pretty vectorial

[kingjewel1]

Hi, i'm having problems with this question and would appreciate it very much if someone would help me out. I've got my C4 exam in a month and none of my teachers has a clue about vectors...

 

Referred to the origin we have 2 points points A (i-5j-7k) and B (10i+10j+5k). P is a point on AB

a) Find a vectr for the line passing through AB(i got i-5j-7k+t(9i-15j-12k))

b)Find the position vector of P such that OP is perpendicular to AB.

I really don't know how to do b).

c) find area of triangle OAB ( i can do when i have b)

d) find ratio in which P divides AB (not sure)

 

Can anyone tell me where i can get a good guide to vectors (book or internet, etc)? It's my weakest topic

 

Cheers guys,

Phi

[Johnny5]

Hi' date=' i'm having problems with this question and would appreciate it very much if someone would help me out. I've got my C4 exam in a month and none of my teachers has a clue about vectors...

 

Referred to the origin we have 2 points points A (i-5j-7k) and B (10i+10j+5k). P is a point on AB

a) Find a vectr for the line passing through AB

 

 

[/quote']

 

A=<1,-5,-7>

B=<10,10,5>

 

Point A has coordinates (1,-5,-7)

Point B has coordinates (10,10,5)

 

As I recall, there is a formula with parameter t in it, which gives you the equation of the line.

 

I think the dot product is used to get the formula. Let me see if i can find it on the web.

 

Here is a link to the exact thing i was thinking of:

 

vector formula for a line

 

In the link above you will see the following quote:

 

In order to write down the vector equation of this line, we need to know two things.

We have to know the position vector of some point which lies on the line, like a on my diagram.

We have to know a vector which gives the direction of the line, like b in my diagram. This is called a direction vector.

 

You certainly do know the position vector of a point which lies on the line. In fact you know two:

 

A=<1,-5,-7>

B=<10,10,5>

 

The other thing you need to know is a vector which gives the direction of the line.

 

So now, i have to think. I need to take the given information and come up with the direction.

 

Done.

 

You have two points, and you can construct a vector from one point to the other by definition. The resulting vector will lie on the line.

 

 

Here are your two points:

 

Point A = (1,-5,-7)

Point B = (10,10,5)

 

The vector from point A to point B is given by:

 

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

 

So this may actually be all you were asked to find in part a.

 

Nonetheless, we can continue on, and find the formula for the whole line.

 

The position vector of any point on the line, in this frame, is given by:

 

R = a + tb (notation at the link i posted)

 

Look carefully at their diagram.

 

 

For their a, you are going to use:

 

A= <1,-5,-7>

 

And you already have a vector which lies on the line namely:

 

AB = 9i + 15j + 12 k

 

So you can use this for their b.

 

You could first turn the vector above into a unit vector, but that's not necessary because t is just a scaling factor.

 

 

So you are now ready to write the answer.

 

There is a line, in some reference frame, its not moving in the frame.

 

The position vector of any point on this straight line is given by:

 

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

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

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

[timo]

@Kingjewel: 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).

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.

In case you are not familiar with the scalar product (you defenitely have to look it up if you´re going to take an exam in vector math), it´s (a, b, c)*(d, e, f) = ad + be + cf. At least in your case and in probably all you´ll encounter for the next few years.

As you said you can do c), I guess I don´t have to say much about it.

 

Don´t have any guides on vector math, sry. But if you´ve got special questions I´m alsmost certain than most of the people in this forum will be able and willing to help.

Question d) is rather easy if you used the parameterization for the line that I used above. If you didn´t realize why this parameteritzation is so nice, then try to plug in t=0, t=1 and perhaps even t=0.5 and look what points you get for C(t).

 

btw.: We had your avatar hanging on our door as a poster in the previous flat I lived in (it was a physicists commune)

[Johnny5]

Your answer to a) is wrong.

 

 

 

No its not.

 

Your answer and the one I gave should be equivalent.

 

Let me check.

 

Here is what you wrote' date=' which is probably equivalent:

The solution 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). Guess that´s also what you

[timo]

No need to check, i referred to the thread starter

[Johnny5]

No need to check, i referred to the thread starter

 

Oh you were talking to the thread starter.

 

Geeze...

 

Hold on

 

Kingjewel: i got i - 5 j- 7k + t(9i-15j-12k)

Kingjewel: <1+9t,-15t-5,-12t-7>

 

Now I got...

 

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

 

So these answers are not equivalent. Therefore one of us has made an error.

 

I don't remember making one. Nonetheless, i will go back and double check.

 

 

Given by thread starter:

A (i-5j-7k) and B (10i+10j+5k). P is a point on AB

 

A,B are position vectors. They go from the origin of the frame (0,0,0) to a field point (x,y,z).

 

P1=(1,-5,-7)

P2=(10,10,5)

vector from P1, to P2 = <10-1,10--5,5--7>= <9,15,12>

 

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

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

[Dave]

Yes, they're both equivalent. I got the same answer from doing it quickly in my head

[timo]

Referred to the origin we have 2 points points A (i-5j-7k) and B (10i+10j+5k). P is a point on AB

a) Find a vectr for the line passing through AB(i got i-5j-7k+t(9i-15j-12k))

Ok, let´s try it the slow way for old, slow-minded people like me. The argument behind the t is certainly going to be a multiple of A-B. As the 1st entry is a 9 this fixes the multiple to -1, or simply B-A. Hence, we get B-A = (10, 10, 5)-(1, -5, -7) = (10-1, 10-(-5), 5-(-7)) = (9, 15, 12) != answer above.

Man, you got me confused people.

[kingjewel1]

Thank you both very much for your help.

although i am familiar with the dot product (which is what i tried to apply), i fail to see how it yields an appropriate answer. ie. (B-A)*P=0 since

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

thus giving t=-1 for i, and other answers for j and k. namely 1/3 and 7/12.

Oder habe ich etwas nicht kapiert?

Thanks

[kingjewel1]

Ah thanks for spotting the error

[Johnny5]

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

[kingjewel1]

So now that we have t=-1 as you said, i should now substitute it into the r equation

i-5j-7k-(9i+15j+12k)

=-8i-20j-19k ?

4i-3k is the right answer according to the back of the book

[timo]

Dunno where you got the -1 from, the only -1 I mentioned was in conjunction with clearing up the mess with your direction vector B-A that Johnny and Dave started (ok, I also had a little part in it). It didn´t really have much to do with your problem so it´s best to completely ignore my 2nd and 3rd post now that you´ve got the correct direction vector (you can also PM me if the english is a problem - I´ll be awake for about one hour from now).

Part b) is simply about finding a point on line C so that the vector from the origin to this point (C-O) is perpendicular to the direction of the line (B-A), so (C(x)-O).(B-A)=0. Solve this equation for x and you have the point. Perhaps the answer will be x= -1 but I don´t know that without actually doing the calculation. Post yours if you´re stuck.

 

EDIT: Johnny, 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.

[Johnny5]

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.

[timo]

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 ...

[Johnny5]

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.

[timo]

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

[Johnny5]

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.

[timo]

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

[Johnny5]

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.

[kingjewel1]

Thank you!

I didn't know that it was possible to add together the different x,y,z terms of the column matrix:s

ie (9).(1+9t)=0

(15).(-5+15t)=0

(12).(-7+12t)=0

i had previously worked out t for each row, hence getting the wrong answer.

so we added ALL like terms and worked out the t.

we then substitute it into the initial equation for AB and get 4i+0j-3k

 

Thank you guys!!!

I didn't realise it was actually an intersection problem.

[timo]

Oh, so it was good that Johnny worked out the numbers. But for clearification: There is no matrix in there at all. Your three equations

(9).(1+9t)=0,

(15).(-5+15t)=0,

(12).(-7+12t)=0

are plainly wrong. It´s really just one.

 

If you take a close look at the scalar product in my 1st post, you´ll notice that it´s only the equation

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

[kingjewel1]

Thank you for clarifying that Atheist.

For the area of the triangle

i can use two methods?

a) cross product of modulus AOxOB

b)cosine rule

 

Which one would you prefer to use?

[timo]

Spntaneously, I´d use the cross product but that does not nessecarily say you have to do so. Perhaps try both methods and see which you like more. You´re preparing for an exam, after all.

[kingjewel1]

Okay. I used 1/2|AxB|=

det u=|i, j, k|

|1,- 5,- 7|

|10, 10, 5|

 

1/2sqrt(45^2+75^2+60^2)=53.0 u^2 correct

 

or A.B=|A||B|COSf

10-50-35=(sqrt75)(sqrt225)COSf

f=54.77

1/2sqrt75.15.sinf

=53.03 correct

 

I like the cross product, it seems much cleaner and more efficient