HL maths question help!!!

[alexalexalex]

Hey guys, could anyone help me with this question because right now I really really dont get it and I hate it when I dont understand stuff so I thought I could ask here hahah

 

tysm!!!

[Vioh]

I can think of 2 ways to solve this problem. The first is the determinant method (which can be justified by Cramer's theorem). This is not part of the syllabus (as far as I know), but it is super easy. Note that IB doesn't punish people for using methods that are outside of the syllabus. The second way is to use the standard row-reduction algorithm (or Gauss-Jordan elimination), which is much more complicated than the first method. I'm not sure this is part of the syllabus, but my IB maths teacher taught us that before the exam, so I assume that it is necessary to know it. If you are not familiar with either of the techniques, it's good to look at Khan Academy for more details.

[kw0573]

@Vioh Although I cannot locate the exact question, I have seen this on one of the paper 1 exams, so can't use rref function on calculator , on or after 2014, or in the specimen, but I forgot where.
@alexalexalexI proposed a third way. Here is some logic as to the nature of planes' normals of a system not having exactly 1 root. 

1) Find direction of line intersection of two known planes. 2) Find k such that the line of intersection is perpendicular to the normal of the unknown plane.
Line intersection has direction vector perpendicular to both normals of 2 planes the line is on. direction vector = (0, -1, 4) × (3, 4, 2) = (-18, 12, 3), which can be simplified to (-6, 4, 1)

From the link above, I briefly gone over about the triple scalar product, (n₁ × n₂) • n₃, which return 0 if the planes either have infinite or no solutions; in all other cases there is always a unique point of intersection. This holds regardless of the constants (D in Ax + By + Cz = D) in the equations of planes. I would like to point out that the determinant of a 3x3 matrix is incidentally a calculation of triple scalar product. 

So take (k, 1, 2) • (-6, 4, 1) = 0. -6k + 4 + 2 = 0, k = 1. [Original question is solved here. The following is an extension]

The most complicated question I have seen, from the link above, is the constant (right hand side) of plane 1 is also unknown, along with a parameter of the normal. 

For example to find line intersection of the 2 known planes in your example, we need a point on the line intersection of 2 planes. Often a point can be obvious, but if not just assume a value for one of the variables and solve a system of 2 equations, either way (1, -1, 1) in this case is an easy point to work with. 

Since we know that there won't be exactly 1 point of intersection, the entire line intersection of just the 2 given planes is either entirely on or not on the plane we solved for. That is, if (1, -1, 1) is on the plane, infinite (a line of) solutions; otherwise no solution.

1(1) + 1(-1) + 2(1) = 2 ≠ 4. Hence no solution. So as I was saying, if the question gave instead kx + y + 2z = m, where k and m were real numbers, then at this point you could say if m = 2 for infinite solutions, and m ≠ 2 for no solution.

When m = 2, the solutions to the system of 3 equations are described by r = (1, -1, 1) + t(-6, 4, 1). Infinite solutions almost always mean a line of solution, except for the trivial case of 3 identical planes.

Summary: normals of planes determine if there is one unique solution or not. The constants of the planes determine whether "not" means infinite or no solution. 

Edited October 16, 2016 by kw0573

[Vioh]

@kw0573 Using 3D vectors like what you've done is definitely a good way to solve the problem. In fact, it might be easier to understand compared to the 2 methods that I proposed since matrices have unfortunately been taken out of the IB syllabus. However, I would urge HL students to learn how to perform row-reduction algorithm because it's much less error-prone as we don't have to visualize the 3D space in our head. Besides, the row reduction algorithm can be applied to any numbers of equations/variables (though it's often the case that the IB will only give students systems with 3 variables). Calculator use is actually not necessary here. With a bit of practice, row reduction algorithm can be done efficiently by hands.

[kw0573]

@Vioh Since CAS calculators are not allowed on the exam, a calculator won't help with either determinant or row reduction. I was just trying to show the geometric meaning behind using a matrix determinant. On the exam hopefully one can just remember the formulae without having to derive them. Ultimately triple scalar product or determinant are using a geometric interpretation and row reduction is a pure algebraic interpretation. Students can choose the ways they are most comfortable with. 

One argument I would make to use geometric interpretation is that we only need the normals and it is closer to syllabus. The algebraic method (row reduction or otherwise) would need also the equation constants, and rewriting the the matrix multiple times. One benefit of the row reduction is that by having the last row [0 0 2-2k | 4]. Set 2-2k = 0, k = 1 would yield the impossible equation 0x + 0y + 0z = 4, characteristic of systems of no solutions. So doing row reduction alone can distinguish among no and infinite solutions (the latter case being 0x + 0y + 0z = 0), of which the geometric method would need an additional point to find out. 

[alexalexalex]

16 hours ago, kw0573 said:

@Vioh Since CAS calculators are not allowed on the exam, a calculator won't help with either determinant or row reduction. I was just trying to show the geometric meaning behind using a matrix determinant. On the exam hopefully one can just remember the formulae without having to derive them. Ultimately triple scalar product or determinant are using a geometric interpretation and row reduction is a pure algebraic interpretation. Students can choose the ways they are most comfortable with. 

One argument I would make to use geometric interpretation is that we only need the normals and it is closer to syllabus. The algebraic method (row reduction or otherwise) would need also the equation constants, and rewriting the the matrix multiple times. One benefit of the row reduction is that by having the last row [0 0 2-2k |

16 hours ago, Vioh said:

@kw0573 Using 3D vectors like what you've done is definitely a good way to solve the problem. In fact, it might be easier to understand compared to the 2 methods that I proposed since matrices have unfortunately been taken out of the IB syllabus. However, I would urge HL students to learn how to perform row-reduction algorithm because it's much less error-prone as we don't have to visualize the 3D space in our head. Besides, the row reduction algorithm can be applied to any numbers of equations/variables (though it's often the case that the IB will only give students systems with 3 variables). Calculator use is actually not necessary here. With a bit of practice, row reduction algorithm can be done efficiently by hands.

4]. Set 2-2k = 0, k = 1 would yield the impossible equation 0x + 0y + 0z = 4, characteristic of systems of no solutions. So doing row reduction alone can distinguish among no and infinite solutions (the latter case being 0x + 0y + 0z = 0), of which the geometric method would need an additional point to find out. 

Thank you guys so much!!

It really helped and I really appreciate you taking time to explain it to me!  cheers!

[IBTopper]

Wow, 3 methods to solve one problem! Thanks so much @Vioh and @kw0573 for your solutions. I personally feel that cramer's rule is the easiest way to start off and check for the value of k. This is because it is just 5 marks (as in the picture). 

I would additionally want to request you both to share your insights on how we could approach problems when using matrices. This is because I think it would not be in IB's new mark schemes and thus the steps won't match (M1 etc) 

Thanks   

[kw0573]

2 minutes ago, IBTopper said:

Wow, 3 methods to solve one problem! Thanks so much @Vioh and @kw0573 for your solutions. I personally feel that cramer's rule is the easiest way to start off and check for the value of k. This is because it is just 5 marks (as in the picture). 

I would additionally want to request you both to share your insights on how we could approach problems when using matrices. This is because I think it would not be in IB's new mark schemes and thus the steps won't match (M1 etc) 

Thanks   

In all fairness, what I proposed is mechanically identical to Cramer's Method, with justification using syllabus content. I don't know what you mean, the only time you need matrix (including determinants) in IB HL Core is when you are asked to solve problems of intersection of 3 planes. IB accepts all systematic approaches (guessing and checking do not warrant method marks) and allocate method marks accordingly. Unless you can offer us some specific questions, I feel we had covered the main approaches. I will only add that you should practice a lot with whichever method you use.