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pastyear:
how to do CIE Paper 3 May/June 2008 Question 10 (ii) the part of show that 3t square + 7t +2 =0
thanks.

Question at below:
The points A and B have position vectors, relative to the origin O, given by
OA = i + 2j + 3k    and       OB = 2i + j + 3k.

The line l has vector equation
r = (1 ? 2t)i + (5 + t)j + (2 ? t)k.

(i) Show that l does not intersect the line passing through A and B. [4]

(ii) The point P lies on l and is such that angle PAB is equal to 60?. Given that the position vector
of P is (1 ? 2t)i + (5 + t)j + (2 ? t)k, show that 3t2 + 7t + 2 = 0. Hence find the only possible
position vector of P. [6]

pastyear:
anyone ? Plz write your explanations and workings.
Thanks

Twinkle Charms:
they say anglePAB=60 which means the two vectors which we'll use are AP and AB

AP=OP-OA and AB=OB-OA

then apply scaler product, ull reach to -3(t+1)=[rootof (6t^2+8t+10)][1/root2] , square both sides and simplify ull get 3t^2+7t+2=0

to get the only position vector of P, u solve this equation, ull get t=-2 and t=-1/3, t=-1/3 is rejected since itll give u obtuse, whereas we have 60 degrees, so substitute t=-2 in OP ull get teh only position vector of P.

pastyear:
How do you get this:

then apply scaler product, ull reach to -3(t+1)=[rootof (6t^2+8t+10)][1/root2]

Twinkle Charms:
look when you get the two vectors, you use them in the scaler product thing,

AB.AP=|AB|.|AP|costheta

get it?

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