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FP4 - vectors and matrices

Find the Cartesian equation of the plane Π \Pi that the matrix M \mathbf{M}
[210010420] \begin{bmatrix} 2 & 1 & 0 \\ 0& 1& 0 \\ 4& 2& 0 \end{bmatrix}
maps all points to under the transformation.
Find the Cartesian equations of the two possible spheres of radii 3 that the plane Π \Pi is tangent to at the origin.
[The equations should be in exact form not involving trigonometric functions.]

The last bit is aarrrd, hard to visualise.
Original post by Ano123

hard to visualise.


Essentially you have a plane that goes through the origin.

Then, there are two spheres - one either side of the plane - that touch the plane at the origin. You could imagine one sitting on the plane at that point, and the other in the mirror image position - treating the plane as a mirror.
Original post by Ano123
Find the Cartesian equation of the plane Π \Pi that the matrix M \mathbf{M}
[210010420] \begin{bmatrix} 2 & 1 & 0 \\ 0& 1& 0 \\ 4& 2& 0 \end{bmatrix}
maps all points to under the transformation.
Find the Cartesian equations of the two possible spheres of radii 3 that the plane Π \Pi is tangent to at the origin.
[The equations should be in exact form not involving trigonometric functions.]

The last bit is aarrrd, hard to visualise.


Okay so for the first part I got the plane 2x+y=0-2x+y=0, is that right? Been a while since FP4 now I'm trying to remember it. :smile:

The last bit is not too bad as far as visualisation goes once you have the plane equation, therefore straight forward to do.
(edited 7 years ago)
Reply 3
Original post by RDKGames
Okay so for the first part I got the plane 2x+y=0-2x+y=0, is that right? Been a while since FP4 now I'm trying to remember it. :smile:

The last bit is not too bad as far as visualisation goes once you have the plane equation, therefore straight forward to do.


Equation of the plane is z=2x z=2x .
What about the next part?
Original post by Ano123
Equation of the plane is z=2x z=2x .
What about the next part?


Ah yeah, slipped up in my working and hadn't noticed I used y instead of z.

For the next part, you imagine that the two spheres are tangent on either side of the plane. They both have the same radius of 3 and are tangent at the origin.

So from here you can just construct a line, which is normal to plane and goes through the origin, with scalar λ\lambda. You can take the magnitude of this line to be 3 and solve for the two values of lambda. Sub each in and get 2 different points, these are the centres of the spheres, then proceed to put them into the sphere equation format of (xa)2+(yb)2+(zc)2=32(x-a)^2+(y-b)^2+(z-c)^2=3^2 where the centre is (a,b,c)

That's my thinking anyway. I like the question. :smile:
(edited 7 years ago)

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