VCE Specialist 3/4 Question Thread!

[keltingmeith]

well the thing is you cant split it up any further as the numerator is 1. generally in specialist you turn expressions into partial fractions when the numerator is of a higher degree than the denominator. This technique is commonly used so integration techniques can be applied (calculus).

Other way round - it's normally when the top is of a lower degree than the bottom (and a substitution won't work). If it's the other way round, you do polynomial division. Also, nothing stops partial fractions from working if the numerator is one - consider :



EDIT: Also, there is a VERY good reason as to why we wouldn't split the up (other than we can't. ) You see, that function is actually one we know how to graph - and we call that a "reciprocal function", which you'll learn about later in the course. (plus, we can quite easily differentiate and integrate it, so there's no need to split it up in terms of those either)

[brightsky]

The first thing to bear in mind is that the adjectives 'linearly dependent' and 'linearly independent' are used to describe SETS of vectors rather than vectors themselves. Technically, it makes no sense to say 'this vector is linearly dependent' or 'this vector is linearly independent'. It is more correct to say 'this set of vectors is linearly dependent' or 'this set of vectors is linearly independent'.

In layman's terms, a set of vectors is said to be linearly dependent if there is some way of obtaining the zero vector by adding scalar multiples of the vectors in the set. In practice, obtaining the zero vector means forming a closed shape, or finding one's way back to where one started.

Consider now the example that you provided. In this case, we have a set consisting of two vectors: a = -3i - 5j and b = 6i + 10j. The first step is to visualise these vectors in your head. Recall that a vector is defined as a geometric object with a magnitude and a direction, so what you should be visualising in your head are two arrows. Now, ask yourself: is it possible form the zero vector by adding scalar multiples of a and b? The answer in this case is yes, as 2a + b = 0. Hence, the set of vectors is said to be linearly dependent.

[keltingmeith]

Thanks, that clears it up for me!
But im having troubles with this question, the set of vectors are linearly dependent but by working it out manually I keep getting x=y=0. Could you please show me how to do:
a= i - j + k
b = -2 + 2j +3k

In my book the first step is to convert the vectors in the form of: xa+yb = 0 and equate coefficients, so if you can, please do it that way so I can understand better. Thank you!

Probably because this set of vectors are linearly independent.
Y'see, using the equality xa+yb=0, then there must exist some m and n such that ma=nb - which means that one is a scalar multiple of the other.
HOWEVER, if you multiply by 2, only the k value matches up. If you multiply by -2, then the i and j match up, but not the k. So, they MUST be linearly independent.

(note: what I've done here is pretty hand-wavey and won't work for a university level mathematics course, but should be fine for VCE)

[keltingmeith]

Oh sorry I didnt see that, but could you show me how to do it by hand, without visualising it? Like for example, say the question asked you to prove or show that vector a and b are independent, could you show me how to do it please?

Do it as I did above.
The whole point of vectors is being able to visualise them - and there's nothing wrong with proving something with explanations. Maths isn't all equals signs and graphs.

All you have to do is show that no p, q or r exists such that pa+qb+rc=0 - whether you do that by explanation or some weird algebra is irrelevant.

[brightsky]

From the definition of linear dependence, the vectors a and b are linearly dependent if and only if xa + yb = 0, where x and y are scalars not all 0. This means that if there exists scalars x and y not all 0 such that the equation above is satisfied, then the set of vectors consisting of a and b is linearly dependent. Let's see whether such scalars do indeed exist...

First, assume that there does exist such scalars. Then:

xa + yb = 0
x(i - j + 3/2 k) + y(-2i + 2j + 3k) = 0
2xi - 2xj + 3/2x k - 2yi + 2yj + 3yk = 0
(2x - 2y) i + (2y - 2x) j + (3/2x + 3y)k = 0

As the textbook says, the next step is to equate coefficients. However, what many textbooks fail to explain is why it is legitimate in this case to simply equate coefficients to determine solutions for x and y. Normally, equating coefficients gives you only one of many solutions. However, in this case, equating coefficients gives you all the solutions. The reason is a little complicated, but worth getting your head around. Suppose you had a box containing three vectors that have an i, j and k component and there was no possible way in which you could form the zero vector by adding scalar multiples of the three vectors in the box. By definition, we say that the set of three vectors is linearly independent. Believe it or not, just by adding scalar multiples of the three vectors in the box, we can form every single vector in R^3. Another way of saying this is that every vector in R^3 can be represented as a linear combination of the three vectors in the box. But that's not all! Since the three vectors form a linearly independent set of vectors, we can go one step further and say that every vector in R^3 can be represented UNIQUELY as a linear combination of the three vectors in the box, a result which can be readily proven using proof by contradiction. This is significant. This means that if I wanted to write i + 2j - 3k as a linear combination of the three vectors in the box, there is only ONE way that I can do so. The implication of this is precisely that, for a linearly independent set of vectors, we CAN equate coefficients without losing any solutions (since there is always only ONE way of representing a particular vector in R^3 as a linear combination of the vectors in the set). Note in our case that the standard basis vectors i, j and k are linearly independent. Hence, we can legitimately equate coefficients...

2x - 2y = 0 and 2y - 2x = 0 and 3/2x + 2y = 0

Solving the three equations above simultaneously yields x = 0 and y = 0.

Recall the definition of linear dependence. The vectors a and b are linearly dependent if and only if xa + yb = 0, where x and y are scalars not all 0. From the working out above, it is clear that the only way in which we can satisfy the equation above is if we set x = 0 and y = 0. Hence, there does not exist scalars x and y not all 0 such that the equation above is satisfied. Hence, the vectors a and b are NOT linearly dependent, which means that they are linearly independent.

Hope this makes sense!

[brightsky]

Draw rectangle ABCD. Let a = AB and b = AD. Note that a and b are linearly independent. Recall that two linearly independent vectors spans R^2, which means that a and b are the only vectors that you need to introduce; there is no need to introduce a third vector.

Draw diagonals AC and BD. Let X be the midpoint of AC and Y be the midpoint of BD. All that remains for us to do is to prove that vector AX = vector AY (i.e. that X and Y are the same point). This is easily done.

AX = 1/2 AC = 1/2 (a + d)
AY = AB + 1/2 BD = a + 1/2 (d - a) = 1/2 (a + d)

Clearly, AX = AY, so X and Y are the same point, and the diagonals of the rectangle ABCD bisect each other. In fact the diagonals of any parallelogram bisect each other. Recall that a rectangle is a type of parallelogram.

[brightsky]

Q7. It is often useful to consider a vector, not simply as a geometric object with a magnitude and a direction, but as a path from one point to another point. Consider the vector AB, which takes you from the point A to the point B. From the diagram, it is clear that AB is not the only way in which you can get from point A to point B; you can go from point A to point D, then from point D to point C, and finally from point C to point B. The second path is much longer, but it is still a way of getting from point A to point B. Hence, we conclude that the vector AB = vector AD + vector DC + vector CB. In other words, a = -d - c - b, which implies a + c = - b - d. Hence, E is the correct answer. Another way to look at it is this: suppose you want to start from point A and end at point A. One way in which you can do so is to go from point A to point B, point B to point C, point C to point D, and then finally from point D back to point A. Since you've started and ended at the same point, you've essentially formed the zero vector (recall our discussion earlier regarding the formation of the zero vector). In other words, AB + BC + CD + DA = 0. This means that a + b + c + d = 0, which means that a + c = - b - d. Again, we get E as the answer.

Q8. This is quite a tricky question but recall that inscribed angles subtended by the diameter are right (see: http://www.cut-the-knot.org/Outline/Geometry/AngleOnDiameter.shtml). This is one of the circle theorems that you are expected to know. It may be instructive to think about why the theorem is true, rather than just accept it as gospel. Recall that the dot product of two vectors that are perpendicular to each other is 0. This means that BA . BC = 0. Let us try express BA and BC in terms of a and b.

BA . BC = 0
(BC + CA) . (BC) = 0
(BC + 2OA) . (BC) = 0
(b + 2a).b = 0
b.b + 2a.b = 0
2a.b = -b.b

Hence, the answer is E.

Hope this makes sense! The topic of vectors may seem very arcane at first glance, since it is nothing like anything you've seen in Methods. However, once you've got the fundamentals down pat, you'll begin to see just how useful and, in some sense, incredible the topic really is.

[keltingmeith]

Can someone confirm if I am doing this right please, I dont have answers to it

Looks good to me.

[keltingmeith]

Also, what does it mean by colinear? Say the question asks to prove that two vectors are colinear, what does it mean and how could you do it? Thanks

Colinear means that they run on the same path - easiest way to prove it is to show that they have the same direction (ie, are parallel) and that they share a point in common.

[brightsky]

In fact, none of the answers are correct. You were right in suggesting that the answer should be -5/14 b = -5/14 (-i + 2j + 3k). Since the scalar product of a and b is negative, we know that the angle between a and b is obtuse. If you draw a and b out on a piece of paper, and then draw the vector resolute of a in the direction b, you'll find that the vector resolute is in fact in the opposite direction of b (although it still lies along the same line as b). None of the answers provided reflect this, so none of the answers are right.

[keltingmeith]

Knew it... thats a practice test my school supplies, says a lot about my school!

Can you help with this question as well, I got stuck after finding position vector AB

You haven't found a position vector, you've found a unit vector. A unit vector has magnitude one - now, let's consider the unit vector i. If we multiply it by a, we get the vector ai - finding its magnitude shows that it has magnitude a.

So, you can guess that multiplying any unit vector by a will give it the magnitude of a - in this case, what do you think we want "a" to be?

[brightsky]

You haven't found a position vector, you've found a unit vector. A unit vector has magnitude one - now, let's consider the unit vector i. If we multiply it by a, we get the vector ai - finding its magnitude shows that it has magnitude a.

So, you can guess that multiplying any unit vector by a will give it the magnitude of a - in this case, what do you think we want "a" to be?

EulerFan101 has pretty much said it all, but I'd just like to add that this question highlights one of the reasons why unit vectors are so useful. Suppose you had a random vector v = 2i + 3j - 4k. Suppose you wanted to find a vector that points in the direction of v but has magnitude 5 instead of sqrt(2^2 + 3^2 + 4^2) = sqrt(4 + 9 + 16) = sqrt(29). All you need to do is shrink the vector v down to the unit vector of v and then multiply the unit vector by the desired magnitude (in this case, 5) to get the vector that you want.

Recall that a vector is defined as a geometric object with a magnitude and a direction. Naturally, in order to 'build' a vector, you need a magnitude and a direction. The process above demonstrates how you can build a vector by mashing together magnitude and direction. The direction of the desired vector is given by the unit vector of v. Then, to build the desired vector, all you need to do is multiply the direction of the desired vector (unit vector of v) by the magnitude of the desired vector (in this case, 5).

[keltingmeith]

Alright i see! So, by working out the unit vector of AB which is are you saying to multiply it by 4 to get the answer? I did it and it worked. Basically instead of being infront of vector b, it should be the specified magnitude, 4? as b

Perfect. Except that 1/13 isn't the unit vector - is the unit vector. Remember, 1/13 is just a scalar as it has no direction - a vector requires both a magnitude AND the direction.

[keltingmeith]

All you need to do is shrink the vector v down to the unit vector of v and then multiply the unit vector by the desired magnitude (in this case, 5) to get the vector that you want.

This is legit 1/3 of one of the first year maths assignments at Monash... Darned cube assignment. :x

[brightsky]

Looks good to me.