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Plane equation Help

S6xWY1E1A1I.jpg
coordinates of B(6, 4.5, 3)
(edited 6 years ago)
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Original post by pereira325
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I hope it will show
20170604_210803.jpg513.6KB
Original post by fireworks659
I hope it will show


Oh yeah I see it. Actually I see it in the OP one too now. Maybe it was my internet or the server.

Can I see how you did part ii?
I'm on edexcel and these questions look a lot harder like how do you even work out the equation of a plane?
Original post by pereira325
Oh yeah I see it. Actually I see it in the OP one too now. Maybe it was my internet or the server.

Can I see how you did part ii?
I'm on edexcel and these questions look a lot harder like how do you even work out the equation of a plane?


Well for part ii) you can just insert coordinates of A,B,C and they should give 0 , which is easier method than finding this equation
Original post by fireworks659
Could you plese help
coordinates of B(6, 4.5, 3)


Well parametrically you would have r=a+λb+μc\displaystyle {\bf r}={\bf a}+\lambda{\bf b}+\mu{\bf c} with parameters λ,μ\lambda, \mu where a{\bf a} can be the position vector of O, the origin, as it must be on the plane, then b{\bf b} and c{\bf c} must be two non-parallel vectors lying in the plane. Can you see what they would be?

If you want the Cartesian equation, you would then need to transform the above into r⋅n=d{\bf r}\cdot {\bf n}=d where n=b×c{\bf n}={\bf b} \times {\bf c} (ie the perpendicular vector the plane) and dd is some scalar quantity for you to find by simply plugging in a point on the plane for r{\bf r}
(edited 6 years ago)
Original post by RDKGames
Well parametrically you would have r=a+λb+μc\displaystyle {\bf r}={\bf a}+\lambda{\bf b}+\mu{\bf c} with parameters λ,μ\lambda, \mu where a{\bf a} can be the position vector of O, the origin, as it must be on the plane, then b{\bf b} and c{\bf c} must be two non-parallel vectors lying in the plane. Can you see what they would be?

If you want the Cartesian equation, you would then need to transform the above into r⋅n=d{\bf r}\cdot {\bf n}=d where n=b×c{\bf n}={\bf b} \times {\bf c} (ie the perpendicular vector the plane) and dd is some scalar quantity for you to find by simply plugging in a point on the plane for r{\bf r}


So if vector a is (0,0,0,) then vector b is OB and c is OC?
Do i need to solve it simultaneously with 3 variables, To use n.a=D?
Original post by fireworks659
So if vector a is (0,0,0,) then vector b is OB and c is OC?
Do i need to solve it simultaneously with 3 variables, To use n.a=D?


c{\bf c} in my equation is not the vector OC. OC does not lie in the plane DOBE. Try again.

Once you have those your b{\bf b} and c{\bf c}, then you can work out n{\bf n}, and you dont need to solve anything simultenously for d. You simply need to sub in a point on the plane for x,y,z since thats what r{\bf r} is. A point on the plane (x,y,z). Doing so leaves you with a single variable d for which you can solve.
Original post by fireworks659
Sorry , I wanted to say OD as in vector c
So i will end up with lambda and mu equation which of the two vectors will I need to use for n?


Yes OD is the vector c. So you have your parametric equation of the plane.

The Cartesian one, as I've said previously, is r.n=d where n is the perpendicular vector to b and c. You don't need to do anything with lambda and mu.
Original post by RDKGames
Yes OD is the vector c. So you have your parametric equation of the plane.

The Cartesian one, as I've said previously, is r.n=d where n is the perpendicular vector to b and c. You don't need to do anything with lambda and mu.


So i end up with x=6(lambda) y=4.5(lambda)+20mu z=3(lambda)
The only substitution i can think of is x=2z the answer in the book is x-2z=0
Why does y=4.5(lambda)+20(mu) is left out, is it because it is only variable with mu?
Original post by fireworks659
So i end up with x=6(lambda) y=4.5(lambda)+20mu z=3(lambda)
The only substitution i can think of is x=2z the answer in the book is x-2z=0
Why does y=4.5(lambda)+20(mu) is left out, is it because it is only variable with mu?


I've just said you don't need to do anything with lambda and mu...

n=b×c{\bf n}={\bf b} \times {\bf c} - this gives you a well defined vector, yes?

Then r⋅n{\bf r} \cdot {\bf n} for any r{\bf r} on the plane, so say it is r=(0,0,0){\bf r}=(0,0,0), gives you a scalar quantity, which in this case is 0 if you pick the origin as the point. So then you have r⋅n=0{\bf r} \cdot {\bf n}=0.

Since n=(n1,n2,n3){\bf n}= (n_1,n_2,n_3) for whatever n1,n2,n3n_1,n_2,n_3 you find, then r⋅n=0{\bf r} \cdot {\bf n}=0 is the same as n1x+n2y+n3z=0n_1x+n_2y+n_3z=0

Get it??

If you do not understand how to derive the Cartesian equation of the plane, I suggest you look back in the textbook as it should tell you how to.
(edited 6 years ago)
Original post by RDKGames
I've just said you don't need to do anything with lambda and mu...

n=b×c{\bf n}={\bf b} \times {\bf c} - this gives you a well defined vector, yes?

Then r⋅n{\bf r} \cdot {\bf n} for any r{\bf r} on the plane, so say it is r=(0,0,0){\bf r}=(0,0,0), gives you a scalar quantity, which in this case is 0 if you pick the origin as the point. So then you have r⋅n=0{\bf r} \cdot {\bf n}=0.

Since n=(n1,n2,n3){\bf n}= (n_1,n_2,n_3) for whatever n1,n2,n3n_1,n_2,n_3 you find, then r⋅n=0{\bf r} \cdot {\bf n}=0 is the same as n1x+n2y+n3z=0n_1x+n_2y+n_3z=0

Get it??

If you do not understand how to derive the Cartesian equation of the plane, I suggest you look back in the textbook as it should tell you how to.


I dont want to bother you a lot but:
b=OB (6, 4.5, 3) c= OD(0, 20, 0) b×c= (0, 90, 0)
I do understand how to derive cartesian equations , but just in this question I can't
Original post by fireworks659
I dont want to bother you a lot but:
b=OB (6, 4.5, 3) c= OD(0, 20, 0) b×c= (0, 90, 0)
I do understand how to derive cartesian equations , but just in this question I can't


OK. First of all, b×c≠(0,90,0){\bf b}\times {\bf c} \neq (0,90,0). What you've done is the DOT product and in a wrong way. What you need to do is the CROSS product. Have you learnt this?

I'm not doing to work it out for you, but say for example that you find n=(1,2,3){\bf n}=(1,2,3)

So then r⋅n=(x,y,z)⋅(1,2,3)=1x+2y+3z=d{\bf r}\cdot {\bf n}=(x,y,z) \cdot (1,2,3)=1x+2y+3z=d as here you take the DOT product between the general point and the perpendicular vector to the plane. Then you find dd by simply plugging in a point you know is on the plane, so (0,0,0)(0,0,0) for example.
(edited 6 years ago)
Original post by RDKGames
OK. First of all, b×c≠(0,90,0){\bf b}\times {\bf c} \neq (0,90,0). What you've done is the DOT product and in a wrong way. What you need to do is the CROSS product. Have you learnt this?

I'm not doing to work it out for you, but say for example that you find n=(1,2,3){\bf n}=(1,2,3)

So then r⋅n=(x,y,z)⋅(1,2,3)=1x+2y+3z=d{\bf r}\cdot {\bf n}=(x,y,z) \cdot (1,2,3)=1x+2y+3z=d as here you take the DOT product between the general point and the perpendicular vector to the plane. Then you find dd by simply plugging in a point you know is on the plane, so (0,0,0)(0,0,0) for example.



Cross product is not in my syllabus, i think it is in FP3, not in C4
But i think i know how to do cross product i will try it out
Original post by fireworks659
Cross product is not in my syllabus, i think it is in FP3, not in C4
But i think i know how to do cross product i will try it out


If it is not on your syllabus, how do you work out the perpendicular vector to two vectors? Or how do you usually find the Cartesian eq. of a plane?
(edited 6 years ago)
Original post by RDKGames
If it is not on your syllabus, how do you work out the perpendicular vector to two vectors?


Well I presume write x,y,z with labda and mu and try to eliminate lambda and mu?
Which i tried to do above, but im not sure
This is why i asked this question on here
(edited 6 years ago)
Original post by fireworks659
Well I presume write x,y,z with labda and mu and try to eliminate lambda and mu?
Which i tried to do above, but im not sure
This is why i asked this question on here


Well what does it say in your textbook on finding the Cartesian equation??

The general approach is to use the cross product but if that's not in your syllabus and you're required to be able to derive eqs. of planes then there must be an alternative that they provide.

EDIT: OK. I just derived it myself via substitution for parameters - a method I never really use.

Okay, so you have r=λ[6923]+μ[0200]\displaystyle {\bf r}=\lambda \begin{bmatrix} 6 \\ \frac{9}{2} \\ 3 \end{bmatrix}+\mu \begin{bmatrix} 0 \\ 20 \\ 0 \end{bmatrix}

For simpler numbers, im just gonna say the plane eq is r=λ[432]+μ[010]\displaystyle {\bf r}=\lambda \begin{bmatrix} 4 \\ 3 \\ 2 \end{bmatrix}+\mu \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}

So then:
x=4λx=4\lambda
y=3λ+μy=3\lambda+\mu
z=2λz=2\lambda


You can add all of these equations to give:
x+y+z=9λ+μx+y+z=9\lambda+\mu

Now you want λ,μ\lambda,\mu in terms of x,y,zx,y,z

Say λ=z2\lambda=\frac{z}{2} and μ=y−3λ=y−3z2\mu=y-3\lambda=y-\frac{3z}{2}

Then you have x+y+z=9(z2)+(y−3z2)\displaystyle x+y+z=9 \left( \frac{z}{2} \right)+ \left( y-\frac{3z}{2} \right)

Then simplify.
(edited 6 years ago)

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