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Mathematics: Post your doubts here!

VinnCh

VinnCh

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LMGD33 said:
Hello. I solved it long ago, and i'm checking my copybook, I hope you solved part (i) correctly.. when you sketch it, you have the angle and the opposite side is (10/a^2 +4) and the adjacent side is (5a/a^2 + 4) so to find the tan of the angle which they provided, you need to write tan(3pi/4) = opp/adj
This gives : tan (3pi/4) = 10/5a
[shift] tan(3pi/4) = 10/5a
-1 = 10/5a
a= -2
Good luck!!
Click to expand...

Thank youu!! I confused the tangent formula with the distance formula. I must be really sleepy now. But thanks to you I got it now! =)
 
LMGD33

LMGD33

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SararaIH said:
So after we integrate this completely, we still have to multiply the -1 at the end, right?
Click to expand...
Sorry if it's unclear, I just relaized!! I multiplied by negative one outside the integral, and the numerator is supposed to be -10 u
which would give us +10 u and the answer they are looking for , sorry again!
 
SararaIH

SararaIH

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LMGD33 said:
Sorry if it's unclear, I just relaized!! I multiplied by negative one outside the integral, and the numerator is supposed to be -10 u
which would give us +10 u and the answer they are looking for , sorry again!
Click to expand...

It's okay! You don't have to apologize
But I'm confused :/
I mean, do we have to keep the -1 outside the integration sign?
 
LMGD33

LMGD33

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SararaIH said:
It's okay! You don't have to apologize
But I'm confused :/
I mean, do we have to keep the -1 outside the integration sign?
Click to expand...

Not really, I just show it as a step to acknowledge the flipping of the limits.. since the upper limit was smaller than the lower, that is why they asked us to flip them and as a general rule you multiple by negative one, doesn't matter where really :D
 
applepie1996

applepie1996

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PhyZac said:
applepie1996
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

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]
Click to expand...
xhizors
credits to PhyZac
Pray for him :D
 
LMGD33

LMGD33

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xhizors said:
Q10 iii
working with little explanation plz, anyone!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf
Click to expand...

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...

OR

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...
 
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