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alright, fortunately i only know the answer to the first part and third part but i am sorry i can't find the correct solution of the second part..http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf
Please help with entire Q10
thank you
alright, fortunately i only know the answer to the first part and third part but i am sorry i can't find the correct solution of the second part..
(i) u substitute the equation of line l in the equation of plane p since they both intersect so
(i+6j-3k+s(i-2j+2k)-3i)(2i-3j+6k) = 0
((-2+s)i+(6-2s)j+(-3+2s)k)(2i-3j+6k) = o
do the scalar product of these 2 and u will get s = 2
now x = 1+2 y = 6-2(2) z=-3+2(2)
so the position vector of A is 3i+2j+k
part iii do uk this common prependicular normal thingy..apply it to these 2 i-2j+2k and 2i-3j+6k to find the direction vector of the line which lies in P
the direction vector u will get is -6i-2j+k and since it passes through A it will have a position vector of 3i+2j+k
now the equation of the line will be 3i+2j+k+t(-6i-2j+k)
now i am off to bed peace!!
Lamda will be denoted as t
You know that P is a point on L , so it has x,y,z coordinates as follows : (1 + 2t), (1 + t), (-1 + 2t) , and m = x + 2y -2z = 1 and n = 2x - 2y + z = 7
Using the formula they provided, you find the distance of P to m and of P to n keeping in mind that they should be equal, so distance of P to m
| (1 + 2t) + 2 (1 +t) -2(-1 + 2t) - 1 | / square root ( 1^2 + 2^2 + (-2)^2 ) to get |4|/3
distance of P to n using the same equation will get you |4t - 8|/3
These two answers are equal, their numerators are equal so you square both sides to get rid of the modulus, leaving you with an equation such as :
4^2 = (4t - 8)^2
16 = 16t^2 - 64t - 64, solve for t and obtain t = 3 or t =1
substitute those two value in the original coordinates of P to get the two position vectors that are (7i + 4j + 5k) and (3i +2j + k) and then find the distance between these two points using your skill from P1! Good luck!
Chem champ having prob with maths???
Here u go.... This was done by Physac
We have to find the point p where the perpendicular distance to the two planes is same.
first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3
since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|
now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)
therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB
BA (or AB, same thing) = OA - OB
= (4 , 2, 4)
now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6
ask if u dont get
Can anyone share a document which contains all the important/key notes for Math P3. Thanks
http://www.s-cool.co.uk/a-level/maths/integration/revise-it/differential-equationsAny notes on differentiation and integration?
I am not the ryt guy to do that..
I just visualise situations of draw them to understand what is going on in vectors question and then act accordingly..
Integration notesAny notes on differentiation and integration?
here yu go ..haha it's okay then!! Don't worry!
btw, can you help me solve this one :
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_3.pdf
question 9 ii) please!!
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