Vector discussion...

[A star]

can any one tell me te formulae to find the distance etween two planes in vectors????

 

[daredevil]

I got lamda=3 but not lamda=1 :/

also in the first isnt it enough to show that the scalar product of DV of l and normal of m equals 0??
what have they written here in the marking scheme...

 

[Rutzaba]

View attachment 42579

I got lamda=3 but not lamda=1 :/

also in the first isnt it enough to show that the scalar product of DV of l and normal of m equals 0??
what have they written here in the marking scheme...


View attachment 42580

gimme full paper links not pics

 

[Rutzaba]

we had this question in the cie xD

 

[daredevil]

oh sorry here u go http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

 

[Rutzaba]

wch part

 

[daredevil]

wch part

part iii

 

[Rutzaba]

part iii

ive gone out of order... mujhse kuch nhi hora is waqt
come bak later insha Allah

 

[Rutzaba]

part iii

found it!
done by sumone else
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

 

[daredevil]

oohh okaa

found it!
done by sumone else
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

yy.. i was ignoring the modulus and simply solving it..
thankss a lott man!!
sooo many prayers for!! :')

 

[Rutzaba]

oohh okaa

yy.. i was ignoring the modulus and simply solving it..
thankss a lott man!!
sooo many prayers for!! :')

WESE KIA KISI AUR NE HAI

 

[Rutzaba]

oops srry fr capslock

 

[daredevil]

WESE KIA KISI AUR NE HAI

doesnt matter... u told it to me so #gratitude...

haha no prob ^_^

 

[Rutzaba]

doesnt matter... u told it to me so #gratitude...

haha no prob ^_^

jjust keep saying ameen to my signature evrytym ^_^

 

[shinnyyy]

can any one tell me te formulae to find the distance etween two planes in vectors????

Find the distance of each plane from the origin and then subtract the D1 by D2 to get the distance b/w two planes .

 

[Rutzaba]

ZaqZainab

 

[daredevil]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_31.pdf

Q8

I found the general point 'N' for PN
PN is the perpendicular line from P to l

I dont know wat i'm doing wrong after that... i have taken the scalar product as zero

[(2+2n)i + (n-1)j + (3n-15)k] * (2i+j+3k) = 0

but my value for n is not coming out to be n=3
it is n = -48/14

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf

Q8

I found the general point 'N' for PN
PN is the perpendicular line from P to l

I dont know wat i'm doing wrong after that... i have taken the scalar product as zero

[(2+2n)i + (n-1)j + (3n-15)k] * (2i+j+3k) = 0

but my value for n is not coming out to be n=3
it is n = -48/14

if you dont stop posting question after question your gona scare me to the moon -_-

 

[panoramafolks]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf

Q8

I found the general point 'N' for PN
PN is the perpendicular line from P to l

I dont know wat i'm doing wrong after that... i have taken the scalar product as zero

[(2+2n)i + (n-1)j + (3n-15)k] * (2i+j+3k) = 0

but my value for n is not coming out to be n=3
it is n = -48/14

here u go full solution... rem me in yah prayers...

 

[A star]

Find the distance of each plane from the origin and then subtract the D1 by D2 to get the distance b/w two planes .

thanks i had it written down but couldnt remember what D was xD