A few Specialist Problems

[d0minicz]

did u mean cis on the right hand side ?

[kamil9876]

Show that



Just showing me how to do the first one should be helpful, thanks

expand the RHS:






which should render once latex gets working...

[evaporade]

Quicker this way:

cis(pi/2-@) = cos(pi/2 - @) + isin(pi/2 - @) = sin@ + icos@

[GerrySly]

Now for the first part I am pretty sure how to get that answer. appears to be exactly in between , hence . But can you just deduce that from the graphic or is there something else you can use?

Now for part b I have no idea how to begin... just began working with vectors today

[Mao]

The easiest way to think about this is to have two trapezium make a parallelogram (flip the second one horizontally). In this case, you will have

top edge = b + a
bottom edge = a + b
middle line is exactly in the centre since the line is from X to X', both are midpoints. since the line is XYX', the length XY is also half of top or bottom edge, i.e.

To prove it is parallel, since DC is parallel to AB, any sum of a and b will be parallel. hence XY is parallel to AB.

[But this is a badly written question.]

[kamil9876]

Let's travel from X to D to C to Y. This gives vector XY:



However, traveling from A to X to Y to B gives AB=a:



Subtract the two equations to get rid of and

[GerrySly]

Ok just a clarification question for this question, I can't remember how our teacher showed us to show they are linearly dependent / independent, but wikipedia says I can arrange them in a 3x3 matrix and find the determinent and if it equals 0 then it is linearly independent. Is this the way we are supposed to show it? I have been ridiculed for using methods found in wikipedia before...

[dcc]

Essentially, you want to find out the nature of the solutions to the equation



which can also be expressed as:



Now, if the matrix is INVERTIBLE (that is, the determinant is non-zero), then you can immediately see that our solution will be of the form:

, that is, the matrices are linearly independent.

If the matrix is not invertible, then you can do ninja shit and they should be linearly dependent.

Applying this to the examples you have provided:

a) Our 'big' matrix is

Notice that , so these vectors are linearly indepedent.

I'm sure the rest are doable in much the same manner (unless you can see straight off the bat that one of the provided vectors is a scalar multiple of another)

[Mao]

for the first case only:

Your teacher's method:

will give you three simultaneous equations

4a + 2b = -4
a - b = 2
3a + 3b = 6

if the above has a solution , then they are linearly dependant, otherwise independant.

The proper method used in uni is as dcc has shown above.

[GerrySly]

Ok, 6.a.i and ii are easy enough (6.a.ii and 6.a.ii is just of that according to the ratio) but I always seem to have trouble with . I thought was just , but using that position vector gives me something weird.

Just helping me understand what the are asking for should be enough, I should be able to work it out from there.

BTW the answer is

[TrueTears]

is just the position vector of M



or



Same thing.

[GerrySly]

Thanks TrueTears, vectors are kicking my ass here...



The answer is just out of my reach, I know it had something to do with using but I am not sure how to relate that to

[TrueTears]

Find using the information provided.

Now pretend you have a triangle namely OCD.

Using the cosine rule to work out length of CD.

You know OC = 5, OD = 7.

[GerrySly]

I always have problems with these "solution" questions, not sure how to approach them. I have problems visualising it, I can visualise the other formula questions where there is an easy exit and entry amount.

By the way that is Chapter 11 Question 8 Cambridge Essential

[TrueTears]

http://vcenotes.com/forum/index.php/topic,9192.msg121038.html#msg121038

That would help immensely.