VCE Specialist 3/4 Question Thread!

[chisel]

Having trouble with vector proofs 
anyone keen to give these a shot?

1) Prove that if the midpoints of a rectangle re joined, then a rhombus is formed.
2) Prove that the median to the base of an isosceles triangle is perpendicular to the base.
3) Prove that the midpoint of the hypotenuse of a right-angled triangle is equidistant from the three vertices of the triangle.

And also, do any of you have any general tips on how to approach these proof questions?

Thank you so much for any help whatsoever 

[polar]

find vectors that represent objects given in the question (e.g. sides, angles...) and then use properties of the shape named to relate these objects

question 2

let the two sides of equal length be vectors and . since they are of the same length, . next, we should find a vector for the median and a vector for the base. the base is the vector joining the two sides hence, . the vector for the median is (the midpoint of the base). finally, we want to show that they are perpendicular. to do this, we can show that the dot product of the two vectors is 0:
and since , hence, the median to the base of an isosceles triangle is perpendicular to the base

also, sometimes it's easier to work with side lengths when you write them as dot products

question 3

let the two shorter sides be represented by vectors and . we know that these are perpendicular, so . the hypotenuse is the vector joining the tail of a with the head of b. hence, . the midpoint of the hypotenuse is . now, with all this information we can do the question
the distance from the first vertex to the midpoint is (I used the origin as the first vertex). now, the length of the hypotenuse is which is also the length from the origin to the second vertex. subtracting this from the distance from the midpoint to the first vertex gives you . the final vertex is at (a,0). the vector joining this vertex to the midpoint is how do we find the length of this? write it as a dot product. hence, the length of this vector is (since ) as well. hence, all vertices are the same distance from the midpoint of the hypotenuse

[Will T]

Essentials Exercise 3C question 8c.

Help please, unless the answer is wrong.

[zvezda]

Would help you will if i knoew how to get graphs onto here.
Would anyone be able to explain to me the significance of the dot product? As in, what is the point of it? and how would be represent such an operation visually?

[507]

Draw your two vectors in two dimensions like so:



essentially represents the amount of in the direction of
So you could think of it as the amount of in the direction of , multiplied by .
Note that it's a scalar.

[sin0001]

While you're at it, can you please explain the significance of the cross product too

[pi]

While you're at it, can you please explain the significance of the cross product too

Re: BORED already? Want to learn a bit of maths? :P

[~T]

Question 25b) of the Chapter 1 review. I did it a while ago and found it interesting, but never thought to post it here.

Given from the previous part that the sum of a sequence is the question asks:
Show that the sum of the sequence is divisible by 14 only when p is odd.

Now, the first thing I did, purely to make it easier in my head, was to state that for to be divisible by 14, must be divisible by 2. But then what? It is easy to see that it will be divisible by 2 when p is odd, and not when it is even, but how would one go about "proving" that mathematically?

Thanks

[~T]

Actually, just had a thought. Would letting and showing that it turns out odd (not divisible by 2) and then letting and showing that it is even (divisible by 2) work?

[Will T]

Pretty sure for 3D question 5j. of the essentials the answer is wrong.
Solve for x





Let , where


But the answer at the book of the book has it as

Is the book wrong or am I losing my mind?

[pi]

Book looks fine to me

For example: sin(pi) + cos(pi) = 0 - 1 = -1, which isn't the "1" required by the question

(i'm assuming sin(x) + cos(x) = 1 is the question)

edit: typo oops

[Will T]

So what's the method of solving this? Or do we have to discount solutions that don't work?

[Bad Student]

You need to reject solutions that don't satisfy the first equation.

[Conic]

So what's the method of solving this? Or do we have to discount solutions that don't work?

For this question you could collect into one funcition, , and solve that for 1, which gives you , , and  .

[zvezda]

question from Ex 2C in the essentials textbook:
expand and simplify modulus(a+b)^2 minus modulus(a-b)^2
By the way, this is to do with the scalar product so a and b are vectors.
Help is much appreciated, because im simply stumped

edit: figured it out.