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Mathematics: Post your doubts here!

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.
 
Alice123

Alice123

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PhyZac said:
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.
Click to expand...
r=(x y z) and n is the direction. a is point on the plane. hope this example helps
 

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PhyZac

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Alice123 said:
r=(x y z) and n is the direction. a is point on the plane. hope this example helps
Click to expand...
OOOhhhhh okaaayYY!
a small question ( i am sorry it is kind of silly) what does a represent?

Thanks alottttttttttttt!! Jazaki Allah khairan....!! May Allah S.W.T reward you with highest results, and have mercy on you and your family , Aameen!!! Thank you soooo muuucccchh!!!
 
Alice123

Alice123

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PhyZac said:
OOOhhhhh okaaayYY!
a small question ( i am sorry it is kind of silly) what does a represent?

Thanks alottttttttttttt!! Jazaki Allah khairan....!! May Allah S.W.T reward you with highest results, and have mercy on you and your family happy, Aameen!!! Thank you soooo muuucccchh!!!
Click to expand...
a is a point on the plane(whose equation u're finding) , for instance position vector of A that lies on the plane. Ask if u dont get. i'l do a qs for u. u help others so much, i'll b glad if i'm of sum help to u
 
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