VCE Specialist 3/4 Question Thread!

[alsheriff]

G'day
Quick question:
                           -->                    -->
In triangle OAB, OA=3i+4k and OB=i+2j-2k
        -->
Find OP where P is:
the point where the bisector of angle AOB intersects AB

And another question; can you find the intersection of two vectors by equating them?

Thanks

[VanillaRice]

G'day
Quick question:
                           -->                    -->
In triangle OAB, OA=3i+4k and OB=i+2j-2k
        -->
Find OP where P is:
the point where the bisector of angle AOB intersects AB

Find the unit vectors for OA and OB, and then add them up. This will give a vector that is in the direction of OP.

Now, how do we find out the length of OP? I'll let you have a go at this yourself. Hint: roughly draw out the triangle, with points O, A, B and P. Think vector projection/resolution.

Post if you get stuck

And another question; can you find the intersection of two vectors by equating them?

Thanks

Remember that in general, vectors can move freely in space. So, they cannot have a unique intersection.


Hope this helps

[brightsky]

With regard to finding the intersection of two vectors, I should add that if the vectors you are considering are bound vectors, rather than free vectors, then the general method used is to find two different pathways from the start of one of the vectors to the point of intersection, and then to solve for the unknowns by equating the two different pathways. I'll illustrate using an example.

Consider the triangle OAB. Let P be the midpoint of the line segment OA, and Q be the midpoint of the line segment AB. The line segment PB and the line segment OQ will therefore be medians of the triangle OAB. Suppose PB and OQ intersect at the point X. Let OA = a, and OB = b. Our aim is to find OX in terms of a and b. We shall do this by constructing two different pathways from the point O to the point X.

The most direct pathway from O to X is obtained by noticing that OX is a scalar multiple of OQ; that is, OX = κ OQ = κ (a + 1/2 AB) = κ (a + 1/2 (b - a)) = κ (1/2 a + 1/2 b) = κ/2 a + κ/2 b.

A second pathway from O to X is obtained by first travelling from O to B, and then from B to X; that is, OX = b + BX = b + λ BP = b + λ (1/2 a - b) = b + λ/2 a - λ b = λ/2 a + (1 - λ)b.

Now, because both pathways start from O and end at X, we can say:

κ/2 a + κ/2 b = λ/2 a + (1 - λ)b

In other words:

κ/2 = λ/2
κ/2 = 1 - λ

From the two equations above, we can conclude that κ = 2/3 and λ = 2/3. Hence, OX = 1/3 a + 1/3 b.

Hope this helps!

P.S. Note that we have just proven that the centroid of a triangle divides each median into segments in a ratio of 2:1. This is the first step to proving that the three medians of a triangle are concurrent.

[Onyx]

Need help with complex numbers
1) show that sin(theta) + icos(theta) = cis(pi/2 - theta)
2) show that cos(theta) - isin(theta) = cis(-theta)
3) show that sin (theta) - icos (theta) = cis (theta-pi/2)
Can someone explain this and show how to do it
thanks.

[brightsky]

First, recall the definition of cofunction: https://en.wikipedia.org/wiki/Cofunction. Given that sine and cosine are cofunctions of each other, sin(theta) = cos(pi/2 - theta) and cos(theta) = sin(pi/2 - theta).

For the first question:

sin(theta) + i cos(theta) = cos(pi/2 - theta) + i sin(pi/2 - theta) = cis (pi/2 - theta), as required. Note that here all we've used is the definition of cofunction.

For the second question:

cos(theta) - i sin(theta) = cos(-theta) + i sin(-theta) = cis(-theta), as required. Note that here we've used the symmetry properties cos(-theta) = cos(theta) (since cosine is positive in the 4th quadrant) and sin(-theta) = -sin(theta) (since sine is negative in the 4th quadrant).

For the third question:

sin(theta) - i cos(theta) = cos(pi/2 - theta) - i sin(pi/2 - theta) = cos(theta - pi/2) + i sin(theta - pi/2) = cis(theta - pi/2), as required. Note that here we've used both the definition of cofunction, and the symmetry properties cos(-something) = cos(something) and sin(-something) = -sin(something).

Note that the concept of cofunctions can be extrapolated to other circular functions as well. Just by looking at the names, we can see that cosine and sine are cofunctions of each other; cotangent and tangent are cofunctions of each other; and cosecant and secant are cofunctions of each other.

[TheAspiringDoc]

Need help with complex numbers
1) show that sin(theta) + icos(theta) = cis(pi/2 - theta)
2) show that cos(theta) - isin(theta) = cis(-theta)
3) show that sin (theta) - icos (theta) = cis (theta-pi/2)
Can someone explain this and show how to do it
thanks.

For the second one, just expand our cis(-t):
Cos(-t)+isin(-t)=cos(t)-isin(t)
This works because cos(-t) = cos(t), and sin(-t) = -sin(t)

I'll get back to you about the other two

[Onyx]

Thanks ! Got it
Also how do I convert to mod arg form
1+i tan theta
i get the mod but whats the arg

[VanillaRice]

Thanks ! Got it
Also how do I convert to mod arg form
1+i tan theta

When calculating the argument, be sure to do a quick sketch of your complex number on an Argand diagram to check you have selected the right angle
Knowing this, have a go at converting 1+i to modulus/argument form. Post if you get stuck

EDIT: Just saw your edit - think I misread your question.


Hope this helps

[Onyx]

When calculating the argument, be sure to
do a quick sketch of your complex number on an Argand diagram to check you have selected the right angle

Knowing this, have a go at converting 1+i to modulus/argument form. Post if you get
stuck

Got it was just a bit confused with the arctan(tan(x))
Cheers

[VanillaRice]

Got it was just a bit confused with the arctan(tan(x))
Cheers

Apologies - seems I misread your question, but I've edited by original comment (copy-pasted the edit below).

EDIT: Just saw your edit - think I misread your question.


Hope this helps

[TheAspiringDoc]

Hi, this is very related to Onyx's previous question.

How do you convert 1+icot(x) to polar form?
It is easy enough to find the modulus:
Mod=sqrt(1+cot²x)=sqrt(cosec²x)=cosecx
But as for the argument..
arg=arctan(cotx)...
but now what? I dug deep into Wikipedia to find that arctan(1/x)=Pi/2-arctan(x), which will allow the argument to be found relatively easily, but I'm not sure how else to find the argument without using Wikipedia.. and I don't understand where Wikipedia formula comes from?

[brightsky]

tan and cot are cofunctions of each other. Hence, arctan (cot (x)) = arctan (tan (pi/2 -x)) = pi/2 -x.

[TheAspiringDoc]

tan and cot are cofunctions of each other. Hence, arctan (cot (x)) = arctan (tan (pi/2 -x)) = pi/2 -x.

Thanks
How did you know to subtract x from Pi/2 rather than add x or subtract Pi/2, etc?

[brightsky]

Thanks
How did you know to subtract x from Pi/2 rather than add x or subtract Pi/2, etc?

From the definition of cofunction.

We call f(x) and g(x) cofunctions of each other if f(pi/2 - x) = g(x) and g(pi/2 - x) = f(x).

From their names, we know tan and cot are cofunctions of each other, which means cot(pi/2 - x) = tan(x) and tan(pi/2 - x) = cot(x). The same applies to sin and cos, and sec and cosec.

[keltingmeith]

Thanks
How did you know to subtract x from Pi/2 rather than add x or subtract Pi/2, etc?

Adding on to brightsky, if you didn't know the two were cofunctions but did know they were related by a transformation (you can also do this with sin and cos, or sec and csc) you can also do it by imagining what the graph of tan(x) looks like compared to tan(x+pi/2), tan(x-pi/2), etc. If the graphs look the same, at this level you can just assume that they're equivalent.