**Update (16/02/20)**

In general form, the (intended) scenario of the exam question is

*The* *vector resolute of in the direction of is ,*

Of course, given and it is standard to find . After a bit of trig and unit vectors, we have (in must useful form)

The exam question, however, is different: the question is, given and , how to find .
The problem with that is, unless the vectors and are appropriately related, the scenario simply cannot occur, meaning ** **cannot exist. Most obviously, the length of must be no greater than the length of . This requirement is clear from the triangle pictured, and can also be proved algebraically (with the dot product formula or the Cauchy-Schwarz inequality).

This implies, of course, that the exam question is ridiculous: for the vectors in the exam we have , and that’s the end of that. In fact, the situation is more delicate; given the pictured vectors form a right-angled triangle, we require that be perpendicular to . Which implies, once again, that the exam question is ridiculous.

Next, suppose we lucked out and began with perpendicular to . (Of course it is very easy to check whether we’ve lucked out.) How, then, do we find ? The answer is, as is made clear by the picture, “Well, duh”. The possible vectors are simply the (non-zero) scalar multiples of , and we’re done. Which shows that the mess in the intended solution, Answer A, is ridiculous.

There is a final question, however: the exam question is clearly ridiculous, but is the question also stuffed? The equations in answer A come from the equation for above and working backwards. And, these equations correctly return no solutions. Moreover, if the relationship between and had been such that there *were* solutions, then the A equations would have found them. So, completely ridiculous but still ok?

Nope.

The question is framed from start to end around definite, existing objects: we have THE vector resolute, resulting in THE values of m, n and p. If the VCAA had worded the question to find *possible* values, on the basis of a *possible* direction for the resolution, then, at least technically, the question would be consistent, with A a valid answer. Still an utterly ridiculous question, but consistent. But the VCAA didn’t do that and so the question isn’t that. The question is stuffed.

…

what even is a “vector resolute”??

Vector projection.

Thanks — I’ve never heard it called that before.

The question is just asking if students have memorised the formula properly, I suppose. The last three options are bizarre.

Nineteenth century? I’m not sure that either of us are quite that old…

Speak for yourself!

Is the issue here that a formula (which is found in a popular textbook) that students who answered this “correctly” had probably memorised is actually meaningless/wrong?

Because I will freely admit to immediately ruling out four answers, selecting the one that remained and moving on with the paper.

Nope.

lol!! Surely it wouldn’t be something as obvious as the fact that no solution for m, n, p exists ….. lol!! Uh oh …..

https://www.wolframalpha.com/input/?i=solve+m%28m%2Bn-p%29%2F%28m%5E2+%2B+n%5E2+%2B+p%5E2%29+%3D+2+and+n%28m%2Bn-p%29%2F%28m%5E2+%2B+n%5E2+%2B+p%5E2%29+%3D+-3+and+p%28m%2Bn-p%29%2F%28m%5E2+%2B+n%5E2+%2B+p%5E2%29+%3D+1

Whoops. It is ….

What a great question to ask the Assessors – How exactly does one solve the equations given in the ‘correct’ option to find the solutions ….??

Just when you think the Exam writers can’t do anything dumber, they still somehow manage it …. lol!!! Brilliant!

The MAV solutions would

nevermention this – it’s too embarrassing for the VCAA for the MAV solutions to ever mentionanyVCAA error. It’s not part of the writing brief. You must close your eyes, blindly calculate the ‘correct answer’ (even when the error means that there isn’t a correct answer) and pretend that everything VCAA is awesome. Just who is writingand vettingthe VCAA exams? The Yr 10 work experience kid?? And where’s the accountability? (NOT the MAV solutions)Don’t cheat. Why are there no solutions?

Geometric intuition isn’t my strong suit, but when I project a vector, I don’t expect it to get longer.

That’s right!

So not only does the question (with its silly options) not really check if students know the formula, it doesn’t even make sense.

Well, it’s strong enough. That’s it. But, there’s more.

(forgive me if this misses the point entirely, it has been a long day…)

But IF solving said equations gave a negative value for any of m, n, p then the solution is wandering around in complex space, whereas the question explicitly says m, n, p are real. This is contradictory at best.

HI, RF. You mean complex values of m etc? Or negative values of m^2 etc? Anyway, pretty much yes.

It was a thought bubble and I pressed “Post Comment” before I’d properly thought it through.

If only the VCAA examiners thought before they pressed.

I think you mean ‘thought before they went to press’ (pun intended).

We should include the vetters and reviewers in this too. After all, to err is human but to find the err is the job of the vetter (I don’t have it in me to be divine when it comes to VCAA).

Let’s try and be realistic here, and not enter too much into the world of fantasy…

For a bit of light relief I got distracted by this question in the New Scientist in accelerating frame of reference

https://www.newscientist.com/article/mg24532671-200-new-scientist-puzzle-44-elevator-pitch/

Answers on line once you have made your decisions using Newton’s laws,Hookes law and Archimedes principle.

Steve R

OK, well done, everyone. And, believe it or not, you’re not done.

Just to summarise where we are. JF used the big gun to note that there are no solutions for m, n and p. Then Alan gave an intuitive reason: the (orthogonal) projection/resolute of a vector cannot be longer than the original vector.

So, what’s left to to do? Three things. First, can you prove/”prove” Alan’s intuitive observation? That is, can you give a solid reason (or an air-tight proof) that the resolute of a vector will not be longer? Second, there are two more ways in which the question is completely ridiculous. Find them! Finally, and either least or most important depending on how you think: is the question wrong? That is, although the question is (triply) absurd, does the question have a correct answer among the available options?

On the first thing: my first instinct is to argue that the projection of v onto u must be shorter than v, since the (magnitude of the) cosine of the angle between v and u is less than one

Yep, that’ll do it! And there’s various versions of that, both algebraic and geometric.

When I imagine a new version of the question, in which the original vector i+j-k has been changed to something with a more reasonable length, I can see the next stupid thing: assuming there is a solution, 2i+3j-k itself must be a solution. Because the projection goes in the same direction as the thing it’s projected onto. So there’s no need to build a complicated system of equations to solve. And maybe the third stupid thing is that the solution is not unique. Every non-zero multiple of 2i+3j-k is a solution.

I messed up copying the numbers from the image. I meant 2i-3j+k of course, not 2i+3j-k

Alan, you have thirty minutes from posting to correct your comment.

I don’t see an edit button. Probably because I’m a drive-by commenter with no connected google/facebook/whatever.

But play around with the numbers, I found another reason the projection is implausible: the vectors i+j-k and 2i-3j+k point away from each other (i.e. the angle between them is greater than pi/2) which means scaling up the length can’t fix the problem. One of them needs to be flipped.

(I had to do some calculating before realizing the angle was a problem. If I did it right, -7i-7j+7k as the only multiple of i+j-k that makes the question work.

Ah, yes, maybe you have to be a human to edit.

And that would be a “yes”.

I tried writing this out myself by sketching a diagram first. My reasoning in going through this may be incorrect but proj(a)_b (the projection of a onto b) is defined by (a.b)/(b.b) *B (where B is the vector for B, I’m not too sure how I’d indicate this since I’m not sure if LaTeX is enabled for comments). The magnitude of A is then |a|. The inequality we wish to show is then |proj(a)_b| <= |a|. By the formula, this simplifies to (a.b)/|b|^2 *|b| <= |a| which then becomes (a.b) <=|a||b|. Finally take the absolute value of both sides to get |a.b| <=||a||b|| which is the Cauchy Schwartz inequality. So that means because of the CS inequality, the vector resolute has to be shorter.

As for the actual question, both C and E will give you nothing but contradictions then the solutions of D should lie on a line. However these last three occur if you switched the order of projection so they’re not answering the question. Finally looking to A, which is the intended set of equations, you can either divide equation 1 by 2 or 3, so that you’re left which m/n=-2/3,n/p = -3 and p/m=1/2 –> m =-2/3 n,p=-n/3 and 2p=m. Then we have m=-2/3 n and m = -2/3 n….Which is a complete waste of time. This seems to suggest that the solutions lie on a line but I’m almost certain that a stage in between invalidates the manipulations equally, however I am not certain as to what it is. I’d say because the variables can’t be “found” from any of the equations that none of the answers are right however I feel as though its deviating from what you intended.

Thanks, MK. Cauchy-Schwarz is the natural algebraic approach to this, though it also really amounts to SRK’s trig. For your second paragraph, see Alan’s most recent comments.

Re: ” … is the question wrong? That is, although the question is (triply) absurd, does the question have a correct answer among the available options?”Provocative question, Marty.

An argument might be made for Option D: It’s not hard to show that m = n = p = 0 is the unique solution. So the question becomes whether it makes sense to talk about a vector resolute in the direction of the zero vector …. (in which case there’s an infinite number of them, including the one proffered by VCAA ….)

Except… if n=m=p=0 then I think it is wrong to say THE vector resolute is … since, as you say, there are an infinite number. THE implies a unique solution, to me at least.

RF, I think two different ideas are getting tangled together, but I agree that the “THE” is an issue. The expression “The values of m, n and p ….” suggests a unique solution, and definitely implies at least one solution.

JF, unique solution to what? I don’t think it really makes sense to project “in the direction of” the zero vector, although one can argue for it: the transformation that zaps everything to 0 is indeed a “projection”. But, even if allowed, that wouldn’t give the projection claimed in the question.

Re: “unique solution to what?”The unique solution to the equations in Option D.

So the projection proffered in the question would therefore be one of the infinite number of possible projections. Yes?

A weak argument I know, but it’s

yourcan of worms that your suggestion (does the question have a correct answer among the available options) opened lol!Personally, I’m happiest with the conclusion that the question is defective for many reasons and VCAA (or maybe it’s the work experience kid) has screwed up again.

Well, no. If we allow projection in the direction of (or onto) the zero vector, then only possible result of that projection is the zero vector. But this attempt to make any sense of the question is my fault, and too far removed from reality.

Thanks to Everyone for commenting, and thanks to the 150 year old Damo for bringing the question to our notice. That pretty much wraps it up, though there’s one clarifying point I don’t think anyone has yet made. I’ll try to update with a summary very soon, although there’s another off-shoot of the MAV-VCAA post to deal with, and the Incredibly Evil Mathologer has just handballed 600 slides (sic) for me to go through.

For what it’s worth, if you take the equations bunged into Wolfram Alpha above by John Friend above, and multiply each side by m^2+n^2+p^2, you get three polynomial equations which are solvable (one solution will be (0,0,0) which we can discount). The others are all complex numbers. ‘Tis a stupid question, and the examiners, if they had any quality of reflection at all, should be deeply embarrassed by it. A standard SNAFU for VCE “mathematics”. (I use scare quotes because the mess that goes into, and comes out of schools, barely deserves the term.)

Yes, the writer’s should have known better, but I can overlook this. We all make mistakes. That’s why this stuff is meant to get scrutinised by vetters ….

It’s the vetters and reviewers that should be drawn-and-quartered and then sacked. They are asleep at the wheel. Every. Single. Year. Why VCAA continues to employ these people I do not know.

Yes, I’m close to agreeing. Anyone can have a cute idea that turns out to be intrinsically flawed, or is just flawed in execution. I’d still suggest it’s pretty poor for a writer to have not spotted

someissue: the single standard diagram of projections should tell you the question makes no great sense, even if one picked vectors a and c that worked.But in any case, I don’t really care which stage of the process one wishes to label flawed. The process as a whole is undeniably flawed, and that’s enough.

Interesting, as we have now had vectors introduced into the NSW HSC syllabus (Extension 1 and 2). Here, it is called the projection. It almost feels like our new syllabus has been copied and posted from the VCE!

That’s unfortunate.

Two genuine questions:

What has been taken out (if anything)?

Do you (as a teacher, I presume) feel this is a positive move or doesn’t it matter?

I don’t know if you are familiar with our old syllabus but it contained: complex numbers, polynomials, integration, conics, volumes,mechanics and harder extension 1 work. The new course has: complex numbers (now with complex exponentials), integration, vectors (in 3D and vector equations of lines), mechanics and proof. I like the fact that vectors are in there, but we go as far as the scalar product. Personally, I would like to see matrices, then the cross product also. Initially, I thought it was a positive move, but now we are teaching it, I think it was rushed out and there is little common between the topics (some). As we are now being told to show real world examples (my background is physics and pure maths), I thought this would be good to have a course that fitted well with the physics course. This doesn’t appear to have happened.

Interesting that conics have gone, although if I remember they were mainly Extension 2?

So, that would mean a lot of the pure geometry work is probably gone as well to make room for vectors. I do like the idea of vector equation of a line (totally absent in VCE and really needs to be there I feel).

The IB syllabus cut matrices back in 2010(ish) due to over-crowding of the curriculum, so maybe it is just something that doesn’t quite fit.

As to “real world examples” I share the opinion of Marty and many others that the exam questions on these are so insanely contrived and don’t even vaguely resemble the “real world”…

Yes, always with the real fucking world. Numeracy poisons everything.

Brett, thanks for the information. I know very little about the NSW system except that the highest level (extension 4?) has had some very good and seriously hard exam problems. I have the general sense that NSW is sensible (albeit getting worse), in a way that the other states are not. If so, I assume that is because competent mathematicians have a big say, but that is just a guess.

As for topics, of course some will be better or worse, and there will be personal preferences. Mostly I don’t think it matters that much (except for statistical inference, which is insane). What one wants is depth and, as much as possible, topics that reinforce each other. Currently, Victoria has neither.

I am teaching numeracy to Year 11 students at present. Next week they will get a gentle introduction to Ramsey theory.

That might be the definition of overly resolute. (But good on you.)