Question

Determine the equation of the spheres S1 and S2 of radius 2sqrt(3) that are tangent to the plane pi:x-y+z=1 at the point P=(0,0,1)

Answer 1

I just solved it :D sorry =X

Answer 2

but u can show urs too :D

Answer 3

First the vectors :
\[\Large \vec n=\begin{pmatrix}1\\-1\\1\end{pmatrix}\]
and :
\[\Large \vec m=-\vec n=\begin{pmatrix}-1\\1\\-1\end{pmatrix}\]
are two normal vectors of the plan !
Let
\[\Large A(x_A,y_A,z_A)\]
The center of the sphere S1
So we can say that :
\[\Large \overrightarrow{PA}=a\overrightarrow n~~~~\text{where } a>0\]
To find a we have :
\[\Large\|a\overrightarrow n\|=\|\overrightarrow{PA}\|=PA=R=2\sqrt3 \]
So :
\[\Large a=\frac{2\sqrt3}{\|\vec n\|}=\frac{2\sqrt3}{\sqrt{3}}=2\]
hence :
\[\Large \overrightarrow{PA}=2\overrightarrow n \]
So :
\[\Large \begin{cases}x_A-0&=2\\y_A-0&=-2\\z_A-1&=2\end{cases}\]
So :
\[\Large x_A=2~~y_A=-2~~z_A=3\]
So :
\[\Large A(2,-2,3).\]
And then, the equation of S1is :
\[\Large S_1 : ~~(x-2)^2+(y+2)^2+(z-3)^2=(2\sqrt3)^2=12\]

Answer 4

great, i solved it in another way but this way is very neat too!!! congratz!

Answer 5

To find the equation of s2, we can do the same but insread of using the vector n we use the vector m like this :
\[\Large\text{Let }B(x_B,y_B,z_B)\text{ the center of } S_2\]
We can say that :
\[\Large \overrightarrow{PB}=b\overrightarrow m\]
And we complete the solution as above !

Answer 6

yep thats perfect
thx a lot!