- Thread starter
- #1

#### 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/

(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/

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