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Prove that three vectors are coplanar

linear dependence and coplanar vectorsHow do I find out of these vectors are coplanar?Proving that three vectors in $mathbb R^3$ are coplanar if one is a linear combination of the other twoCoplanar Vectors ProofVectors in a planeProof Vectors are coplanarIf $veca,vecb,vecc$ are three coplanar vectorsIf $veca,vecb,vecc$ are coplanar vectors, the prove required result.Condition for coplanar vectorsIs a vector making equal angles with three non-zero coplanar vectors perpendicular to that plane of three vectors?

4

1

$begingroup$

Three vectors are given: $u,v,w$. It is given that:
$|u|=|v|=|w|= sqrt2$; $ucdot v=ucdot w=vcdot w=-1$.
Prove that vectors $u,v,w$ are coplanar (on the same plane).

I have a few ideas, but I don't know if they are helpful in this case:

I know that three vectors are co-planar if $ucdot(v x w)=0$.
In addition, I assume that you can prove it with linear dependence, but I don't know how to use it here.

In addition, I thought that maybe the angle between the vectors can be of help- $120$ degrees between every $2$ vectors- but does that necessarily mean that they are on the same plain- co-planar?

linear-algebra vector-spaces vectors

share|cite|improve this question

edited Mar 28 at 21:14

J. W. Tanner

4,3651320

asked Mar 28 at 20:21

noam Azulaynoam Azulay

235

New contributor

noam Azulay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

4

1

$begingroup$

Three vectors are given: $u,v,w$. It is given that:
$|u|=|v|=|w|= sqrt2$; $ucdot v=ucdot w=vcdot w=-1$.
Prove that vectors $u,v,w$ are coplanar (on the same plane).

I have a few ideas, but I don't know if they are helpful in this case:

I know that three vectors are co-planar if $ucdot(v x w)=0$.
In addition, I assume that you can prove it with linear dependence, but I don't know how to use it here.

In addition, I thought that maybe the angle between the vectors can be of help- $120$ degrees between every $2$ vectors- but does that necessarily mean that they are on the same plain- co-planar?

linear-algebra vector-spaces vectors

share|cite|improve this question

edited Mar 28 at 21:14

J. W. Tanner

4,3651320

asked Mar 28 at 20:21

noam Azulaynoam Azulay

235

New contributor

noam Azulay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Three vectors are given: $u,v,w$. It is given that:
$|u|=|v|=|w|= sqrt2$; $ucdot v=ucdot w=vcdot w=-1$.
Prove that vectors $u,v,w$ are coplanar (on the same plane).

I have a few ideas, but I don't know if they are helpful in this case:

I know that three vectors are co-planar if $ucdot(v x w)=0$.
In addition, I assume that you can prove it with linear dependence, but I don't know how to use it here.

In addition, I thought that maybe the angle between the vectors can be of help- $120$ degrees between every $2$ vectors- but does that necessarily mean that they are on the same plain- co-planar?

linear-algebra vector-spaces vectors

share|cite|improve this question

edited Mar 28 at 21:14

J. W. Tanner

4,3651320

asked Mar 28 at 20:21

noam Azulaynoam Azulay

235

New contributor

noam Azulay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 28 at 21:14

J. W. Tanner

4,3651320

asked Mar 28 at 20:21

noam Azulaynoam Azulay

235

New contributor

noam Azulay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 28 at 21:14

J. W. Tanner

4,3651320

asked Mar 28 at 20:21

noam Azulaynoam Azulay

235

New contributor

noam Azulay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 28 at 21:14

J. W. Tanner

4,3651320

edited Mar 28 at 21:14

J. W. Tanner

4,3651320

asked Mar 28 at 20:21

noam Azulaynoam Azulay

235

New contributor

noam Azulay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 28 at 20:21

noam Azulaynoam Azulay

235

1 Answer
1

active

oldest

votes

6

$begingroup$

The sum of three angles between three pairs of non-coplanar angles is always less than 360 degrees. However instead of proving this general statement one can arrive at the result at once, observing that the vectors in your particular problem form an equilateral triangle. An easy check shows:

$$(u+v+w)cdot(u+v+w)=|u|^2+|v|^2+|w|^2+2ucdot v+2vcdot w+2wcdot u=0\
implies u+v+w=0.$$

As the three vectors are linearly dependent, they are coplanar.

share|cite|improve this answer

edited Mar 28 at 23:40

answered Mar 28 at 21:12

useruser

6,19211031

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did you think of adding the the vectors to one another in the first place? Why does "u+v+w=0" mean that they are co-planar? Did you write that they are linearly dependent because 1∙u+1∙v+1∙w=0, the coefficients are not all 0?
  $endgroup$
  – noam Azulay
  Mar 28 at 21:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice answer!
  $endgroup$
  – mjw
  Mar 28 at 22:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @noamAzulay It is not hard to notice that the vectors form equilateral triangle. The non-coplanar vectors are linearly independent and therefore cannot add to 0. And you are right $1ne0$ quite certainly.
  $endgroup$
  – user
  Mar 28 at 22:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did you start the proof with this multiplication (u+v+w)(u+v+w)? Is it because you knew in advance that your goal is to reach u+v+w=0 in order to prove this?
  $endgroup$
  – noam Azulay
  Mar 28 at 23:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @noamAzulay As I wrote in another comment, the equality $u+v+w=0$ is evident, as soon as one realizes that the angles between the vectors are 120 degrees. The same method can be applied in fact to any three vectors such that the angles between them satisfy $alpha+beta+gamma=2pi$ (or $alpha+beta-gamma=0$). The only complication will be in finding the correct coefficients. As soon as they are found one proceeds with the same demonstration $(c_1u+c_2v+c_3w)^2=0$.
  $endgroup$
  – user
  Mar 28 at 23:33
  
  

  
  
  
  

 | 
show 1 more comment

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6

$begingroup$

The sum of three angles between three pairs of non-coplanar angles is always less than 360 degrees. However instead of proving this general statement one can arrive at the result at once, observing that the vectors in your particular problem form an equilateral triangle. An easy check shows:

$$(u+v+w)cdot(u+v+w)=|u|^2+|v|^2+|w|^2+2ucdot v+2vcdot w+2wcdot u=0\
implies u+v+w=0.$$

As the three vectors are linearly dependent, they are coplanar.

share|cite|improve this answer

edited Mar 28 at 23:40

answered Mar 28 at 21:12

useruser

6,19211031

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did you think of adding the the vectors to one another in the first place? Why does "u+v+w=0" mean that they are co-planar? Did you write that they are linearly dependent because 1∙u+1∙v+1∙w=0, the coefficients are not all 0?
  $endgroup$
  – noam Azulay
  Mar 28 at 21:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice answer!
  $endgroup$
  – mjw
  Mar 28 at 22:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @noamAzulay It is not hard to notice that the vectors form equilateral triangle. The non-coplanar vectors are linearly independent and therefore cannot add to 0. And you are right $1ne0$ quite certainly.
  $endgroup$
  – user
  Mar 28 at 22:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did you start the proof with this multiplication (u+v+w)(u+v+w)? Is it because you knew in advance that your goal is to reach u+v+w=0 in order to prove this?
  $endgroup$
  – noam Azulay
  Mar 28 at 23:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @noamAzulay As I wrote in another comment, the equality $u+v+w=0$ is evident, as soon as one realizes that the angles between the vectors are 120 degrees. The same method can be applied in fact to any three vectors such that the angles between them satisfy $alpha+beta+gamma=2pi$ (or $alpha+beta-gamma=0$). The only complication will be in finding the correct coefficients. As soon as they are found one proceeds with the same demonstration $(c_1u+c_2v+c_3w)^2=0$.
  $endgroup$
  – user
  Mar 28 at 23:33
  
  

  
  
  
  

 | 
show 1 more comment

6

$begingroup$

The sum of three angles between three pairs of non-coplanar angles is always less than 360 degrees. However instead of proving this general statement one can arrive at the result at once, observing that the vectors in your particular problem form an equilateral triangle. An easy check shows:

$$(u+v+w)cdot(u+v+w)=|u|^2+|v|^2+|w|^2+2ucdot v+2vcdot w+2wcdot u=0\
implies u+v+w=0.$$

As the three vectors are linearly dependent, they are coplanar.

share|cite|improve this answer

edited Mar 28 at 23:40

answered Mar 28 at 21:12

useruser

6,19211031

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did you think of adding the the vectors to one another in the first place? Why does "u+v+w=0" mean that they are co-planar? Did you write that they are linearly dependent because 1∙u+1∙v+1∙w=0, the coefficients are not all 0?
  $endgroup$
  – noam Azulay
  Mar 28 at 21:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice answer!
  $endgroup$
  – mjw
  Mar 28 at 22:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @noamAzulay It is not hard to notice that the vectors form equilateral triangle. The non-coplanar vectors are linearly independent and therefore cannot add to 0. And you are right $1ne0$ quite certainly.
  $endgroup$
  – user
  Mar 28 at 22:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did you start the proof with this multiplication (u+v+w)(u+v+w)? Is it because you knew in advance that your goal is to reach u+v+w=0 in order to prove this?
  $endgroup$
  – noam Azulay
  Mar 28 at 23:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @noamAzulay As I wrote in another comment, the equality $u+v+w=0$ is evident, as soon as one realizes that the angles between the vectors are 120 degrees. The same method can be applied in fact to any three vectors such that the angles between them satisfy $alpha+beta+gamma=2pi$ (or $alpha+beta-gamma=0$). The only complication will be in finding the correct coefficients. As soon as they are found one proceeds with the same demonstration $(c_1u+c_2v+c_3w)^2=0$.
  $endgroup$
  – user
  Mar 28 at 23:33
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

The sum of three angles between three pairs of non-coplanar angles is always less than 360 degrees. However instead of proving this general statement one can arrive at the result at once, observing that the vectors in your particular problem form an equilateral triangle. An easy check shows:

$$(u+v+w)cdot(u+v+w)=|u|^2+|v|^2+|w|^2+2ucdot v+2vcdot w+2wcdot u=0\
implies u+v+w=0.$$

As the three vectors are linearly dependent, they are coplanar.

share|cite|improve this answer

edited Mar 28 at 23:40

answered Mar 28 at 21:12

useruser

6,19211031

$endgroup$

share|cite|improve this answer

edited Mar 28 at 23:40

answered Mar 28 at 21:12

useruser

6,19211031

answered Mar 28 at 21:12

useruser

6,19211031

answered Mar 28 at 21:12

useruser

6,19211031