Math Is Fun Forum

!nval!d_us3rnam3

Member

Registered: 2017-03-18

Posts: 46

The Penultimate Vector Problem

Let $A, B$ and $C$ be points in $3$-space:
[asy]
size(200);
import three;

currentprojection = orthographic(3, 1, 0.5);
real big = 6;
triple O = (0,0,0);
triple X = (1, 0, 0);
triple Y = (0, 1, 0);
triple Z = (0, 0, 1);
triple A = (0, 2, -1);
triple B = (2, 0, 3);
triple C = (3, -1, 5);

draw(O--big*X, Arrow3(size = 0.2cm));
draw(O-- -big*X, Arrow3(size = 0.2cm));
draw(O-- big*Y, Arrow3(size = 0.2cm));
draw(O-- -big*Y, Arrow3(size = 0.2cm));
draw(O-- big*Z, Arrow3(size = 0.2cm));
draw(O-- -big*Z, Arrow3(size = 0.2cm));

label("$x$", big*X, S);
label("$y$", big*Y, S);
label("$z$", big*Z, W);

dot("$A$", A, S);
dot("$B$", B, S);
dot("$C$", C, S);
[/asy]Prove that if $A, B$ and $C$ are collinear, then
\[\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}.\]

I'm stuck, can anyone explain how to solve this?

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"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: The Penultimate Vector Problem

hi !nval!d_us3rnam3

Not able to follow your diagram instructions but it looks like this is just

" Prove that A = (0, 2, -1);  B = (2, 0, 3);  C = (3, -1, 5);   are collinear. "

Sufficient to show vector AB = a multiple of vector BC as these go in the same direction and have a point in common. 

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

!nval!d_us3rnam3

Member

Registered: 2017-03-18

Posts: 46

Re: The Penultimate Vector Problem

No,  that's not really the question.
I'm trying to prove that for some collinear points A, B, and C,

Last edited by !nval!d_us3rnam3 (2019-07-03 11:29:47)

---

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: The Penultimate Vector Problem

hi !nval!d_us3rnam3

That's much clearer.  I wonder why your first post just looks like code to me.  When I want maths expressions I use LaTex.  Mostly that works ok.  You have to start your code with square brackets math and end with square brackets /math.  Does your first post look ok to you?  Maybe you have some software on your machine that displays it for you.  I'd like to try and get to the bottom of this for future posters.

Anyway, back to your question.  Those 'x' signs must be cross products (also called vector products).  You can read about them here:

https://www.mathsisfun.com/algebra/vect … oduct.html

and there's a calculator here:

https://www.symbolab.com/solver/vector- … pmatrix%7D

When you use that three times for the three products you get:

(6i -2j -4k)  (3i - j -2k) and (-9i +3j +6k)

These add up to (0, 0, 0) as required.

Hope that's what you wanted.

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

!nval!d_us3rnam3

Member

Registered: 2017-03-18

Posts: 46

Re: The Penultimate Vector Problem

Not really, actually. I'm looking for a proof of that for any collinear points A, B, and C. If you could show me that, that would be great.

Last edited by !nval!d_us3rnam3 (2019-07-04 13:27:45)

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"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: The Penultimate Vector Problem

Ok, sorry for the misunderstanding.  I can do that too.

To save time with typing I'm going to leave out the overlines on the vectors.

If A, B and C are colinear then AB = kBC => AO + OB = k(BO + OC) which re-arranges to OA = kOB + OB -kOC

So AxB + BxC + CxA becomes

(kOB + OB -kOC) x OB + OB x OC + OC x (kOB + OB - kOC)  = -k (OC x OB) + (OB x OC) + k x (OC x OB) + (OC x OB) = 0

Hope that is finally what you want.

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

!nval!d_us3rnam3

Member

Registered: 2017-03-18

Posts: 46

Re: The Penultimate Vector Problem

Yes, it is, thank you.
Coud you explain your last line a bit further? How did you get there?

---

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: The Penultimate Vector Problem

I have made use of the following:

i x i = j x j = k x k = 0

i x j = k     j x i = -k and four more like these.

CO = - OC etc.

I've just realised that using k for the multiplier was a poor use of symbol.  Sorry if that is confusing.  Fortunately I haven't actually used k as a unit vector in my proof, so all the ks are multipliers.

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