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Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: pastyear on May 22, 2010, 02:20:47 pm

Title: Math Vectors
Post by: pastyear on May 22, 2010, 02:20:47 pm: 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]
Title: Re: Math Vectors
Post by: pastyear on May 22, 2010, 02:28:00 pm: anyone ? Plz write your explanations and workings.
Thanks
Title: Re: Math Vectors
Post by: Twinkle Charms on May 22, 2010, 04:02:09 pm: 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.
Title: Re: Math Vectors
Post by: pastyear on May 22, 2010, 04:10:35 pm: How do you get this:

then apply scaler product, ull reach to -3(t+1)=[rootof (6t^2+8t+10)][1/root2]
Title: Re: Math Vectors
Post by: Twinkle Charms on May 22, 2010, 04:11:58 pm: look when you get the two vectors, you use them in the scaler product thing,

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

get it?
Title: Re: Math Vectors
Post by: pastyear on May 22, 2010, 04:16:23 pm: 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
Title: Re: Math Vectors
Post by: pastyear on May 22, 2010, 04:19:11 pm: cos 60 = 1/2 = mod 3t+3 / square root of 6t square + 8t + 10 times square root 2 Mod
Title: Re: Math Vectors
Post by: Twinkle Charms on May 22, 2010, 04:25:43 pm: Quote from: pastyear on May 22, 2010, 04:16:23 pm
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
Title: Re: Math Vectors
Post by: pastyear on May 22, 2010, 04:37:48 pm: still cannot get
i get 3t square + 8t +2
Title: Re: Math Vectors
Post by: Twinkle Charms on May 22, 2010, 04:52:59 pm: 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?
Title: Re: Math Vectors
Post by: pastyear on May 22, 2010, 05:02:01 pm: ya
can you type out the next few steps
Thanks
Title: Re: Math Vectors
Post by: Twinkle Charms on May 22, 2010, 05:14:49 pm: yes i mentioned the rest of the steps in my 1st post, read it carefully ull get it, its simple =)
Title: Re: Math Vectors
Post by: CHEMMASTER6000 on May 22, 2010, 05:36:53 pm: wow zara, thought you had issues with vectors
Title: Re: Math Vectors
Post by: Twinkle Charms on May 23, 2010, 09:53:15 am: 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..=)