VCE Specialist 3/4 Question Thread!

[manishmao]

Thanks. I misread the question, thought it was asking for sloping faces angle not sloping edges, which is way easier. Thanks. However, your answer is different to textbook ... which has 45 degrees and a tan^-1(5/3) as well as tan^-1(3/4). Tan^-1 (5/3) is the angle that the sloping face BAD has with horizontal. Dk why they solved for that.

[Syndicate]

If I understood the question correctly..

Apex D is above C not B. Which means angle CAD is 45 degrees, as triangle CAD is a right angle isosceles triangle.

[manishmao]

Yes don't worry I solved it. I misread sloping edge as sloping face for some reason. (Though one of the angles is a the angle of the sloping face for some reason)

[peanut]

When resolving a vector into two vectors, one parallel to another and one perpendicular to another, how do you refer to each? For example, if I resolve a into two vectors, one parallel to b and the other perpendicular, can I refer to one as a_⊥ and a_||? Will I need to state which is which (like "Let a_⊥ be the vector resolute of a in the direction perpendicular to b")

[wyzard]

When resolving a vector into two vectors, one parallel to another and one perpendicular to another, how do you refer to each? For example, if I resolve a into two vectors, one parallel to b and the other perpendicular, can I refer to one as a_⊥ and a_||? Will I need to state which is which (like "Let a_⊥ be the vector resolute of a in the direction perpendicular to b")

The notation  a_⊥ and a_|| works perfectly well for vector resolute of a. Usually it's a good idea to include a statement "Let a_⊥ be the perpendicular vector resolute of a".

Alternatively, you can draw a diagram showing the vector resolutes a_⊥ and a_||. It can be inferred from the diagram you're referring to the vector resolutes.

However if you're pressed for time, simply writing a_⊥ and a_|| is okay as it's pretty obvious what they're referring to.

[peanut]

How did the textbook go from image 1 to image 2?

[Vaike]

Hey guys was just wondering if someone would be able to explain how to do these sorts of questions. I know you it can be solved on cas and/or graphically, but I was trying to do it algebraically and really didn't understand the textbooks worked solutions. Thanks

[Syndicate]

How did the textbook go from image 1 to image 2?

They have just swapped it around

Hey guys was just wondering if someone would be able to explain how to do these sorts of questions. I know you it can be solved on cas and/or graphically, but I was trying to do it algebraically and really didn't understand the textbooks worked solutions. Thanks

So first of all you need to work out the crucial points (these are the points where one of the absolute value part is equal to 0) of the question.

|2x-5| = 0
2x-5 = 0
x = \(\frac{5}{2} \)

and
|4-x| = 0
4-x = 0
x=4

So basically you can separate the domains of the question into three parts.

If the domain is less than 5/2, than both absolute values are negative.
If the domain is in between 5/2 and 4, then only |2x-5| is positive.
If the domain is larger than 4, than both absolute values are positive.

You also need to remember that |-x| = |x|, so if you have a -x in you absolute value than you will need to convert it into a positive one. ie |3-2x| = | 2x-3|
Case 1: x \( \leq \space \frac{5}{2} \)
This means both absolute values are negative.
-(2x-5) - -(x-4) = 10
-2x+5 +x-4 =10
-x = 9 => x = -9 - This is a solution as -9 is smaller than 5/2

Case 2: \( \frac{5}{2} \leq x \leq4 \)
Meaning |2x-5| is positive because for all values larger than 5/2, |2x-5| will yield a positive value, and |x-4| is negative.
2x-5 - -(x-4) = 10
2x-5 +x -4 = 10
3x -9 = 10
x = 19/3 (the book made a mistake, however this is still not a solution, as 19/3 is not a part of 5/2 \( \leq\) x \(\leq\) 4)

Case 3: 4 \(\leq\)x
Meaning both absolute values are positive.
2x-5 - (x-4) = 10
x -1=10
x =11 (This is a solution as 11 is larger than 4)

Therefore the solutions for this question are 11 and -9.

[Vaike]

Therefore the solutions for this question are 11 and -9.

Thank you so much I think I understand. If you've got the time I'd appreciate looking over this to check my understanding

|x-3|+|5-x|=6.
In this case, the crucial points would be x=3 and x=5, and the equation may be rearranged to |x-3|+|x-5|=6.
So the domain can now be split into x ≤ 3, 3≤x≤5, and 5≤x.
If x is less than or equal to 3, both absolute values are negative, and hence solving gives x=1. This solution is valid as 1≤3.
If x is between 3 and 5, |x-3| is positive but |5-x| is negative, and when solving x gets cancelled and hence no solutions present.
If x is greater than or equal to 5, both absolute values are positive, and solving gives x=7. This solution is valid as 5≤7.

I think this is right, however I don't quite understand why a solution must be within the bounds of the domain to be valid, is it because during the solving process 'additional answers' are created? Also, is there any explainable;e reasoning behind the crucial points forming the basis of the 3 domains? Thanks so much again

[Syndicate]

Thank you so much I think I understand. If you've got the time I'd appreciate looking over this to check my understanding

|x-3|+|5-x|=6.
In this case, the crucial points would be x=3 and x=5, and the equation may be rearranged to |x-3|+|x-5|=6.
So the domain can now be split into x ≤ 3, 3≤x≤5, and 5≤x.
If x is less than or equal to 3, both absolute values are negative, and hence solving gives x=1. This solution is valid as 1≤3.
If x is between 3 and 5, |x-3| is positive but |5-x| is negative, and when solving x gets cancelled and hence no solutions present.
If x is greater than or equal to 5, both absolute values are positive, and solving gives x=7. This solution is valid as 5≤7.

I think this is right, however I don't quite understand why a solution must be within the bounds of the domain to be valid, is it because during the solving process 'additional answers' are created? Also, is there any explainable;e reasoning behind the crucial points forming the basis of the 3 domains? Thanks so much again

You have done it correctly 

Whenever you have squares, modulus/ absolute values and domains involved, there is a possibility of getting extraneous solutions. So you always need to check if your solutions are a part of the domain. 

Crucial points are basically the points where a new line is form in a complex absolute value function.

For example:


In the absolute value function above, the crucial points are -1 and 1. An absolute value is basically like a piecewise function, which means you can separate it into different pieces. It would be smart to separate the function above into three different domains, and then find the equations of those parts, in order to solve for whatever you are looking for.

[Gogo14]

What does "describe the subset of the complex plane" mean?

[keltingmeith]

What does "describe the subset of the complex plane" mean?

You should be familiar with what a set is - for example, the real numbers, the set of integers, etc. A subset is just a second set that is contained by the first set. So, the set of integers is a subset of the real numbers. (0, 2) is a subset of [0, 2], etc.

So, they're just asking you to describe your shape, because any curve you draw in the complex plane is going to be a subset of the complex plane.

[wyzard]

What does "describe the subset of the complex plane" mean?

Think about it this way, a complex plane contains the entire complex number on a cartasian coordinate. Each of the sets in the question imposes some condition for the complex numbers to meet. That set then contains all the complex numbers where said condition is met, which will be a collection of points on the complex plane.

Very often the shape will then be some geometric shape, like a circle, a straight line, ellipses etc.

That is what it meant by describing the subset of the complex plane

[Quantum44]

Can someone help me with these three questions?

[zhen]

Can someone help me with these three questions?

Here's my answer to the questions