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Mathematics: Post your doubts here!

PANDA-

PANDA-

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Muhammad Muneeb said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
q9 (i), how do i calculate co ordiates of D :/
Click to expand...

Since you're only asking about D, I'm going to guess you found M... which is (5,2)

M is the midpoint between B and D

The formula for midpoint would be..
(x2-x1)/2 , (y2-y1)/2

So (x2-x1)/2, where x2 is the x value of D, and x1 is the x value of B are equal to 5, the x value of M the midpt. Solve.
So (y2-y1)/2, where y2 is the y value of D, and y1 is the y value of B are equal to 2, the y value of M the midpt. Solve.

You'll get
x2= 7
y2= -2
 
Muhammad Muneeb

Muhammad Muneeb

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PANDA- said:
AB = OB - OA
AB = (2-5)i+(7-1)j+(p-2)k
= -3i + 6j + 2k

Now make it a unit vector, so we can later multiply the magnitude of the vector we need (28) to get the vector itself.
For that you'll need to find the modulus by finding the square root of the sum of i^2, j^2 and k^2...

That gives you
Modulus = sqrt((-3)^2 + 6^2 + 2^2)
= 7

To find the unit vector, we need to divide our vector by the modulus, or multiply it by 1/modulus as so...

Unit vector = 1/7(-3i+6j+2k)

Multiply the magnitude (28) by the unit vector, you'll get
(1/7)(28)(-3i+6j+2k)

4(-3i+6j+2k)
Ans. = -12i+24j+8k
Click to expand...
Thanks man :D
 
littlecloud11

littlecloud11

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tom ed said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_32.pdf
q7
Click to expand...

7i) Find the direction vector of AB-
direction vector AB = B -A
= 3i + 4j - (i + 2j +2k)
= 2i + 2j -2k
line eq. = position vector of A + direction vector
= i +2j +2k + s (2i +2j -2k)

ii) Any point of line AB has the coordinates (1+2s)i + (2+2s)j + (2-2s)k
so let OP be (1+2s)i + (2+2s)j + (2-2s)k
For OP to be perpendicular to AB the scalar product of OP and the direction vector of AB is 0
1+2s 2
2+2s . 2 = 0
2-2s -2

2+4s + 4+ 4s -4 + 4s =0
s= -2/12 = -1/6
Substitute the value of s in OP
therefore p = 2/3 i + 5/3 j + 7/3k

iii) the Plane contains line AB so the normal of the place is perpendicular to the direction vector of AB.
The plane OAB is perpendicular to our plane, so the normal vector of plane OAB is also perpendicular to pour plane.
Find the normal vector of plane OAB by the cross product of the direction vector of OA and OB:
|1 2 2|
|3 4 0|
= (0 - 8) i -(0- 6)j + (4-6)k
= -8i +6j -2 k
Find the cross product of the direction vector of AB and the normal of plane OAB to find the normal of the plane
|2 2 -2|
|-8 6 -2|
= (-4+ 12)i - (-4-16)j + (12 + 16) k
= 8i + 20j +28k
= 4i + 5j +7k (simplified ratio)
for eq of plane-
r.n = a.n
(x y z) . (4 5 7) = (1 2 2). ( 2 5 7)
2x + 5y + 7z = 26
 
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PhyZac

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Soldier313 said:
Aoa wr wb
Just have a little doubt, for 6 ii ) i am getting the angle as -5.1 degrees, while the ms says +5.1 degrees. Have i made an error somewhere, or do i just have to ignore the -ve sign??

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_33.pdf

Thank you :)
Click to expand...
I am sure the mistake was yours, degrees can not be negative you know.

But I will tell a possible mistake,

when you take acute angle make sure to take mod of the scalar product!
 
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PhyZac

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littlecloud11 said:
7i) Find the direction vector of AB-
direction vector AB = B -A
= 3i + 4j - (i + 2j +2k)
...................
for eq of plane-
r.n = a.n
(x y z) . (4 5 7) = (1 2 2). ( 2 5 7)
2x + 5y + 7z = 26
Click to expand...
I understood every step here except one.

How did you go from

r.n = a.n
to
ax + by + cz = d

I am really sorry for disturbing but I have big trouble with r.n = a.n form of plane.

iKhaled or anyone please.
 
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