VCE Specialist 3/4 Question Thread!

[keltingmeith]

Last question I promise

I was doing the checkpoints 2014 and I surprisingly got this easy question wrong. It asked to compose two components of vector a, one in the direction and parallel to vector b and the other in the direction and perpendicular to vector b.

I had exactly similar questions in my text book and the book says in order to find a vector component the formula is where both vector 's are the unit vector. Can someone please help with this? I am so confused

Your formula is right - they messed up.

Does your actual vector match theirs? If not, figure out what exam that question came from (most of the questions in checkpoints are past exam questions) and look up the solutions that VCAA provides, because those solutions are never wrong (unless they're less than a few months old, but VCAA tend to fix those reasonably quickly)

[brightsky]

Hahaha sir.jonse you are way to keen!

Before you even begin solving the problem, you should realise that there are two unit vectors perpendicular to both a and b, and that the two unit vectors point in opposite directions.

Let c = xi + yj + zk be the required vector. Since c is a unit vector perpendicular to both a and b:

a.c = x + 2y + 2z = 0......(1)
b.c = 2x + 2y + z = 0......(2)
|c| = sqrt(x^2 + y^2 + z^2) = 1......(3)

Solving (1), (2) and (3) simultaneously yields:

x = - 2sqrt(17)/17, y = 3sqrt(17)/17, z = -2sqrt(17)/17

or

x = 2sqrt(17)/17, y = -3sqrt(17)/17, z = 2sqrt(17)/17

Hence, c = - 2sqrt(17)/17 i + 3sqrt(17)/17 j - 2sqrt(17)/17k or c = 2sqrt(17)/17 i - 3sqrt(17)/17 j + 2sqrt(17)/17k.

[brightsky]

Haha I'm just poking fun at you. Your enthusiasm definitely bodes well for next year!

We have three unknowns: x, y and z. We need three equations in order to solve for the three unknowns. The first two equations make use of the fact that c is perpendicular to both a and b. But we don't just want any vector that is perpendicular to both a and b. We only want vectors that have a magnitude of 1. The third equation makes use of the fact that c is a unit vector.

[brightsky]

Which three equations did you solve simultaneously to get x = -2, y = 3 and z = -2?

As I think you've already figured out, the problem with your answer is that it is not a unit vector. I agree that it is a vector perpendicular to both a and b, but it is not a unit vector perpendicular to both a and b. Also, you've failed to take into account the vector perpendicular to both a and b that points in the opposite direction.

[brightsky]

I'm not exactly sure whether this is what you've done, but bear in mind that sqrt(x^2 + y^2 + z^2) = 1 cannot be simplified to x + y + z = 1. Notice how the expression under the square root in the first equation is the sum of three perfect squares, but is not a perfect square itself. It is not possible to break the square root up into three parts and take the square root of x^2, y^2 and z^2 separately. Even then, sqrt(x^2) = |x| not just x. It is (sqrt(x))^2 that equals x. There is a subtle difference.

[brightsky]

It works because you have essentially done the following:

1. Found a vector with a random magnitude that is perpendicular to both a and b.
2. Found the unit vector of the vector obtained in Step 1.

Of course, you obtain the same, correct answer, but you've introduced an extra step, whereas you could have directly found the unit vector in Step 1 had you made the third equation sqrt(x^2 + y^2 + z^2) = 1.

I'm also a little perplexed as to how you got x = -2, y = 3 and z = -2 when these three values don't satisfy the third equation: x + y + z = 0. Even if you tweak the third equation so that it becomes: x + y + z = 1, you don't get x = -2, y = 3, z = -2. In any case, do you see how the third equation, as it stands, is a little random?

[keltingmeith]

It is (sqrt(x))^2 that equals x.

You also gotta remember there that said statement only works for x>=0, and the function isn't defined for x<0. In fact, you could still define that as |x|, since for x>=0 |x|=x

I would also like to add that this is a wholly stupid question for VCE and not to get your hopes up for having discovered a new method, for there are far better ways of answering this question. (which aren't in VCE, which many people still maintain to be rather annoying and stupid)

[brightsky]

To solve the system of equations by hand, you could use the Substitution Method. Rearrange the first equation to make z to subject, and then substitute into the second and third equation so that you end up with two equations in the variables x and y only. Next, rearrange the first of the two equations to make y the subject, and then substitute into the second of the two equations so that you end up with one equation in the variable x only. Solve that equation for x, and then plug back the value of x that you obtain back into the preceding equations to obtain y and z. It is a tedious process, which means that chances are you will be permitted to use a CAS calculator if ever such a question pops up in the end of year exams.

As EulerFan101 alluded to in his post above, there is another method that you could use to solve this particular problem, which does not involve solving a system of three equations. On your CAS calculator, type in the following:

Define a = [1, 2, 2]
Define b = [2, 2, 1]
unitV(crossp(a,b)) ---> This should give you one of the unit vectors.
unitV(crossp(b,a)) ---> This should give you the other unit vector.

This method is not taught in Spesh but if you're interested and have some spare time up your sleeve, I'd recommend that you read up on the cross product: http://en.wikipedia.org/wiki/Cross_product. Judging from what you have posted, you've already encountered the dot product (also called the scalar product). The cross product is simply another kind of product. There are several other products as well, which you will study if you plan to take some maths courses at uni, such as the box product (also called the scalar triple product), and what have you. Of course, none of these products are in the Spesh course, but again, if you're interested, it mightn't be a bad idea to search them up.

[Zues]

what does "The letter C is used to denote the set of complex numbers where z is an element of C"

im not sure how to process this, can someone simplify this down into what i need to know, perhaps with an example would be awesome! (if you can)

[keltingmeith]

what does "The letter C is used to denote the set of complex numbers where z is an element of C"

im not sure how to process this, can someone simplify this down into what i need to know, perhaps with an example would be awesome! (if you can)

Remember in methods, sometimes you write "" to specify the domain of x.

When you write that, you're simply saying that x can be any real number - such as 0.1, pi, 0.728687308763..., and so on.

Well, what if instead of being a real number, we wanted to say that x was a complex number? Well, we would write "" instead - because just as R means real numbers, C means complex numbers. Often, instead of writing x, we use z.

Consider the equation . We know from the discriminant, there are no REAL solutions. So, if , then there is no solution to this equation. HOWEVER, if instead we decide to look for some x such that , then there IS a solution - we can have .

z=x+yi

So, if z is an element of C, that means z= x+yi is a complex number. That is, its not a real number i.e

I'm just going to be picky and say you shouldn't bold font the i, because that implies that i is a vector, which it isn't. (well... sort of, but not really, something something quaternions/beyond the scope of specialist)

[Zues]

thanks for that guys 

why did they multiply both sides by "i". how do we know we have to do this?

[Zues]

yeah defiantly mate. good intuition!

[kinslayer]

When you are dividing by a complex number like this, the thing to do is to multiply top and bottom by the complex conjugate of the denominator. This is because , which is a real number. Like sir.jonse said, it's similar to "rationalising" expressions with surds in them. Once you are dividing by a real number, it's easy to express the complex number in the form of because you can distribute the division operation over the sum.

The complex conjugate of is , and , so the whole fraction is equal to .

[keltingmeith]

How in the living earth is the answer not D. Isnt the whole point of complex numbers that the numbers are part of the system? I see why its E though, because . BUT HOW ON EARTH ARE THEY REAL NUMBERS??

This is a bad question, because nowhere does it state that a,b CAN'T be complex.

However, the general consensus is that if z=a+ib, then . It's just easier to define, because if a and b were complex themselves (say a=c+id and b=e+if, ), then the whole thing gets a little difficult to work with:



Which holds particularly if a,b,c,d,e,f are all variables and not constant.

Sidenote: the \bar{} command will put a bar over whatever's in the curly brackets.

[keltingmeith]

OH! So youre saying the is the complex number, and are real numbers?

Got it in one. (Y)