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Mathematics: Post your doubts here!

iKhaled

iKhaled

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surinr said:
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

Please help with entire Q10
thank you
Click to expand...
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!! :sleep:
 
Alice123

Alice123

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Anika Raisa and rosogolla993
Sorry for the late reply.. Net was too slow last night


Let Q be the closest point on the line to P... Then PQ is perpendicular to the direction of the line.....
as Q lies on the line, the position vector of Q
OQ=(1 3 -4) - t(2 1 3)= (1+2t 3+t -4+3t)

Now PQ=OQ-OP
=(1+2t 3+t -4+3t)- (-1 4 11)=(2+2t -1+t -15+3t)
Since PQ is perpendicular to the line
PQ* direction of l=0

(2+2t -1+t -15+3t)(2 1 3)=0
u get the value of t ie 3
sub value of t in PQ
(2+2t -1+t -15+3t)= (8 2 -6)
lPQl=sqroot(8^2+2^2+6^2)= 10.2
 

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surinr

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iKhaled said:
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!! :sleep:
Click to expand...

how did you get s=2 exactly?
 
Jiyad Ahsan

Jiyad Ahsan

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LMGD33 said:
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!
Click to expand...

thnx ! there was so much going on i sort of phased out :p
 
Jiyad Ahsan

Jiyad Ahsan

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Alice123 said:
Chem champ having prob with maths??? :p
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
Click to expand...

yeah.. i mean did you look at the stupid question.. there was so much going on ! i looked at it and phased out.. literally i've never seen such a .. dash.. question
awesome explanation btw .. Thnx ! :)
 
Jiyad Ahsan

Jiyad Ahsan

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Mubariz Ahmad said:
Can anyone share a document which contains all the important/key notes for Math P3. Thanks :)
Click to expand...

umm i have some on complex numbers and vectors that some people shared here
 

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layapaxx

layapaxx

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i need help in this vector questionss
 

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