wildareal's questions thread

[wildareal]

A position vector in two dimensions has a magnitude 5 and its direction measured anticlockwise from the x-axis, is 150 degrees. Express the vector in i and j form.

[kamil9876]

Just use the definition of in terms of the unit circle.

[wildareal]

A and B are defined by position vectors a=2i -2j-k and b=3i+4k

Find the unit vector which bisects angle AOB.



Why is it (a hat) +(b hat) will bisect AOB?

Is it because you can resolve it into a triangle?

[Mao]

Because and will form a rhombus (draw this)

will be a diagonal of this rhombus, which always bisects the angle.

[kamil9876]

let A' and B' be points such that OA'=a-hat and OB'=b-hat

you see that

angle AOB = angle A'OB'

Since |OA'|=|OB'| the triangle A'OB' is isosceles hence the line that bisects the angle also bisects the line, hence OM bisects the angle where M is the midpoint of A'B'. We see that OM=(OA'+OB')/2 which is parralel to a-hat + b-hat (it is half of it hence a scalar multiple of it).

[wildareal]

An orienteer runs five kilometres due north to a marker A. She then turns and runs nine kilometres on a bearing of S50 degree E to a marker B. She then turns six kilometres due east to a marker C. If i is a unit vector due east and j is a unit vector due north, then the position vector of marker C relative to the starting point is?

Thanks.

[Martoman]

draw a picture and call A the origin.

The only movement to the *right* horizontally is 9cos(40) and 6.

So

The movement up is a bit tricky but you can see that the only movements are the 5 going up and the 9sin(40) going down. So its:

Combining: -> magic.

[wildareal]

|z – 6| – |z + 6| = 3 Why is it that the domain is x < –3/8?

[wildareal]

Complex Complex Question

a)Find the roots of z^2 -6z+25=0 where zEC, and hence find the sum of the roots.
b) Let u and v be the roots of the equation z^2+bz+c=0 where, b,c zEC.
i) show that u+v=-b and uv=c
ii) Hence show that if u=p=iq where p,q ER and u and v are complex conjugates, then b and c are real.
c) Find the quadratic equation in z which has the roots 2+root(5)i and -2+root(50i
d) The tricky bit:

A quadratic equation in z has roots u and v. The sum of the roots is -3 and the product of the roots is 4. Find a quadratic equation in z which roots (u+v) and (u-v).

[pHysiX]

a) Quadratic formula:






Hence, sum of roots is 6.

b) i)
Apply quadratic formula to find that

Does not matter what you choose as u & v. Just make sure you get ur signs right.


as required.






as required.


ii) Brain Fart T_T


c)
or
. Expand:

[dcc]

a)Find the roots of z^2 -6z+25=0 where zEC, and hence find the sum of the roots.

Completing the square of this gives you and as the roots, hence their sum is (the sum is going to be a real number, as a simple consequence of the conjugate roots theorem).

b) Let u and v be the roots of the equation z^2+bz+c=0 where, b,c zEC.
i) show that u+v=-b and uv=c

We can rewrite a monic polynomial in terms of its roots, that is as .  Expanding this out and equating coefficients (since our rewrite is true for all ), you find the desired result.

ii) Hence show that if u=p=iq where p,q ER and u and v are complex conjugates, then b and c are real.

I don't quite understand what you've written here - you might need to retype this, unless I'm having a really bad day.

c) Find the quadratic equation in z which has the roots 2+root(5)i and -2+root(5)i

We can find the polynomial using our result b)i), which gives us , unless you've mistyped the roots or I've misread them

d) A quadratic equation in z has roots u and v. The sum of the roots is -3 and the product of the roots is 4. Find a quadratic equation in z which roots (u+v) and (u-v).

Since the sum of the roots of our quadratic is real, then we know that u and v are complex conjugates (this might be what we wanted to prove in b)ii)?).  This means that the real part of both of our roots will be .  Hence taking the product of our roots, we get the equation , where is the imaginary part of (where I just chose to say that has a non-negative imaginary part).  From this, we gather that .  Hence we say that one possible equation is .

[wildareal]

Asymptotes of ((y-1)^/4)-((x+2)^2/9)=1

[kamil9876]

|z – 6| – |z + 6| = 3 Why is it that the domain is x < –3/8?

Let's do the algebraic work with care:

write with real then our condition is equivalent to::



This is equivalent to:



Now this is equivalent to:



Now here is your problem, this is NOT equivalent to:



Why is it not equivalent? because what if-24x-9<0 and it satisfies this equation, then does it satisfy the former? no! because  You see that what is really equivalent to our problem is.

AND

In general what many people forget is that solving an equation means to find the EXACT set of values (x,y) that satisfy the equation. It is more than just algebraically manipulating things by doing things to both sides until you get some magic equation.

[wildareal]

Asymptotes of ((y-1)^/4)-((x+2)^2/9)=1

would you do:
(x-2)=+-3/2(y+1) as it is the inverse?

Thanks.

[Mao]

Asymptotes of ((y-1)^/4)-((x+2)^2/9)=1

would you do:
(x-2)=+-3/2(y+1) as it is the inverse?

Thanks.

not sure what you mean by 'inverse', but those are the correct asymptotes