What is the Correct Method to Determine the Coplanarity of Vectors?

[GreenGoblin]

(1,-1,a)
(1,2,2a)
(3,-a,9)

are three vectors. We have to find 'a' that makes them coplanar

The strategy suggested to me is that this is done by... choosing two of the vectors and taking a crossproduct, then (IT SHOULDNT MATTER WHICH TWO SINCE THEY DONT REALLY COME IN A SET ORDER) dottying with the third. This should give... ZERO as the outcome, right? If this is the wrong strategy please tell me because I have done the question wrong.

I get 3 and 9 as my values of a as the triple product I make a^2 - 12a + 27. This is a question where I am fine with the calulcations but I can't find any material that says, yes, this is definitely the correct method. Even wikipedia confuses me more. But am I using the right idea? IS my answer ok can someone say? Please?

ALSO NEXT QUESTiON (not related to these same vectors this is general)

show that:
(i) if a.c = b.c for all c, then a = b
(ii) if a x (b x c) = (a x b) x c ,then (a x c) x b = 0

for (i) i haven't a clue because all i see is something obvious that already 'shows' itself. what can you do to show it apart from rearrange but then that is still the same.

for (ii) i am thinking of triple scalar expansion and showing equality but then that doesn't 'show' anything more than the orignal statement does...? does it/

a x (b x c) = (a.c)b - (a.b)c
(a x b) x c = -(c.b)a + (c.a)b

equality implies (a.b)c = (c.b)a
but what to do from there/

 

[Amer]

take any two vectors find the cross product the resulting vector would be perpendicular to the plane containing these two vectors
then find the dot product of the result of the cross with the remaining vector this should be zero since the dot product of any two perpendicular vectors equal zero

 

[GreenGoblin]

take any two vectors find the cross product the resulting vector would be perpendicular to the plane containing these two vectors
then find the dot product of the result of the cross with the remaining vector this should be zero since the dot product of any two perpendicular vectors equal zero

am i correct

 

[CaptainBlack]

(1,-1,a)
(1,2,2a)
(3,-a,9)

are three vectors. We have to find 'a' that makes them coplanar

If these are coplanar then the third is a linear combination of the first two.

CB

 

[Opalg]

(1,-1,a)
(1,2,2a)
(3,-a,9)

are three vectors. We have to find 'a' that makes them coplanar

The strategy suggested to me is that this is done by... choosing two of the vectors and taking a crossproduct, then (IT SHOULDNT MATTER WHICH TWO SINCE THEY DONT REALLY COME IN A SET ORDER) dottying with the third. This should give... ZERO as the outcome, right? If this is the wrong strategy please tell me because I have done the question wrong.

I get 3 and 9 as my values of a as the triple product I make a^2 - 12a + 27. This is a question where I am fine with the calulcations but I can't find any material that says, yes, this is definitely the correct method. Even wikipedia confuses me more. But am I using the right idea? IS my answer ok can someone say? Please?

Your method and answer are both correct.

ALSO NEXT QUESTiON (not related to these same vectors this is general)

show that:
(i) if a.c = b.c for all c, then a = b

You know that (a–b).c = 0 for all c, therefore in particular when c = a–b ... .

 

[GreenGoblin]

Your method and answer are both correct.You know that (a–b).c = 0 for all c, therefore in particular when c = a–b ... .

i think i see...
we are considering a counter example, if a =/= b then for the case c= a-b, then (a-b)(a-b) = 0 but, a-b =/= 0 so this is not possible since a nonzero vector dottyed with itself... cannot give 0! (Coffee)

 

[CaptainBlack]

If these are coplanar then the third is a linear combination of the first two.

CB

These give us the three equations:

\( \begin{array}{ccccc} \lambda & + & \mu & = & 3 \\ -\lambda&+&2 \mu & = & -a \\ \lambda a&+&2 \mu a & = & 9 \end{array} \)

Adding the first and second gives: \(\mu=1-\frac{a}{3} \),

Substituting into the first gives: \(\lambda =2+\frac{a}{3} \)

then substituting these into the third gives: \(a^2-12 a+27=0 \), so \(a=3\) or \(9\)

The \(a=3\) solution is degenerate (if gives \(\mu=0\) ) and the third vector is a multiple of the first (but they are trivially coplanar), which leaves \( a=9 \) as non-degenerate

CB

 

[Amer]

am i correct

yea you are correct 3,9 a values

 

[HallsofIvy]

In fact, the dot product of vector u with the cross product of vectors v and w is simply a determinant:
If u= <a, b, c>, v= <d, e, f>, and w= < g, h, i> then
$$ w \cdot (v\times w)= \left|\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i\end{array}\right| $$

so here you want to find a so that
$$ \left|\begin{array}{ccc}1 & -1 & a \\ 1 & 2 & 2a \\ 3 & -a & 9 \end{array}\right|= 0 $$

 

[GreenGoblin]

These give us the three equations:

\( \begin{array}{ccccc} \lambda & + & \mu & = & 3 \\ -\lambda&+&2 \mu & = & -a \\ \lambda a&+&2 \mu a & = & 9 \end{array} \)

Adding the first and second gives: \(\mu=1-\frac{a}{3} \),

Substituting into the first gives: \(\lambda =2+\frac{a}{3} \)

then substituting these into the third gives: \(a^2-12 a+27=0 \), so \(a=3\) or \(9\)

The \(a=3\) solution is degenerate (if gives \(\mu=0\) ) and the third vector is a multiple of the first (but they are trivially coplanar), which leaves \( a=9 \) as non-degenerate

CB

how relevant is degeneracy? as far as i am aware it has not been covered in my course and i don't thinkis needed for my purpose but thanks for the details. does this make a=3 invalid? degenerate sounds like a pejorative term i am guessing it is not good in other contexts it is not good... but then again callimg someone irrational is not usually good and irational numberrs are fine

 

[Deveno]

in this case "degenerate" means: "not fully general". for example, two lines (in a plane) don't always intersect in a point, because they might be parallel. this is a "degenerate" (special) case. the reason why a = 3 is degenerate, is it makes two of the vectors collinear (namely, (1,-1,3) and (3,-3,9)), so that all 3 vectors cannot help but be co-planar.

often, a "degenerate" case is one where "something" is 0: for example a point is a degenerate circle, because it has 0 radius.

see: http://en.wikipedia.org/wiki/Degeneracy_(mathematics)

 

[GreenGoblin]

in this case "degenerate" means: "not fully general". for example, two lines (in a plane) don't always intersect in a point, because they might be parallel. this is a "degenerate" (special) case. the reason why a = 3 is degenerate, is it makes two of the vectors collinear (namely, (1,-1,3) and (3,-3,9)), so that all 3 vectors cannot help but be co-planar.

often, a "degenerate" case is one where "something" is 0: for example a point is a degenerate circle, because it has 0 radius.

see: http://en.wikipedia.org/wiki/Degeneracy_(mathematics)

ok, so, is it still a valid solution?

 

[CaptainBlack]

how relevant is degeneracy? as far as i am aware it has not been covered in my course and i don't thinkis needed for my purpose but thanks for the details. does this make a=3 invalid? degenerate sounds like a pejorative term i am guessing it is not good in other contexts it is not good... but then again callimg someone irrational is not usually good and irational numberrs are fine

In this case degeneracy implies one of the vectors is a multiple of the other, so effectively you have only two vectors, which are trivially coplanar since two vectors define a plane.

CB

 

[Deveno]

yes, it is still a valid solution.

 

[HallsofIvy]

Any two vectors determine two lines through the origin which will then define a single plane. The "generic" case is that a third vector will then give a set that will span three dimensions. The case where the three vectors span only two dimensions or one are "degenerate". Note that if all three vectors span a single line (so any vector of the three is a multiple of one other vector) that line lies in one plan so they still lie in a single plane.