VCE Specialist 3/4 Question Thread!

[scribble]

GG. b^3 is a speedy little guy

[Thorium]

EDIT: Added a diagram.

GG. b^3 is a speedy little guy

Thanks to u both

EDIT: Sorry, I didnt get the very last step. (Please let Scribble be first this time )

[scribble]

well |a| (with a squiggle) is the magnitude, or length of the vector a.
i wouldn't think you'd need to explain that though.

alisina, did you get stuck on the last step of my working? i just subbed what i wrote in purple into the a.b on the above line.

[Thorium]

I don't know how to explain this perfectly but here is the same as simply because and are basically line segments. So they are essentially both the same thing in this context and hence, the modulus symbol can be cancelled out.

well |a| (with a squiggle) is the magnitude, or length of the vector a.

Sorry again, I should have explained what I didnt understand in more detail. But anyway, this was the ans that I was looking for.
"i just subbed what i wrote in purple into the a.b on the above line."

Thanks again to you all.

[Nato]

Let

show that:

some hints would be appreciated

[LOLs99]

Let

show that:

some hints would be appreciated

Hint: Start with LHS and "S"

[Conic]

Let

show that:

some hints would be appreciated

Try the approach you would normally use to divide cartesian complex numbers

Spoiler

[revcose]

You could also use De Moirve's theorem

Spoiler

[survivor]

How do you show a solution on an Argand diagram?

For example, solve z^2 + 1 = 0, and show the solution on an Argand diagram.

z=i and z=-i, how do i go from there?

[scribble]

Z= 0+i has the coordinates (0,1)
Z=0-i has the coordinates (0,-1)

[Irving4Prez]

Hey guys,

I need help with answering this question:

The position vectors of points A and B are a=i-j-k and b=2i-2j-2k
Find the unit vetor which bisects angle AOB.

So far I've found the unit vectors of a and b and I've attempted to find the resultant vector by adding the 2 unit vectors together, but I'm not understanding the reason behind this process and I'm having trouble progessing. Thanks.

[lzxnl]

Wait...A, B and O seem to be on the same line as b=2a...is there a typo in the question?

[alchemy]

Possibly, because when I tried calculating the magnitude of the addition of the 2 unit vectors, I got 0 :/

But in general:
-Find the unit vectors of a and b.
-Find the sum of unit vectors a and b (this sum is actually the vector that bisects angle AOB)
-Find the unit vector of the sum vector of vectors a and b.

OR you could try this fantastic alternative method:

-Find the dot product of vectors a and b, and hence find the angle between vectors a and b. This is the angle AOB
-Let another vector (e.g. vector c) be the vector that bisects AOB. This vector would then make 1/2 of the angle AOB with the x-axis. So, divide the value you got in the first step to find the angle vector c makes with the x-axis.
-Let c = xi + yj +zk and it's magnitude equal to 1
-Find the dot product of a and c. Let this value equal to the angle vector c makes with the x-axis (refer to step 2), to form 'equation 1'.
-Find an equation for the magnitude of c and let this be equal to one (refer to step 3). Let this be equation 2.
-Now, the angle between vector b and c must be the same as the angle between vectors a and c. You can place that value for the angle in an equation that finds the dot product of vectors b and c. So, it would look like: b.c = |b||c| cos(angle between vectors a and c). Let this be equation 3.
-Solve the 3 equations to get your answer.

EDIT: Hard to explain the latter method without a working example. I'll update this post, when I have time, with an example

[lzxnl]

Interesting. I always did a much lengthier way for those questions. Your method is much better LOL.

Scale both vectors such that they're the same length. Then, the vectors span an isosceles triangle. Find the position vector of the midpoint of the base of the isosceles triangle. This vector will bisect your angle.

[brightsky]

Yeah the 'formula' most people use is 1/2 (a hat + b hat). The. You can select whatever length you want.