Last part of question 9

[Special At Specialist]

It asked us to find the angle between the two vectors in terms of pi. I used the dot product and ended up with:
cos(θ) = (sqrt(2) - sqrt(6)) / 4
Then I took a guess and said θ = 7pi/12
Was that right??? What answer did you get and how did you get it?

[Special At Specialist]

Also, did you all get 5*sqrt(7)/16 for the last part of question 10?

[mikesguns]

yeh got the same for both. last part of 9 was 7pi/12, but i used the fact that i knew (sqrt(6)-sqrt(2)) / 4 was 5pi/12, and found the negative of that which was 7pi/12. However im not sure you will get full marks for it, not sure if i will either

[Deleted User]

Also, did you all get 5*sqrt(7)/16 for the last part of question 10?

Yes, but I forgot to simplify it. I wrote (10 root7)/32.

[appledesu]

Would you lose a mark if you didn't write 7pi/12?

[TheRajinator]

how did you know that (sqrt(6)-sqrt(2)) / 4 is 5pi/12?

[DJing]

Just having to prove the exact value of pi/12 in the past meant that it was possible to do it by recognition

However, the double angle formula 'recognition' could also be used where

cos(x) = (root(2)-root(6))/4 = (1/2)*(sqrt(2)/2) - (sqrt(3)/2)(Sqrt(2)/2) = cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)=cos(7pi/12)

so x=7pi/12

[#1procrastinator]

Sheeeeeeeeit I just applied cos right away forgot how it has to be right triangle lol my brain was in moronland

[datfatcat]

Also, did you all get 5*sqrt(7)/16 for the last part of question 10?

Yes, but I forgot to simplify it. I wrote (10 root7)/32.

And I didn't simplify the root (mine was not 7, it was 4 sth...)

[Deleted User]

I found Arg of the first vector and subtracted the Arg of the second vector. Answer = 7pi/12

[BigAl]

I think you could use cos double angle formula and get the answer..I didnt have time so I basically draw the angles and take away from pi..as a result 7pi/12

[DJing]

You also could have used vector resolutes to find the right-angled triangle and then simple trig (i think)

But yeah, there were many possible methods, but they all arrive at the same answer

[malekv]

fuuu i got
pi/3

isn't it velocity vector?
gradient at that point?
fuuu what was the exact wording of the question ? GG

Me too. Sheeeeeeeeeeeit.

[WhoTookMyUsername]

It asked us to find the angle between the two vectors in terms of pi. I used the dot product and ended up with:
cos(θ) = (sqrt(2) - sqrt(6)) / 4
Then I took a guess and said θ = 7pi/12
Was that right??? What answer did you get and how did you get it?

took a guess...

[WhoTookMyUsername]

Just double checking,
but am i right in saying to find the angle a particle makes with the ground (projectile motion) you find v(t)
then find the angle between this and the ground?