mreditor16
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Following on from this, after obtaining the augmented matrix in REF it should suggest either a line or no solutions which is where you make your argument as to whether the last row suggests that the value is a parameter, which everything else is dependent on (hence giving a line) or suggesting that the last row is a contradiction so that there is no solutions --> parallel.I guess you could setup an augmented matrix [A|b] and get it to REF. Then examine at the last row and see what all the existing possibilities are for each entry. You will always find that one of the columns will be non-leading and hence one parameter will exist --> solution will be given by a line. If not, then b will be your leading column, implying no solution exists for the system and hence the planes won't intersect i.e. they will end up being parallel. NOTE that the last row will never be the zero row, since the two planes are distinct, otherwise you would be solving for intersections points of two identical planes which is redundant.
Then you probably don't need to use matrices, since it is just a system of two equations in three unknowns, so you could solve it as if you are just solving a set of simultaneous equations and use the given conditions to work from there, since you already know the parametric/Cartesian form of a line in R^3Thanks so much VBN, but this question was in a section before augmented matrices and the like... Like only the topic of "introduction to linear equations" had been introduced before this Q popped up. so then is there a method that doesn't use matrices (not learnt yet, in terms of this Q) etc?
Wait what lol? So what are you suggesting as a method?Then you probably don't need to use matrices, since it is just a system of two equations in three unknowns, so you could solve it as if you are just solving a set of simultaneous equations and use the given conditions to work from there, since you already know the parametric/Cartesian form of a line in R^3
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You can also do it without setting up an augmented matrix. Just solve for the variables and show the relationships as VBN said above.Thanks so much VBN, but this question was in a section before augmented matrices and the like... Like only the topic of "introduction to linear equations" had been introduced before this Q popped up. so then is there a method that doesn't use matrices (not learnt yet, in terms of this Q) etc?
Let the first plane be , and the second plane be , where A,B,C,D do not share a common factor (otherwise we would divide through by that), and similarly for E,F,G,H. Also, as the planes are distinct, E/A, F/B, G/C and H/D cannot all be equal.
Oh okay, thanks so so much guys.You can also do it without setting up an augmented matrix. Just solve for the variables and show the relationships as VBN said above.
Another method is to find the "points" of intersection through a relationship (which is kinda what I said above). The relationship you obtained will give all possible solutions sets for the "points" of intersection. Then try to deduce that the relationship suggests either a line or no points of intersection (hence parallel) as the only possibilities.
No points of intersection doesn't necessarily imply parallelism though. It is true in , but for planes in higher dimensions, it's not true in general, so this question seems to be requiring you to prove algebraically they are indeed parallel when there are no solutions in .Another method is to find the "points" of intersection through a relationship (which is kinda what I said above). The relationship you obtained will give all possible solutions sets for the "points" of intersection. Then try to deduce that the relationship suggests either a line or no points of intersection (hence parallel) as the only possibilities.
Question did ask for inNo points of intersection doesn't necessarily imply parallelism though. It is true in , but for planes in higher dimensions, it's not true in general, so this question seems to be requiring you to prove algebraically they are indeed parallel when there are no solutions.
I know, I just don't know whether they allow you to assume that in , no solutions imply parallelism, because otherwise, all they're asking you to prove for the "no solutions" part is that it's possible to have no solutions, which is surely obvious?Question did ask for in
There are numerous ways to do this question. One could even do this using a dot product (too tedious however).
Your solution above however is a lot simpler, though it isn't as obvious.
@ mreditor16, this solution is probably the best one to follow and is most likely how they want you to deliver the proof without further knowledge.Let the first plane be , and the second plane be , where A,B,C,D do not share a common factor (otherwise we would divide through by that), and similarly for E,F,G,H. Also, as the planes are distinct, E/A, F/B, G/C and H/D cannot all be equal.
Multiply equation of by E, and that of by A, to get: and .
For intersection points, we have these two equations as simultaneous equations. So subtract the two equations to get:
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No solutions exist (i.e. planes do not intersect) iff the following is true: AND AND . Because otherwise, we would have an LHS which has a non-zero coefficient, in which case would could always solve for y and/or z. In this case, BE = AF, CE = AG, AE = AE (obviously), so the normal to the two distinct planes point in the same direction, so the planes are parallel.
Now, otherwise, we will have NOT all of the following true: AND AND . Then we can always solve for z or y and we get a line (I have to go for a bit, I'll explain further if this isn't resolved).
tbh I don't get his solution after the third paragraph :/@ mreditor16, this solution is probably the best one to follow and is most likely how they want you to deliver the proof without further knowledge.
Once you got to the (4) - (3) line, you could say this: