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Planes question

Two planes have vector equations r.(2i+j-3k)=3 and r.(2i+j-3k)=9

Find the distance between them. I'm not too sure how to do this, as I haven't been given a point on the planes? :s-smilie:
Reply 1
Well, note that the plane's normals are the same XD
EDIT: Oh, derp, of course the normals are the same. I obviously need to start reviewing vectors!

Also, if not having a point were the problem, you could just find one. For the first one, there's (0,0,-1), (1,1,0), etc...
(edited 11 years ago)
Original post by Music99
Two planes have vector equations r.(2i+j-3k)=3 and r.(2i+j-3k)=9

Find the distance between them. I'm not too sure how to do this, as I haven't been given a point on the planes? :s-smilie:

^As aznkid66 said. Or, you also have the distance of the planes from the origin and you could find the difference between this as the planes are parallel.
Reply 3
Original post by Felix Felicis
^As aznkid66 said. Or, you also have the distance of the planes from the origin and you could find the difference between this as the planes are parallel.


Makes sense. So with the distance of the plane from the origin its 9 and 3, so the difference is 6, but then what?
Original post by Music99
Makes sense. So with the distance of the plane from the origin its 9 and 3, so the difference is 6, but then what?

The distance from the origin isn't 9 and 3. If you have a plane in scalar product form rn=D\mathbf{r} \cdot \mathbf{n} = D then the distance from the origin is Dn\dfrac{D}{| \mathbf{n} |}
Reply 5
Original post by Felix Felicis
The distance from the origin isn't 9 and 3. If you have a plane in scalar product form rn=D\mathbf{r} \cdot \mathbf{n} = D then the distance from the origin is Dn\dfrac{D}{| \mathbf{n} |}


Ah yeah, I keep getting confused and thinking i'm dealing with the unit normal vector :P.

Also how would you convert this into scalar product form?

x=-1 +2K + 3D
y=2-K-3D
z=-2+4K+5D

I tried to eliminate K and D but I must be doing it wrong...
Reply 6

Spoiler



EDIT: Apparently I have no idea what I'm doing.
EDIT2: Ah, I see, it was a different problem. My bad, my bad. ^^
(edited 11 years ago)
Original post by Music99
Ah yeah, I keep getting confused and thinking i'm dealing with the unit normal vector :P.

Also how would you convert this into scalar product form?

x=-1 +2K + 3D
y=2-K-3D
z=-2+4K+5D

I tried to eliminate K and D but I must be doing it wrong...

You have two lines in the plane:

(xyz)=(122)+K(214)+D(335)\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix} + K \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + D \begin{pmatrix} 3 \\ -3 \\ 5 \end{pmatrix}

Use the cross product to find the normal vector and use the dot product of the position vector of one point on plane to find D

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