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C4 Stucks - vector + trig questions

Hi could do with some guidance on these 2 questions:

1. relative to fixed origin O, the points A and B have position vectors

1 and 6
5 3
-1 -6 respectively.

find in exact simplified form

(i) cosine of angle AOB. Am familiar with using cos theta = dot product of a and b divided by mod A x mod B but this is 3D so what do i do with the position vectors I have been given to find cosine AOB?
(ii) the area of triangle AOB. prob 1/2absinC
(iii) the shortest distance from A to the line OB. no idea

2. (i) use the derivatives of sin x and cos x to prove that

d/dx (tan x) = sec(^2)x

(ii) the tangent to the curve y = 2xtanx at the point where x=pi/4 meets the y axis at the point P. find the co-ordinate of P in the form kpi^2 where k is a rational constant. btw only the pi is squared - not the k.
Reply 1
the first one is a bit boring...

2i) d (tanx) / dx = d (sinx/cos) dx = [cosx(cosx) - sinx(-sinx)]/(cos^2 x) = sec^2 x

you should probably show a few more steps when you're given the answer. but its just the quotient rule...

ii) dy/dx = 2xsec^2 x + 2tanx
(x=pi/4) = (pi/2)2 + 2 = 2+pi

y = pi/2

so:

y = (2+pi)x + c
(pi/4,pi/2)
pi/2 = pi/2 + pi^2 / 4 + c
c = -pi^2 /4

so k = -1/4
Reply 2
thank u. yes i shud be able to do vector qs but unfortunately i can't yet - first past paper we doing so not got hang of the tricks :smile:
Reply 3
i) A.B= (6 x 1) +(5 x 3) + (-6 x -1) = 27

l Al = (1+25 +1)^0.5 = 27^0.5
lBl= (36+36+9)^0.5 = 81^0.5 = 9

cos AOB= 27/ 9(27)^0.5= 27^0.5/9= 3(3)^0.5/9= 3^0.5/3
Reply 4
ii) 0.5 x 27^0.5 x 9 x sinAOB (which you can work out from my answer above, cause i haven't got a calc to hand)
Reply 5
can i ask, how come u don't use the Z co-ordinate???
Reply 6
iii) the shortest line from A to OB must be perpendicular to OB. Say it touches at X.
AX.OB=0
Reply 7
i did use the Z coordinate. i don't think my workings clear- sorry :smile:
Reply 8
thanks a lot. quite impressed with a 15 yr old helping me out
Reply 9
u dont know for sure hes 15 he cud have lied when signing up.

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