Math Vectors

[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?

[pastyear]

ya
P of AB =( 1  -1  0)
P of PA  =( 2t    -3-t    1+t)
P AB  dot P PA = 3t+3

Mod P AB = square root 2
Mod P PA = square root 6tsquare + 8t + 10

[pastyear]

cos 60 = 1/2  = mod 3t+3 / square root of  6t square  + 8t + 10 times square root 2 Mod

[Twinkle Charms]

ya
P of AB =( 1  -1  0)
P of PA  =( 2t    -3-t    1+t)
P AB  dot P PA = 3t+3

Mod P AB = square root 2
Mod P PA = square root 6tsquare + 8t + 10

look thts where ur makin mistake , its not PA, its AP

[pastyear]

still cannot get
i get 3t square + 8t  +2

[Twinkle Charms]

AP=(-2t  3+t  -1-t) and AB=(1  -1  0)

did you get these vectors??

(-2t   3+t   -1-t).(1  -1  0)= |AP|.|AB| cos60

|AP| means modulus of AP = rootof [(-2t)^2  +  (3+t)^2  +  (-1-t)^2] and similarly for |AB|.

did you do it this way?

[pastyear]

ya
can you type out the next few steps
Thanks

[Twinkle Charms]

yes i mentioned the rest of the steps in my 1st post, read it carefully ull get it, its simple =)

[CHEMMASTER6000]

wow zara, thought you had issues with vectors

[Twinkle Charms]

yea buh its simple one right?

i dunno i was studin vectors in the morning, no doubt yet, if i do ofc ill come here..=)