Specialist 1/2 Question Thread!

[Shadowxo]

Hi I've been able to do this vector question and get the correct answer, however, I've been doing under the assumption the angle between the two is 0 (going in same direction). Can someone explain why this happens? Or am I wrong in my working out?
http://i.imgur.com/Q0ce9bp.jpg

Great answer by VanillaRice (beaten by 1 minute haha)
a.(a+b) = a.a +a.b
It's one of the properties of the dot product
So then you end up with (a.a +a.b -a.b) / |a| = a.a / |a| which results in the answer - you don't need to assume theta = 0

[Sine]

Great answer by VanillaRice (beaten by 1 minute haha)
a.(a+b) = a.a +a.b
It's one of the properties of the cross product
So then you end up with (a.a +a.b -a.b) / |a| = a.a / |a| which results in the answer - you don't need to assume theta = 0

I really hate to do this but it's the dot product not cross. Cross product is not studied in spec maths.

[VanillaRice]

I really hate to do this but it's the dot product not cross. Cross product is not studied in spec maths.

Nevertheless, distributivity is still a property of both cross and dot products 

[Shadowxo]

I really hate to do this but it's the dot product not cross. Cross product is not studied in spec maths.

Oops! Accidentally wrote cross instead of dot (been doing a lot of dot and cross product in maths lately). Applies to both but meant to say dot

[A TART]

You don't need to substitute a dot b in this case (you can, but a there is a simpler method). The dot product works very much like multiplication, so we can use that property in this question.
i.e.
With this in mind, have another go at the question. The most likely reason you are able to substitute theta = 0 is that when you do so, you are assuming the angle between a and b is 0, and also that the angle between a and (a+b) is 0 - note the two dot products are not the same.

Hope this helps

Thanks Makes it a whole lot easier

[lilhoo]

Hey guys, I'm a year 11 student wondering if it is ok to do spesh 3/4 without 1/2.
I have been recommended to do spesh 3/4 on MANY occasions by my methods and recently, physics teacher (both also teach spesh 3/4).
If i can arrange "catch-up lessons" with either of the above teachers as well as studying by myself, what are my chances of surviving Spesh 3/4? Thanks

[VanillaRice]

Hey guys, I'm a year 11 student wondering if it is ok to do spesh 3/4 without 1/2.
I have been recommended to do spesh 3/4 on MANY occasions by my methods and recently, physics teacher (both also teach spesh 3/4).
If i can arrange "catch-up lessons" with either of the above teachers as well as studying by myself, what are my chances of surviving Spesh 3/4? Thanks

Your chances are good
I don't think that 1/2 is required to have a good basis to do 3/4 (although this may now be different with the new study design..). The most important thing is that you're willing to put in the work    You'll learn a few new concepts (complex numbers, vectors), so it might be worth focusing on these if you're wanting to do any work before next year. Given that it's been suggested to you many times by the teachers that actually teach the subject, I think it might be worth considering (especially if you like maths  ).

[KiNSKi01]

These questions seem really straightforward but how do you figure out the different vectors and write them in correct format?

[Eric11267]

These questions seem really straightforward but how do you figure out the different vectors and write them in correct format?

Ok so unit vector i represents one unit in the positive x direction, whilst the unit vector j represents one unit in the positive y direction. Moving from C to D is going 6 units in the negative x direction, so it is -6i. Moving from C to A is going 7 units in the negative x direction and 3 units in the positive y direction, so it is -7i+3j. Going from C to B is moving one unit in the negative x direction and three units in the positive y directoin, so it is -i+3j.

[KiNSKi01]

Cheers Eric, I'm familiar with how to calculate the scalar product of vectors in two dimensions but I'm not completely sure on how to tackle questions with 3 dimensions. I assume its a similar process but I don't have answers to the questions 

[Eric11267]

Cheers Eric, I'm familiar with how to calculate the scalar product of vectors in two dimensions but I'm not completely sure on how to tackle questions with 3 dimensions. I assume its a similar process but I don't have answers to the questions 

Yes its a similar process, just add the products of the corresponding coefficients for each unit vector. So for part a, it would ge -2 x 4 + -5 x -1 + 1 x 3 = 0

[Shadowxo]

Cheers Eric, I'm familiar with how to calculate the scalar product of vectors in two dimensions but I'm not completely sure on how to tackle questions with 3 dimensions. I assume its a similar process but I don't have answers to the questions 

Yep, Eric's right. Just a bit of background knowledge:
We only multiply the parallel components (that is, i with i, j with j, k with k) because theta = 0 so cos(theta)=1. The perpendicular components (eg i and j) are perpendicular so cos(theta)=0. You should remember that a.b=|a||b|cos(theta) which is why we do this.
So (2i+3j).(2i+3j)=2i.2i+2*2i.3j+3j.3j = 4+0+9 = 13. (You don't have to do this in the exam or anything)
So just multiply the coefficients for each unit vector, as Eric said
You can do this for vectors in as many dimensions as you like, even higher than 3

[KiNSKi01]

So (2i+3j).(2i+3j)=2i.2i+2*2i.3j+3j.3j = 4+0+9 = 13. (You don't have to do this in the exam or anything)
So just multiply the coefficients for each unit vector, as Eric said

I'm not exactly sure what you have done here    Don't u just multiply the coefficients in front of i and j so you get 2*2 + 3*3 = 14.
I know it is the same answer but where does the zero come from?

[Shadowxo]

I'm not exactly sure what you have done here    Don't u just multiply the coefficients in front of i and j so you get 2*2 + 3*3 = 14.
I know it is the same answer but where does the zero come from?

Sorry it's not exactly necessary to show it just thought some background info would be nice
What I did was expand it out, and just wanted to show that i.j =0 as they're perpendicular (same with i.k and j.k) which is why you only multiply the coefficients in front of the same unit vectors, not the different ones.

[Eric11267]

I'm not exactly sure what you have done here    Don't u just multiply the coefficients in front of i and j so you get 2*2 + 3*3 = 14.
I know it is the same answer but where does the zero come from?

Adding up the products of the coefficients only works in unit vector notation because the vectors involved have magnitude 1 and each unit vector is perpendicular to the other two. Later you'll encounter questions that will involve the dot product of vectors which are not in unit vector notation (such as in vector geometry proof questions). Its important to know about the properties of the dot product.
For example if I have vectors a,b,c then:
 c· (a+b)= c·a+c·b   or
(c+a)·(c+b)= c·c+c·b+a·c+a·b or
c·c=|c|²   etc
I think its important you learn the definition of the dot product, because it will be useful in the future