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# P3 Vectors Question watch

1. The plane of euqation

r = (3,1,2) + lamda(2,1,2) + mew(1,-2,-1)

goes though the points (3,1,2) , (5,2,4), and (4,-1,1)

how would i then find the cartesian equation of the plane?

[I know its somewhere along the lines of eliminated the lambda and mew parameters but im not too familiar with this.]

any help would be appreciated..thanks
2. r = (3,1,2) + lamda(2,1,2) + mew(1,-2,-1)
goes though the points (3,1,2) , (5,2,4), and (4,-1,1)
how would i then find the cartesian equation of the plane?
One way would be to write the plane in the form , where , is a vector perpendicular to the plane(and so perpendicular to the two lines) and is any vector in the plane.
Note comes from as represents any vector in the line and it's scalar product with is, of course, .
The plane of euqation

r = (3,1,2) + lamda(2,1,2) + mew(1,-2,-1)

goes though the points (3,1,2) , (5,2,4), and (4,-1,1)

how would i then find the cartesian equation of the plane?

[I know its somewhere along the lines of eliminated the lambda and mew parameters but im not too familiar with this.]

any help would be appreciated..thanks
what exam board is this for , aqa ?

Edit: do u have the answers for the question?
4. exam board is AQA maths spec a

i have the answer but this is not the whole question . the question i am stuck on requires you to find the angle between 2 planes [ 1 is given in hte final part in cartesian form] and the other plane which you have to work out before you gotta convert it into catesian form to calculate the angle
5. what past paper is this from?
6. i hope you check this thread b4 2morow, cos i wanna get this sorted out what - if we get a similar question in the exam, worth 6 marks
ill be screwed !!!
7. june03 its from if i remember correctly

yeah its a dody concept , this mite help

parametric equation of plane is x = 6lamda + 3 mew

y = 6 - 3mew

z = 2mew

hte answer for the cartesian eqn is 2y + 3z =12
june03 its from if i remember correctly

yeah its a dody concept , this mite help

parametric equation of plane is x = 6lamda + 3 mew

y = 6 - 3mew

z = 2mew

hte answer for the cartesian eqn is 2y + 3z =12
where did you get those parametrics from?
9. sounds more like p6 to me
edexcel neway
The plane of euqation
r = (3,1,2) + lamda(2,1,2) + mew(1,-2,-1)
goes though the points (3,1,2) , (5,2,4), and (4,-1,1)
-> (x, y, z) = (3, 1, 2) + lm(2, 1, 2) + mw(1, -2, -1)
-> (x, y, z) = [(3 + 2lm + mw), (1 + lm - 2mw), 2 + 2lm - mw)]
-> x = 3 + 2lm + mw, y = 1 + lm - 2mw, z = 2 + 2lm - mw

Need to eliminate parameters lm and mw.

x + z = 3 + 2lm + mw + 2 + 2lm - mw = 3 + 2lm + 2 + 2lm = 5 + 4lm
x + y - z = 3 + 2lm + mw + 1 + lm - 2mw - 2 - 2lm + mw = 2 + lm

Hence: x + z = 5 + 4lm and x + y - z = 2 + lm
-> (x + z - 5)/4 = lm and x + y - z - 2 = lm

-> Equate expressions for lm: (x + z - 5)/4 = x + y - z - 2
-> x + z - 5 = 4x + 4y - 4z - 8

-> Cartesian Equation: 3x + 4y - 5z = 3
11. Here's the method Gaz suggested:
n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)

Let r = (x, y, z), then:
(x, y, z) . (-3, -4, 5) = (3, 1, 2) . (-3, -4, 5)
- 3x - 4y + 5z = - 9 - 4 + 10
3x + 4y - 5z = 3, which agrees with Nima's answer.
12. (Original post by Nima)
-> (x, y, z) = (3, 1, 2) + lm(2, 1, 2) + mw(1, -2, -1)
-> (x, y, z) = [(3 + 2lm + mw), (1 + lm - 2mw), 2 + 2lm - mw)]
-> x = 3 + 2lm + mw, y = 1 + lm - 2mw, z = 2 + 2lm - mw

Need to eliminate parameters lm and mw.

x + z = 3 + 2lm + mw + 2 + 2lm - mw = 3 + 2lm + 2 + 2lm = 5 + 4lm
x + y - z = 3 + 2lm + mw + 1 + lm - 2mw - 2 - 2lm + mw = 2 + lm

Hence: x + z = 5 + 4lm and x + y - z = 2 + lm
-> (x + z - 5)/4 = lm and x + y - z - 2 = lm

-> Equate expressions for lm: (x + z - 5)/4 = x + y - z - 2
-> x + z - 5 = 4x + 4y - 4z - 8

-> Cartesian Equation: 3x + 4y - 5z = 3
cheers, does actually all make sense!
13. (Original post by dvs)
Here's the method Gaz suggested:
n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)

Let r = (x, y, z), then:
(x, y, z) . (-3, -4, 5) = (3, 1, 2) . (-3, -4, 5)
- 3x - 4y + 5z = - 9 - 4 + 10
3x + 4y - 5z = 3, which agrees with Nima's answer.

that is so wrong, look at your first line:

n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)

(5-3) x (5-4) does not equal -3
14. (Original post by ram)
that is so wrong, look at your first line:
n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)
(5-3) x (5-4) does not equal -3
It isn't wrong, it's perfectly correct.
He has used the vector product - not the scalar product (hence the 'x' rather than '.').
The vector product gives a vector perpendicular to both and .
Alternatively you could find a perpendicular vector by setting it to be i+aj+bk and solving the simultaneous equations you get from setting the scalar product with the two vectors in the plane to zero.
15. (Original post by ram)
that is so wrong, look at your first line:

n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)

(5-3) x (5-4) does not equal -3
This x is the cross or vector product - not multiplication - the calculation is correct
16. from what paper is this question from.....i can always get the answers to it from thw aqa website
17. its from june 03.......in the context of the question you dont need to convert into cartesian form. they give a vector that is perpendicular to plane 1. and the normal of plane 2 you can find out just by inspecting the eqn. then u just dot product the two perpendiculars....and find the angle.
18. Equation of plane is r . n = a . n

Find n, by crossing the multiples of lamda and mew
n = (2,1,2) x (1,-2,-1)
n = (3, 4, -5)

a is just a point on the plane. Just use (3,1,2)

so
r . n = a . n
r . (3, 4, -5) = (3, 1, 2) . (3, 4, -5)
r . (3, 4, -5) = 3
3x + 4y -5z = 3

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