it's so hard :(

[asdfqwerty]

The parallelogram OXYZ has O at the origin.The vector joining O to Z is given by 5i while the vector joining O to X is given by 2i + 7j. Let P be a point on the extended line of XY, such that the vector joining P to Z is perpendicular to OY. Find the coordinates of P.

Thanks

[kinslayer]

OY = OX + OZ = 7i + 7j

If P is "on the extended line XY" then it will have j-coordinate 7. Secondly PZ must have gradient -1 since it is perpendicular to OY.

So, we know that PZ will be of the form xi - 7j and also that -7/x = -1. So, x = 7. Since OP + PZ = OZ, we find that the coordinates of point P are -2i + 7j.

[b^3]

This might help you visualise it a bit easier.

[asdfqwerty]

thank you everyone, but how did you get:
"we know that PZ will be of the form xi - 7j"

[alchemy]

thank you everyone, but how did you get:
"we know that PZ will be of the form xi - 7j"

kinslayer has skipped a few steps there. I'll try fit them in.
So first we get OY=OX+XY=OX+OZ=2i+7j+5i=7i+7j
For two vectors to be perpendicular their dot product has to equal to zero.
Hence, PZ . OY = 0.
PZ . (7i+7j) = 0
PZ= PO + OZ
Let the horizontal axis coordinate of OP be xi.
PZ=(-xi-7j) + 5i = (5-x)i + 7j
PZ . OY = ((5-x)i -7j) . (7i + 7j) = 0
7(5-x) - 49 =0
-7x=14
x=-2
So, OP = -2i+7j

[bananasinpjs]

please help me too
the third term of an arithmetic sequence is 5/8 and the sixth term is 7/16. Find the sum of the first 12 terms

[Zealous]

please help me too
the third term of an arithmetic sequence is 5/8 and the sixth term is 7/16. Find the sum of the first 12 terms

Formula for arithmetic sequence:


Fill in the information we know:



Solve simultaneously (using your preferred method):


Formula for sum of an arithmetic sequence:


Substitude our a value, our d value and n=12 as we want the sum of the first 12 terms:



Therefore the sum of the first 12 terms is 39/8.

^

I never got to give b^3 a mention, some of his replies to my and many others' questions whilst I was in year 12 saved me so much stress. Absolute legend.

+1 b^3 is so dedicated in the Mathematics boards, I really appreciate the effort he puts in.

[1MIN]

wow!!! you're good

Please help me too!!!

Question 8c) and d)

Thank you

[Special At Specialist]

please help me too
the third term of an arithmetic sequence is 5/8 and the sixth term is 7/16. Find the sum of the first 12 terms

You don't even need to do this formulaically, just look for a quick pattern:
3rd number = 5/8 = 10/16
6th number = 7/16
So we can fit 2 numbers evenly spaced between them: 9/16 and 8/16. These are the 4th and 5th numbers respectively.
Following on:
7th number = 6/16
8th = 5/16
...
12th = 1/16
Going backwards:
2nd = 11/16
1st = 12/16

So for the sum of the first 12 terms, we want to calculate 1/16 + 2/16 + ... + 12/16
Sum = (1 + 2 + ... + 12) / 16
Sum = 78 / 16
Sum = 39/8

wow!!! you're good

Please help me too!!!

Question 8c) and d)

Thank you

A = r(t) = (2 + t)*i + (1 + t)*j
B = (5, 6) = 5i + 6j

PART A)
AB = AO + OB = OB - OA = B - A
AB = 5i + 6j - (2 + t)*i - (1 + t)*j
AB = (3 - t)*i + (5 - t)*j

PART B)
Distance of AB = sqrt((3 - t)^2 + (5 - t)^2)
Minimum distance at d/dt (AB) = 0
Since AB is a square root, just let the derivative of the argument equal zero.
d/dt ((3 - t)^2 + (5 - t)^2) = 0
-2(3 - t) - 2(5 - t) = 0
(3 - t) + (5 - t) = 0
8 - 2t = 0
t = 4   (seconds)

PART C)
Dog is one second infront of woman, so let y(t) = r(t + 1)
C = y(t) = r(t + 1) = (3 + t)*i + (2 + t)*j
Now to find angle ABC we need to find vector AB and vector BC:
AB = (3 - t)*i + (5 - t)*j   (as calculated in part a)
BC = BO + OC = OC - OB = C - B
BC = (3 + t)*i + (2 + t)*j - 5i - 6j
BC = (t - 2)*i + (t - 4)*j
Now find angle ABC using the dot product:
a.b = |a||b|cos(θ)
θ = arccos(a.b / (|a||b|))
θ = arccos(((3 - t)(t - 2) + (5 - t)(t - 4)) / (sqrt((3 - t)^2 + (5 - t)^2) * sqrt((t - 2)^2 + (t - 4)^2)))
Max at dθ/dt = 0, t >= 0
If d/dt (arccos(f(t))) = 0 then -f'(t) / sqrt(1 - (f(t))^2) = 0 and thus f'(t) = 0
So once again, just let the derivative of the argument equal zero:
d/dt (((3 - t)(t - 2) + (5 - t)(t - 4)) / (sqrt((3 - t)^2 + (5 - t)^2) * sqrt((t - 2)^2 + (t - 4)^2))) = 0
Solve this on your CAS calculator to get:
t = 7/2   (seconds)

I can't see part D anywhere.

[1MIN]

just wanted to make sure isn't BA . BC = 2t^2 -14t +26?

this is what i did:
BA = (-3 +t)i + (-5+t)j
BC = (-2+t)i + (-4+t)j
BA . BC = (-3 + t)(-2 +t) + (-5+t)(-4+t)
             = 6 -5t + t^2 +20 - 9t +t^2
               = 2t^2 -14t +26

but then if i use this it gives me the in correct ans. if i used 2t^2 -14t -14 it gives me the right ans t = 7/2
was there meant to be a negative in front of the 20?

[Special At Specialist]

just wanted to make sure isn't BA . BC = 2t^2 -14t +26?

this is what i did:
BA = (-3 +t)i + (-5+t)j
BC = (-2+t)i + (-4+t)j
BA . BC = (-3 + t)(-2 +t) + (-5+t)(-4+t)
             = 6 -5t + t^2 +20 - 9t +t^2
               = 2t^2 -14t +26

but then if i use this it gives me the in correct ans. if i used 2t^2 -14t -14 it gives me the right ans t = 7/2
was there meant to be a negative in front of the 20?

You're right, but the answer is the same either way.
I did AB . BC when I should've done BA . BC.
If I expand (3 - t)(t - 2) + (5 - t)(t - 4) I get -(2t^2 - 14t + 26), so I get the negative of what you got. But when I let the equation equal zero (or in this case, the derivative equal to zero), having a negative value equal to zero is the same as having a positive value equal to zero, so we end up with the same answer.

I'll show you on Wolfram Alpha that both ways give you t = 7/2 as the answer:
First way (the AB . BC way):
https://www.wolframalpha.com/input/?i=solve+d%2Fdt+%28%28%283+-+t%29%28t+-+2%29+%2B+%285+-+t%29%28t+-+4%29%29+%2F+%28sqrt%28%283+-+t%29%5E2+%2B+%285+-+t%29%5E2%29+*+sqrt%28%28t+-+2%29%5E2+%2B+%28t+-+4%29%5E2%29%29%29+%3D+0
Second way (the BA . BC way):
https://www.wolframalpha.com/input/?i=solve+d%2Fdt+%28%28%28-3+%2B+t%29%28-2+%2Bt%29+%2B+%28-5%2Bt%29%28-4%2Bt%29%29+%2F+%28sqrt%28%283+-+t%29%5E2+%2B+%285+-+t%29%5E2%29+*+sqrt%28%28t+-+2%29%5E2+%2B+%28t+-+4%29%5E2%29%29%29+%3D+0