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**jonny****Guest**

The vector (-1,1,1) bisects the angle between the vectors C(x,y,z) and (3,4,0). determine a unit vector along C(x,y,z).

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,462

hi jonny

You could do this using the dot poduct with vectors

Or use Pythagoras as CD = DB and OC = OB = 5k.

Either way you will only have two equations with three unknowns.

But don't worry because when you find the 'unit vector' you can eliminate the last unknown.

Bob

*Last edited by bob bundy (2011-12-22 00:26:02)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Americanlooser****Guest**

omg! looks like math isn't fun after all!!!

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Americanlooser;

It is loads of fun. I laugh all day and night over it. I am laughing now. Welcome to the forum.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,383

bob bundy wrote:

hi jonny

You could do this using the dot poduct with vectors

I tried to continue this and got stuck at

. Can anyone tell me the next steps to the answer?Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,462

hi Monox D. I-Fly

It looks like my younger self did not actually try this method. Often a poster puts a question and that's the last we hear from them so I don't usually spend too much time on such, until I get into a dialogue and know it's worth the time investment.

I've tried my method and got stuck at the same point you did. Hhmmm! Looks like I need a new approach. This morning I've filled 6 sides of A4 with algebra and I still haven't got anywhere. But I'll continue to try. Back when I have a proper answer.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,462

OK. I have finally found an answer. I'll give you the steps so you can complete it yourself as I know you like to do this.

Firstly a small change to the diagram. Make A (3, 4, 0) D (-k, k, k) and C (x, y, z)

With OCD and OAD congruent right angled triangles where angle OAD = OCD = 90.

(1) Note that OA = OC = 5. So x^2 + y^2 + z^2 = 25 ............................... equation 1.

(2) Use the dot product, AD.OA = 0, to get an equation for k and solve.

(3) Use the dot product, CD.OC to get an equation involving x, y and z ......... equation 2.

(4) O, A, C and D must lie in the same plane so OC can be expressed as a linear combination of OA and AD.

Form a set of simultaneous equations and use a pair to make lambda the subject and then another pair to make lambda the subject again.

Equate these to form another equation connecting x, y, and z ..................... equation 3.

(5) Use equations 2 and 3 to eliminate x and express z in terms of y.

(6) Use equations 2 and 3 to eliminate z and express x in terms of y.

(7) Substitute for x and z in equation 1 so you form a quadratic in y.

This will factorise with (y-4) as one factor as we know A satisfies it. The other factor will give y at C.

(8) Use the equations from steps (5) and (6) to get x and z as well.

(9) Now you know C find a unit vector in the direction of C.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,383

bob bundy wrote:

I'll give you the steps so you can complete it yourself as I know you like to do this.

How do you assume so?

bob bundy wrote:

(2) Use the dot product, AD.OA = 0, to get an equation for k and solve.

I got k = 0. Is that correct?

bob bundy wrote:

(3) Use the dot product, CD.OC to get an equation involving x, y and z ......... equation 2.

Assuming that k = 0 is correct, I got

bob bundy wrote:

(4) O, A, C and D must lie in the same plane so OC can be expressed as a linear combination of OA and AD.

Form a set of simultaneous equations and use a pair to make lambda the subject and then another pair to make lambda the subject again.

How? I don't have any idea.

*Last edited by Monox D. I-Fly (2018-03-02 13:19:42)*

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,462

hi Monox D. I-Fly

I'll give you the steps so you can complete it yourself as I know you like to do this. How do you assume so?.

I've noticed you have been reviewing old posts and attempting the questions. As these are not your homework, I assumed you liked to do this.

(2) Use the dot product, AD.OA = 0, to get an equation for k and solve. I got k = 0. Is that correct? .

As k gives the coordinates of point D, this must be incorrect.

Here's my working (I'm using row rather than column vectors as it saves time) :

(3) Use the dot product, CD.OC to get an equation involving x, y and z ......... equation 2 Assuming that k = 0 is correct, I got

(4) O, A, C and D must lie in the same plane so OC can be expressed as a linear combination of OA and AD.

Form a set of simultaneous equations and use a pair to make lambda the subject and then another pair to make lambda the subject again. How? I don't have any idea.

2 equations are not enough. I needed a third constraint. The first (x^2 + y^2 + z^2 = 25) puts C on the surface of a sphere radius 5. The second puts C on a circle centre on D. I had to think what other geometrical property would help.

The plane defined by O, A and D must also contain C. In vector algebra a point in a plane can be fixed by creating a linear combination of two vectors in the plane that are not parallel. OA and OD are two such vectors. So use lambda and mu as parameters for the combination and we have :

Each of the x, y and z coordinates must be equal so :

........................(4)........................(5)

........................(6)

Eliminating mu from (4) and (5) gives x + y = 7lambda

and eliminating mu from (5) and (6) gives y - z = 4lambda

Now eliminating lambda from these gives

4x - 3y + 7z = 0 which is equation 3.

You can eliminate z from (2) and (3) to get x in terms of y and

you can eliminate x from (2) and (3) to get z in terms of y.

Substituting these into (1) leads to a quadratic in y.

These values of y give points that are on the sphere, on the circle and in the plane. Both A and C satisfy these constraints and I know that y = 4 at A.

So the quadratic must factorise as (y-4)(y .....) so it's fairly easy to get the y we need for C. Then you can easily work out x and z.

So now we know OC. But it's unlikely to be a unit vector. Work out its length and divide each component by this amount to create the unit vector. I haven't yet bothered to complete this step as it's 'textbook'.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 1,383

bob bundy wrote:

How did you get -x + y + z from that?

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,462

hi Monox D. I-Fly

I chose C so that OA = OC and we know OA = √(3^2 + 4^2 + 0^2) so x^2 + y^2 + z^2 = 25

The result follows directly from that.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 342
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… to be continued.

*Last edited by Alg Num Theory (2018-03-05 11:56:27)*

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**Alg Num Theory****Member**- Registered: 2017-11-24
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Okay, let me try something else. I will assume OC to be in the plane formed by OA and OD and let (*r*,*s*,*t*) be a nonzero vector perpendicular to this plane. We have

which leads, on eliminating *s* and *t*, to

Now we can’t have *r* = 0, otherwise *s* = *t* = 0 as well. Therefore:

Now substitute for *z* from my previous post and we can pin down *x* and *y* a bit more.

*Last edited by Alg Num Theory (2018-03-05 11:47:59)*

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**Alg Num Theory****Member**- Registered: 2017-11-24
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Ah, ah, ah. We want a *unit* vector, so we can straightaway take √(*x*²+*y*²+*z*²) = 1 – no need to mess around with ugly expressions! Why didn’t I think of that before?

This means my equation [1] in post #12 becomes

Also from post #12,

From the previous post,

Now we can solve for everything!

*Last edited by Alg Num Theory (2018-03-05 11:56:53)*

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**Alg Num Theory****Member**- Registered: 2017-11-24
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And I get …

… tada! Is it right? Is it? Yes, I’m very sure it is.

Let me check … first:

so it is indeed a unit vector. Now

s0 DÔA = CÔD. And … drum roll …

– so CÔA = 2·DÔA = 2·CÔD!

YES!!

*Last edited by Alg Num Theory (2018-03-05 12:47:20)*

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