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Coplanarity of vectors

GreenGoblin

Member
Feb 22, 2012
68
(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 cant 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 havent 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 doesnt '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/
 
Last edited:

Amer

Active member
Mar 1, 2012
275
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

Member
Feb 22, 2012
68
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

Well-known member
Jan 26, 2012
890
(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

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
(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 cant 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

Member
Feb 22, 2012
68
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

Well-known member
Jan 26, 2012
890
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
 
Last edited:

Amer

Active member
Mar 1, 2012
275

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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

Member
Feb 22, 2012
68
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 dont 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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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

Member
Feb 22, 2012
68
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

Well-known member
Jan 26, 2012
890
how relevant is degeneracy? as far as i am aware it has not been covered in my course and i dont 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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
yes, it is still a valid solution.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.