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August 29, 2025, 03:45:29 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2583993 times)  Share 

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kinslayer

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Re: Specialist 3/4 Question Thread!
« Reply #3480 on: September 06, 2014, 11:15:51 pm »
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Thanks but I'm still confused :/

Sorry about that, which part confuses you?

You should know that if you have two points, say and and draw a line between them, then the line that bisects this line at right angles has equation . Remember |u-z| can be thought of as the distance in the plane between u and z.

You aren't the first person to ask that question in this forum actually. Take a look at this post.. bonus picture!

Re: questions XD


juzza12

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Re: Specialist 3/4 Question Thread!
« Reply #3481 on: September 07, 2014, 12:37:18 pm »
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Sorry about that, which part confuses you?

You should know that if you have two points, say and and draw a line between them, then the line that bisects this line at right angles has equation . Remember |u-z| can be thought of as the distance in the plane between u and z.

You aren't the first person to ask that question in this forum actually. Take a look at this post.. bonus picture!

Re: questions XD
Actually it's okay now thanks!

lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #3482 on: September 07, 2014, 08:37:41 pm »
+1
It's not that. You need BOTH components to make one full cycle. So you need sin(t/3) AND sin(2t/3) to both repeat.
It's not a matter of the i or j component you look at. The calculation would be the same if you swapped the i and j components.

So, sin(t/3) has period 6pi and sin(2t/3) has period 3pi. You can see that after 3pi, the j component has repeated itself but the i component hasn't.
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lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #3483 on: September 07, 2014, 11:10:55 pm »
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Thanks!

Whilst we're at it, can you please get me started on this question as well?

The position vectors of points A and B are:
a= i - j + k  b= 2i - 2j - 2k

Find the unit vector that bisects <AOB

Answer tells me to add a hat and b hat together. Why?

If you add two position vectors a and b, the point that is defined by the sum of the two vectors forms a parallelogram with the origin and the two other points. Now, if a and b are unit vectors, the four points O, a, b and a+b form a rhombus. The diagonal of a rhombus bisects the angles, so a+b will bisect the angle between a and b.

Alternatively, think of it this way. Consider the unit vectors u and v, which are position vectors of U and V respectively. Then consider the midpoint of the line segment that joins points U and V. The position vector of this point is 1/2 (u+v). You can check that yourself.
Now, as u and v are unit vectors, the triangle formed by O, U and V is isosceles. Also, the line from the apex to the base of an isosceles triangle bisects the base AND the apex angle. So yeah.
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BLACKCATT

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Re: Specialist 3/4 Question Thread!
« Reply #3484 on: September 08, 2014, 04:12:21 pm »
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Can anyone please help with 4a or 4b?
« Last Edit: September 08, 2014, 04:27:12 pm by BLACKCATT »

IndefatigableLover

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Re: Specialist 3/4 Question Thread!
« Reply #3485 on: September 08, 2014, 05:41:40 pm »
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A complex number z satisfies the inequality

Find the least possible value of |z|.

Find the greatest possible value of Arg z.

Thorium

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Re: Specialist 3/4 Question Thread!
« Reply #3486 on: September 08, 2014, 06:17:58 pm »
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Can anyone please help with 4a or 4b?

Are you sure it is resting? Coz in that case v=0 for all t
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Zealous

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Re: Specialist 3/4 Question Thread!
« Reply #3487 on: September 08, 2014, 07:35:59 pm »
+1
A complex number z satisfies the inequality

Find the least possible value of |z|.

Find the greatest possible value of Arg z.



The inequality defines a circle with center and radius of 2 units.

Have a look at the diagram. The least possible value of |z| will be the distance from the line connecting the origin to the center of the circle minus the radius of the circle.

So find the magnitude of the origin to the center of the circle, then take away the radius of the circle:


The maximum argument of z will be the argument of the line connecting the origin to the tangent to the circle (purple line) - hopefully the diagram makes that clearer! So you can see we've formed a right angled triangle (using a bit of geometry). We just need to find the angle between the blue line and the horizontal, and the angle between the blue and purple line to find our desired angle.

(this is the angle between the purple and blue line!)

Now we need to find the angle between the blue line and the x axis (horizontal):

To angle between the purple line and the horizontal is just the angle between the blue line and horizontal, minus the angle between the blue and purple line:



Hope that helps! (there may be better ways of doing this... not sure haha)
« Last Edit: September 08, 2014, 07:44:23 pm by Zealous »
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Re: Specialist 3/4 Question Thread!
« Reply #3488 on: September 08, 2014, 07:49:08 pm »
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Haha that's wonderful Zealous! I was working on it at school today and my teacher showed me the graphical way of working it out but I didn't really understand it since it was 2 minutes before the end of class LOL but your working has certainly cleared up things for me :)

Just to clarify, the final answer for the argument is due to the angle being in Quadrant 2 right?

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Re: Specialist 3/4 Question Thread!
« Reply #3489 on: September 08, 2014, 07:55:47 pm »
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Can someone please help me with this exam question? I'm having trouble understanding where to start (it's from exam 1 2012, question 4)

theshunpo

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Re: Specialist 3/4 Question Thread!
« Reply #3490 on: September 08, 2014, 08:56:09 pm »
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Can someone please help me with this exam question? I'm having trouble understanding where to start (it's from exam 1 2012, question 4)

For Question 4B) Start by splitting the tension force into horizontal and vertical components. The vertical component cannot exceed the force due to gravity. The normal force of the mass would be 0N, as the crate is on the verge of leaving the floor.
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i.e. T*sin(30)=50g
« Last Edit: September 08, 2014, 09:16:24 pm by theshunpo »
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Spxtcs

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Re: Specialist 3/4 Question Thread!
« Reply #3491 on: September 08, 2014, 09:05:51 pm »
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Thanks!

How would I begin part b? Would I resolve the components?
Thanks in advance

theshunpo

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Re: Specialist 3/4 Question Thread!
« Reply #3492 on: September 08, 2014, 09:11:24 pm »
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For Part C, first resolve horizontally to determine the force toward the right. This force must be equal to the frictional force. To work out the frictional force you would first have to resolve vertically to work out the normal/reaction force, then multiply this by the coefficient of friction (0.2). The horizontal force exerted on the mass must be exactly equal to the frictional force (this is the point where it would be on the point of moving). This would leave only T to solve for.
Pm me if you want me to send you the working.
« Last Edit: September 08, 2014, 09:16:37 pm by theshunpo »
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keltingmeith

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Re: Specialist 3/4 Question Thread!
« Reply #3493 on: September 08, 2014, 09:14:57 pm »
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For Part B, first resolve horizontally to determine the force toward the right. This force must be equal to the frictional force. To work out the frictional force you would first have to resolve vertically to work out the normal/reaction force, then multiply this by the coefficient of friction (0.2). The horizontal force exerted on the mass must be exactly equal to the frictional force (this is the point where it would be on the point of moving). This would leave only T to solve for.
Pm me if you want me to send you the working.

That's a lot of work for a 1 mark question - reconsider what you're actually trying to find the value of T for, then decide if you're going about it the wrong way.

theshunpo

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Re: Specialist 3/4 Question Thread!
« Reply #3494 on: September 08, 2014, 09:15:42 pm »
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That's a lot of work for a 1 mark question - reconsider what you're actually trying to find the value of T for, then decide if you're going about it the wrong way.
OH SORRY, let me edit that. I meant part C!!
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