get back to us with your contradictions to the proofs of the Fundamental Theorem of Algebra (which states that "Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers.")

 

jesus. you say you've never seen the proofs, but you automatically suspect that they're wrong. this alone is should be a good indicator of how much time we're all wasting in this thread.

 

and here, http://www.google.ca/search?hl=en&q=proof+fundamental+theorem+algebra&btnG=Search&meta=

 

come back when you've got your contradictions.[/quote']

 

Ok this post really does it.

 

Here is what I will now do.

 

I will utilize all of my powers of concentration, all of my skill in reasoning, and access any and all of the General order spatiotemporal modal binary deontic doxastic logic which i can, and prove unequivocally that it is not the case that every algebraic equation of degree n, has n solutions.

 

Give me one day.

You can use my answer to part (e) to immediately write down a potential function that is zero at a point where the electric field is nonzero.

 

Start with the Coulomb potential V®=-ke/r and add a term ke/r₀. So V®=-ke/r+ke/r₀. This gives the exact same E-field as the original V®' date=' and it is zero at r=r[sub']0[/sub] (where E is nonzero).

Well lets see, here is your answer to part e:

(e) The value of the electric potential can be chosen to be zero at any convenient point.

 

No matter what V(x) is' date=' you can always modify it by adding a constant c. This is because V(x) and V(x)+c give the same electric field. Since it is the field that is measureable, that is all we care about. So the two potentials above are physically equivalent. And c can always be chosen to be -V(x0) for an arbitrary x=x0. What does that tell you?[/quote']

 

Actually this part here makes perfect sense:

 

Start with the Coulomb potential V®=-ke/r and add a term ke/r0. So V®=-ke/r+ke/r0. This gives the exact same E-field as the original V®, and it is zero at r=r0 (where E is nonzero).

 

Let V® = -ke/r.

 

Now compute the gradient of V®.

 

Ñ V® = ¶V /¶r r

= ¶(-ke/r) /¶r r

 

= -ke¶(1/r) /¶r r

 

= -ke(-r^-2) r

 

= ke/r²r

 

Since

E = -Ñ v

 

It follows that:

 

E = - ke/r²r

 

Which is the electric field of an electron, at the origin of a reference frame.

 

The force on a positively charged test charge, with charge Q, is given by QE, hence we have:

 

F = - kQe/r²r

 

The negative sign implies the force is attractive, since Q, and e are positive.

 

Ok so we have the potential of an electron, as a function of r, namely:

 

V® = -ke/r

 

Now, suppose we add the constant ke/r₀ to V®, to obtain:

 

 

V® = -ke/r + ke/r₀

 

When we take the partial with respect to r, we get the same gradient function as before, hence the same electric field.

 

But clearly, when r=r0 the potential is zero, that is:

 

If V® = -ke/r + ke/r₀

then V(r₀)=0

 

So r0 is a point in the field where the potential is zero, as you can see.

 

It cannot be the origin, because that is division by zero error, unless the charge unit e, is zero inside the electron.

You can have relative rotation, i.e. you can measure your rotation with respect to anything you want. From a physics perspective rotation is absolute in that you can do measurements that will tell you if you are rotating or have rotated. There are frames that are not rotating.

See this is the kind of answer i want to read.

By saying that there are frames that aren't rotating, you are saying that rotation is absolute. The formulation of the statement, is as if that is the case.

Yes, rotation is absolute.

There is a mathematical relativeness to it, but when you are talking reality, it is as you say. I just don't want to be in the minority here, and don't bail on me.

For quite some time, I've been trying to develop the logic necessary to discuss rotation as an absolute.

There has to be some element which is beyond the mathematical treatment of it.

On Take One you got 1b and 1f right. You're looking good now.

 

But more importantly' date=' do you understand the reasoning process?[/quote']

 

Tom, what would you say to my answer of part c, it's a bit long winded. Here it is:

 

The answer to c is not so easy to understand as the answer to b. Right now I am trying to think of a realistic potential function v(x' date='y,z), which has a value zero at some point, yet the electric field at the point is nonzero.

 

I have already read Tom's answer, and he got me thinking of a paraboloid.

 

Here is the formula for it:

 

z = x2/a2 + y2/b2

 

The above formula, leads to the shape of a satellite dish, a circular paraboloid.

 

As you can see from the scalar function, when x=y=0, z is also equal to zero.

 

So suppose you have this, due to some charge distribution somewhere:

 

 

v = x2/a2 + y2/b2

 

You can see that the potential function is zero, when x=y=0.

 

Let's look at the gradient of v.

 

Ñv = 2x/a2 + 2y/b2

 

As it turns out, this implies that when x=y=0, that the electric field is zero too.

 

But you should now be able to see the following:

 

Had the exponents of x,y been 1, instead of zero, then after the partial derivatives were taken, we would have had a constant for an answer. So, suppose that you have the following potential function:

 

v(x,y,z) = 3x+2y+5z

 

Clearly, you have v(0,0,0) = 3*0+2*0+5*0=0

 

So at the point (0,0,0) in a frame, the voltage is zero, but now take the partials to obtain:

 

gradient v = 3i^+2j^+5k^

 

Which is clearly nonzero everywhere.

 

I'm not saying that the above function for v corresponds to any realistic configuration of charge distribution throughout the universe, but it does make Tom's point, which is that there are mathematical functions which constitute counteraxamples to part © here.[/quote']

 

Specifically, the equation for a plane works.

 

But is there any configuration of charge which gives a plane for the potential function?

 

I couldn't think of any.

Sorry, this may be a little off your main point, but the Wolfram article gave me a massice head-ache...

It's not necessary to start the thread off with a massive amount of mathematics. I was hoping to slowly move from simple things about rotating frames of reference, to the math. Hopefully we can both avoid headaches, although matrices are not that complicated, but certainly if you aren't familiar with linear algebra, you dont have a chance at understanding rotation explained in terms of matrices. However, and this would be my point, I think anyone can develop an intuitive understanding of rotation, without knowing matrix algebra, which is the reason I started off this paragraph by saying, "it's not necessary..." So don't worry about matrices for now.

What's the physical difference between "rotating objects in stationary coordinate systems' date=' and stationary objects in rotating coordinate systems"?

 

The only one I can think of is that a rotating coordinate system would have a non-Euclidean geometry. [/quote']

 

I'm not an expert with non-euclidean geometry, but if you know a bit about that, you can try to show me how it connects.

 

I was thinking more of something like Mach. Suppose nothing in the universe existed, other than a single billiard ball. Make it an eight ball.

 

Now choose a reference frame to analyse the motion in.

 

Suppose you choose a reference frame in which the center of mass of the eight ball is at rest, and furthermore the eight ball isn't spinning. Hence, the number 8 isn't rotating in the frame.

 

Now, suppose that some external force is applied to the eight ball, to get it spinning in the frame, without moving the center of mass.

 

As you can see, Newton's laws of motion weren't violated, and there is now a spinning eight ball. To really make this clear, "There is now an eightball which is spinning in an inertial reference frame."

 

Now, go back to before the external force was applied.

 

The center of mass of the eight ball is at rest in the frame, and the eight ball isn't rotating.

 

Now, add the stars. Thus, we are not ignoring their existence, and Mach talked about this.

 

Now, there are many non-inertial frames we can conceive of, in which the eight ball is spinning... but, and this is the key point, they are non-inertial frames.

 

This takes a bit of explanation, but you have to go into this, otherwise you miss the whole point.

 

Suppose that the number 8 is pointing to the center of galaxy X.

 

Currently, the eight ball isn't spinning, in some frame. Furthermore, let it be the case that there are no external forces acting upon the eightball, so according to Newton/Galileo's first law of motion, the center of mass will be at rest or moving in a straight line at a constant speed, in any inertial reference frame. So imagine a frame in which the CM of the eightball is at rest.

 

Thus, the number eight will continue to face galaxy X, until external forces are applied to it.

 

Mathematically, we can discuss a reference frame which is in orbit about the eight ball.

 

Think of it as if there is a satellite orbiting the eightball. From the satellite reference frame, the number 8 appears to be moving in a circle.

 

Only in reality, it is the satellite in orbit, not a spinning eight ball, because the number 8 is always pointing at galaxy X.

 

If you ignore the existence of the stars, it isn't clear right away, that the satellite frame is a non-inertial frame.

 

But, when you realize that from the satellites point of view, the center of mass of galaxy X is accelerating, in absence of an external force, which violates Newton's third law, it is then you realize the satellite frame is a non-inertial one.

 

So what does this have to do with my question?

 

Well, my question is, "Is rotation absolute or relative?"

 

I am trying to reduce the answer to a problem in binary logic, which amounts to a decision that needs to be made, regarding frames of reference.

 

I am thinking like this:

 

If A is true then "spin is relative."

If not A then "spin is absolute"

 

That is a very rough beginning.

 

In order to know what I mean, it actually helps to look at the spinning cubes in the wolfram article.

 

Suppose that there is a reference frame connected to the cube. The cube has six sides, and your reference frame has three lines, which define six directions relative to the origin of the frame. Let the origin of the frame be permanently located at the center of mass of the cube, and let the axes of the frame permanently emerge through the center of the face of the six sides, no matter what external forces act upon the cube.

 

So,we have a cube of material, and the rest frame of the cube.

 

We can now speak of the orientation of the cube, relative to other objects in the universe.

 

Now, suppose that in the rest frame of the cube, the trajectory of galaxy X is a circle.

 

Thus, when viewed from the rest frame of the cube, the center of mass of galaxy X is moving in a circle.

 

But, when viewed from a reference frame in which the center of mass of galaxy X is moving in a straight line at a constant speed, it appears that the cube is spinning.

 

So is spin absolute or relative?

 

There is a reference frame in which the cube isn't spinning.

There is a reference frame in which the cube is spinning.

 

Relative motion of things in the universe can be viewed from either frame simultaneously.

 

So to understand my question, there is only one more part.

 

Suppose you are standing on one of the faces of a cubical planet.

 

Your eyes are closed, so you cannot see the stars. They are there, but you cannot see them. You are at rest, in the rest frame of the cube.

 

Now, either the cube is spinning relative to galaxy X, or not, but you cannot see galaxy X, because your eyes are closed.

 

Can you from where you are, tell which is the case?

 

If the answer is no, then spin is relative.

If the answer is yes, then spin is absolute.

 

That's where my question is aimed.

 

I am positive the answer is that spin is absolute, but I'm not sure the best way to begin discussing it mathematically, using Newtonian physics.

 

I think something about coriolis force, should be used. The annoying part is that you can never get a clear answer to this from PhD's in physics. Otherwise i'd have understood this a decade ago.

 

When you ask them about it, all you get is "coriolis force is a fictitious force" and you are left with the idea that they don't understand rotational frames of reference.

 

They seem to know that if an ice skater pulls her arms in, she spins faster, but anyone knows that's true, who watches the sports channel.

 

Anyway, hopefully you get the idea of the question at least.

 

Also' date=' what kind of analysis are you doing of precession, purely mathematical or experimental as well?[/quote']

 

Ultimately, I don't see why not both. I have a firm example in mind though.

 

A toy spinning top, like a dradel.

 

Let me see if i can find an animation.

 

Here is one:

 

Precession animation

If you know the moment of inertia around one axis, you can use the theorem to find the moment about another axis - the moments won't be the same. Once the part breaks off the orbital axis will change. Say you wanted to calculate the angular momentum about this new axis. Rather than calculate the moment directly, which would be difficult because you don't have symmetry about the rotational axis, you can use the parallel axis theorem to find the new moment of inertia.

Here is hyperphysics on parallel axis theorem

Here is an article at Wolfram about rotation. In the article you will see mention of a rotation matrix.

I'd like to discuss this particular article, with a view in mind to answer the question, "is rotation absolute or relative."

This is going to help me in my ongoing analysis of precession, and things like Foucalt's pendulum. I am hoping to learn something new, and resolve a hundred or so unanswered questions of mine.

The article starts off with what is a very important statement, as regards the mathematical analysis of rotating frames of reference. The article begins:

When discussing a rotation, there are two possible conventions: rotation of the axes, and rotation of the object relative to fixed axes.

I do not see them as conventions. Something deeper, which involves the very meaning of "inertial reference frame" is going on.

There may be some kind of mathematical equivalence between rotating objects in stationary coordinate systems, and stationary objects in rotating coordinate systems, but there is a physical difference.

My point is this, I don't think the mathematical treatment brings out the physics of rotation.

If you read down into the Wolfram article, which is extremely well written, you will notice the following:

Any rotation can be given as a composition of rotations about three axes (Euler's rotation theorem), and thus can be represented by a matrix operating on a vector,

Somewhere in the Euler rotation theorem lies a connection to the aerospace concepts of: yaw, pitch, and roll.

I very much suspect that this proof would already assume existance of the complex numbers' date=' and as such it doesn't justify very much at all.

[/quote']

 

I would like to see said proof. Do you know it?

 

Kind regards

Actually, a clock can still be useful if (using Johnny's terminology) dG/dt is a constant.

Dr. Swanson, is there any standard terminology for what I was referring to?

"Proper time" doesn't seem quite adequate, since I am referring to a property of clocks, not a property of time.

If I am being unclear, I know how to remedy that.

A Turing machine? I've read a little about those from Stephen Wolfram. Turing machines are computer programs composed of a grid of separate and discrete cells that create results out of initial conditions and a particular set of rules. They are kind of like cellular automata.

 

http://www.turing.org.uk/turing/scrapbook/tmjava.html

That's not what I thought it was' date=' maybe I was thinking of something else.

Let me see if i can find what i was thinking of.

Here is what i was thinking of, i got universal turing machine confused with universal truth machine. When you said Godel that threw me off.

Godel's universal truth machine

There is an error in the reasoning Godel used.

At any rate that doesn't appear to be what you were talking about.

Also, if you have a link to what you read at Stephen Wolfram's site, I'd like to have a look at it. I went to the link you posted, and saw mention of three kinds of "Universal Turning Machines"

Regards

If I may, i think i know what might be constructive.

According to history, the mathematician Gauss proved that every integral rational equation of degree n, has exactly n roots.

Apparently, so the history books say, one of his proofs was his dissertation for his doctorate. He gave three others.

Now, I have never even seen one of these supposed proofs.

So i challenge anyone, to produce a proof of that which is italicized above. If such a proof is successful, it must be the case that:

x^2+1=0

has exactly two roots, and they will be +i, and -i.

So if the proof doesn't contain any errors, then imaginary numbers will be justified.

I suspect though, that there is a logical error in all four of Gauss' proofs.

But the point is time would have no meaning if nothing was changing anyways.

That's right.

This question arose in post 17 of http://www.scienceforums.net/forums/showthread.php?t=11414from

It was off topic so I'm moving it here.

Off topic:

Originally Posted by Johnny5

Isn't the total gravitational field at the center of the universe necessarily zero? Maybe he could distinguish it that way.

 

At the center of universe the gravitational potential should be at maximum' date=' the sum of gravity forces is zero.[/quote']

 

All I really meant was that if you put something at the center of the universe, the net gravitational force on it should be zero, and thus it will remain at rest there.

 

I believe it was Sir Isaac Newton who proved this.

 

In order to prove that at the center of the universe the gravitational potential should be a maximum, you have to use gravitational field theory.

 

If and when Latex is working, perhaps I will check what you say by doing that.

 

First you have define gravitational potential, analagous to electrical potential, and go from there. Mathematically speaking, its really easy.

 

You even come up with the idea of objects which gravitationally repel each other. Mach's idea.

 

Just postulate negative inertial mass.

(I missed this post last time a went to this thread and for some reason I can't edit my last post any more.)

(Strange this post I can edit ?)

The edit time has been changed to six hours. After you make your post' date=' it is editable for only six hours.

As for your last posts, the main thing I see is that the idea of a hypersphere goes against so much classical physics, as to be the source of endless problems in the conceptualization of relative motion.

The sheer number of problems caused, suffices to indicate that the universe is not hyperspherical. The motion of the ship has to take place in some frame. If there is no frame of reference that this can happen in, its over for the idea. If an object travels in a straight line, then the direction of its motion cannot change, and if its speed is constant, it will not return to where it was, [i']intuitively[/i]. If it does manage to return to its starting point, the rational conclusion is that its direction changed, hence it was accelerating.

So therefore, the one who defines a hyperspherical universe, needs to state whether or not the direction of motion changes, in this example problem. That is, they must explain the frame in which the motion is to be analyzed in. They will undoubtedly find that their definition is riddled with contradictions.

The best you can do for a clock is random (aka white) frequency noise. The integral of this is random-walk noise in the phase (i.e. the time). So two synchronized clocks will' date=' at best, do a random-walk away from each other. You always have to resynchronize; it's just a matter of how often.[/quote']

 

This last part here, on Gaussian white noise, the random walk problem covered in probability theory, Markov chains...

 

Were those mathematical things developed for the specific problem of clock synchronization, or did the mathematics develop first, and then got applied to the problem of clock synchronization later?

 

Regards

The EEP assumes the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system. I know this principle is supported by many experiments.

 

But recently i read an unknown article which seems to contradict the equivalence principle. I don't know how seriously to take it. You can read the entire book here.

I haven't had the chance to read the whole article, but I read enough to know that the author intends to be taken seriously.

In the part I read through carefully, he is discussing braking radiation, and comparing light emitted by electrons, when they are accelerated, and comparing inertial accelerations, to gravitational accelerations.

I've been aware of a problem there for a very very long time. I'm not sure if he is mathematically saying something I already know, but the very fact that he is thinking about the issue, shows that he understands the location of a problem with the EEP.

There is something to his article.

Regards

If we live in a discrete universe governed by causal determinism then then universe itself can "run" on a Universal Turing Machine and "now" would merely be the current iterative frame of some underlying state transition engine. I believe such a model of the universe is hypothesized in the form of loop quantum gravity

Hold your horses.

What is meant by "universe can run on a universal Turing machine"?

That's one question I have. I read about this awhile back, and figured out how I would approach the problem. (Godel sentence, and rules of how a UTM works. As I recall, the Universal Turing/truth machine, spits out true sentences when you ask it a question, but there is some kind of sentence which confuses the machine.)

The second question I have is this...

How do you connect the first question, together with the meaning of 'determinism" to loop quantum gravity?

So those are my two questions.

Regards

t=1/3 is correct but I see little point in you working out kingjewel´s problem.

Actually, the dot product method must be correct. But I'm not sure why the other approach didn't work yet. Im gonna figure it out and fix it tomorrow.

Thought experiment: You stare at a clock' date=' and nothing else is moving, nothing changing, no internal clocks either (or imagine all you have is your internal clock). The clock ticks once, and then again but the second time a longer amount of time passes before it ticks. Would you have any way to tell that the second tick took longer before it happened? How could you?

[/quote']

 

The only way, is to have another clock to compare to.

 

And for a clock to function the way we need it to, the amount of actual time between any two consecutive ticks must be equal.

 

The italicized part there, has to be incorporated into the definition of clock.

 

Or, in your example here, you are a comparing a bad clock to a good clock.

 

So a "good clock" is one which ticks out equal amounts of time, in successives ticks, a bad one ticks out different amounts of time sometimes, between successive ticks.

 

Mathematically, you could say that the 'rest rate' of the clock has to be a constant.

 

Denote the rest rate of a clock by G.

 

Then, dG/dt=0

 

Where the derivative is to be taken in an inertial reference frame in which the clock is at rest.

 

As far as human perception of time that is a whole different story. For most of us, eight hours a day, we have no perception of time whatsoever. It's called sleep. Yet time still passes, regardless of our perception of how much of it has passed... or even deeper whether or not we are even aware of it passing.

 

However, obviously there is something we are all endowed with, which allows us to measure time... a clock.

 

It seems to tick fairly regularly to me.

 

Nice post by the way.

I don't fully comprehend his explanation either.

I don't think he does either, so that makes three of us.

I would start of questioning his first sentence

"Quantum mechanics works."

There are different interpretations of QM.

I think that's a source of confusion.

A question I would ask him though, is what is meant by 'determinism.'

But I wanted a correct result . The book is correct, btw (not much of a surprise, though).

I am going to try your suggestion and see what happens.

In question b) you are asked to find a point P on the line C(t) so that P-O is perpendicular to B-A. Two vectors are perpendicular if their scalar product vanishes so your condition for P=C(x) -I´ve renamed t to x here to highlight that you are looking for a definite value instead of having an arbitrary parameter- would be (C(x)-O)*(B-A) = 0. That´s one equation with only one unknown. As x only appears with a power of one, the solution is also unique.

Here is the formula for the dot product of two vectors:

U*V = |U||V| cos(U,V)

If U is perpendicular to V, then the angle between them is 90.

Cosine of 90 degrees is zero.

Now, here is what you wrote for his part a:

The solution to a) is C(t) = A + t*(B-A) where C is a point on the line through A and B (which depends on one parameter, as a line is one-dimensional).

Ok, I drew a picture to help me understand what you wrote up there.

Pretty much the same formula I used.

I have R(t), for position vector, you have C(t), for location of a point.

In your formula there, i take it that is vector B minus vector A, times t.

So they appear to be the same formula.

Now, we need to find a vector from the origin O, to a point P on the line, such that vector OP is perpendicular to the infinite straight line through AB.

Your idea is to use the dot product to do this.

Let me pretend I already know the answer.

So here is the answer:

OP = Xi + Yj + Zk

Now, here is the vector, which lies entirely on the line, through the points A,B:

AB= <10-1,10-(-5),5-(-7)> = <9,15,12> = 9i + 15j + 12 k

(I got that from post #2 in this thread)

So these two vectors will be perpendicular iff their dot product is zero. They are perpendicular, hence:

0 = <X,Y,Z>*<9,15,12>

Therefore:

0 = 9X+15Y+12Z

Now, here is the formula (again) for arbitrary position vectors to points on the line:

R(t) = <9t+1,15t-5,12t-7>

(I took that from post #11 in this thread)

So we have:

X(t) = 9t+1

Y(t) = 15t-5

Z(t) = 12t-7

Hence:

9X(t) = 9(9t+1)

15Y(t) = 15(15t-5)

12Z(t) = 12(12t-7)

So that we must have:

0 = 9(9t+1) + 15(15t-5) + 12(12t-7)

As a constraint on parameter t.

So therefore:

0 = 81t+9 + 225t-75 + 144t-84

So therefore:

0 = (81+225+144)t+9 -75 -84

So therefore:

0=450t-150

So

150=450t

15=45t

3=9t

1=3t

hence t=1/3

hmm ok somewhere there is an error.

The one day I leave my laptop at work is the one I´d need it ... let´s see if I find a piece of paper somewhere ...

Do it in your head, Dave does.

EDIT: Johnny' date=' the idea of what you´re doing is correct (didn´t check the numbers) because the shorstest distance of a line to a point is the line segment from the line to the point that´s perpendicular to the line and vice versa. But it´s complete overkill here. Using the scalar product is much easier, faster and especially more related to vector math.[/quote']

 

Well I didn't get his book answer, so now what.

 

Right now I am checking to make sure the method I used does indeed answer the question.

 

I used the Pythagorean theorem to obtain a formula for the length of an arbitrary position vector, from the origin to random points on the line.

 

Then i differentiated, and set that equal to zero.

 

So that I could find the value of t, which minimized the length of any such position vector.

 

I got t=4/9 for the answer.

 

Then plugged that back into the formula for arbitrary position vectors, to get the unique vector with the shortest length.

 

Unless I made a mathematical error, which he has to check for, the answer in the back of his book appears wrong.

 

As for the dot product, I can now analyze how that could be used to answer the question.

 

Hopefully, i get the same answer both ways, then he will believe that the answer in the back of his book is wrong.

 

Here is what Atheist suggested

 

In question b) you are asked to find a point P on the line C(t) so that P-O is perpendicular to B-A. Two vectors are perpendicular if their scalar product vanishes so your condition for P=C(x) -I´ve renamed t to x here to highlight that you are looking for a definite value instead of having an arbitrary parameter- would be (C(x)-O)*(B-A) = 0. That´s one equation with only one unknown. As x only appears with a power of one, the solution is also unique.

Ah thanks for spotting the error

Its ok.

I don't fully understand the German.

Or have I went not kapiert?

Also, do you want the other questions answered?

Question b is really quite interesting. You want the vector with the minimum length that runs from the origin of the frame, to the straight line in question.

There is a unique answer to this question, for any randomly chosen line.

O is the origin of the frame, and has coordinates (0,0,0)

P is a point on the given line.

Here is a formula for all vectors which extend from the origin to points on the line:

R = <9t+1,15t-5,12t-7> = <x(t),y(t),z(t)>

The above is a position vector, which means that its tail is at the origin of the reference frame, and the tip lies on the line in question.

Using the pythagorean theorem the magnitude of R is given by:

|R| = [ (9t+1)² + (15t-5)² + (12t-7)² ]^1/2

Which implies that:

|R|² = [ (9t+1)² + (15t-5)² + (12t-7)² ]

Differentiating both sides with respect to t yields:

2|R| d R/dt = 2(9t+1)(9) + 2(15t-5)(15) + 2(12t-7)(12)

Minima at dR/dt=0

0=(18t+2)(9) + (30t-10)(15) + (24t-14)(12)

0=(162t+18) + (450t-150) + (288t-168)

0= (162+450+288)t +18-150-268

0= 900t +18-150-268

0= 900t -400

400= 900t

4=9t

t=4/9

R(t) = <9t+1,15t-5,12t-7>

R(4/9) = <9(4/9)+1,15(4/9)-5,12(4/9)-7>

R(4/9) = <5, 5/3, -5/3>

Here is the answer for part b:

5i + 5/3j - 5/3k

Regards