Specialist 1/2 Question Thread!

[KiNSKi01]

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)

Oh whoops sorry. Yeah I get ya know. Thanks Shadow and Eric for the help and advice! 

[KiNSKi01]

When calculating the unit vector do you need to first take out a common factor of the coefficients of 'i' and 'j'

e.g a= 6i + 4j goes to 1/root13 (3i + 2j)

[LifeisaConstantStruggle]

doesn't really matter you can take out the common factor at the end of your calculation and simplify it.

[KiNSKi01]

How do you calculate the magnitude of vector AB when you don't know the individual 'i' and 'j' values of a and b?

[LifeisaConstantStruggle]

How do you calculate the magnitude of vector AB when you don't know the individual 'i' and 'j' values of a and b?

The dot product, a.b=0, which indicates that these two vectors are perpendicular. a (OA) and b (OB) would form the vertices of a right-angled triangle, and AB would be the hypotenuse of it. The magnitude of AB would be the length of the hypotenuse.

[Sine]

How do you calculate the magnitude of vector AB when you don't know the individual 'i' and 'j' values of a and b?

a.b = 0 means that vector a and b are perpendicular
we know the length of a is 40 and the length of b is 9
we can determine the length of AB via tthe pythagorean theorem (AB is just the pathway between A and B )

EDIT: didn't even realise someone posted the solution (must've missed that WARNING message)

[KiNSKi01]

How do you approach a question like this. I have tried to split up A into its perpendicular and parallel components in relation to B but I don't know what to do from there 

[Sine]

How do you approach a question like this. I have tried to split up A into its perpendicular and parallel components in relation to B but I don't know what to do from there 

draw a large diagram to scale it may help you see things you haven't considered before

[KiNSKi01]

I drew out a large diagram. Am I right in saying you want to calculate the component of (i +j) which is perpendicular to (3i + j); meaning you would find the dot product of these two position vectors divided by the magnitude of (3i +j)^2. When I do this I just get a fraction 2/5 instead of root17/5 (the answer to the question). What am I missing here?

[KiNSKi01]

Any help would be great 

[LifeisaConstantStruggle]

Any help would be great 

Tbh I don't think the answers are correct? The shortest distance of the point (1,1) and the vector 3i+j would be the vector resolute of i+j in the direction perpendicular to 3i+j, but no answer seems to be correct.

[VanillaRice]

I drew out a large diagram. Am I right in saying you want to calculate the component of (i +j) which is perpendicular to (3i + j); meaning you would find the dot product of these two position vectors divided by the magnitude of (3i +j)^2. When I do this I just get a fraction 2/5 instead of root17/5 (the answer to the question). What am I missing here?

I agree with LifeisaConstantStruggle - I haven't actually been able to get the answer for this question either. However, I will show you my thought process:
You indeed looking for the vector component of OA which is perpendicular to OB. What you've calculated (2/5) is the scalar component of OA parallel to OB i.e. the vector projection of OA onto OB. Let us call the the component of OA parallel to OB a vector u.
u = 2/5<1,1>
So, the vector component of OA perpendicular to OB must be given by OA - u. The magnitude of this vector should give you the answer, however I get sqrt(10)/5.

[KiNSKi01]

Wouldn't be the first time I've been given work with incorrect answers 

What you said is exactly what I did and wouldn't the answer include 'i' and 'j' in it anyway since it doesn't ask for magnitude?

[VanillaRice]

Wouldn't be the first time I've been given work with incorrect answers 

What you said is exactly what I did and wouldn't the answer include 'i' and 'j' in it anyway since it doesn't ask for magnitude?

If you think about it, the distance (or length) implies magnitude

[KiNSKi01]

lol yeah true. Oh well it was an impossible question anyway