Question #149884
The plane 3x+4y+2z=1 touches the conicoid 3x^2+2y^2+z^2=1
True or false with full explanation
1
Expert's answer
2020-12-17T06:41:19-0500

The plane 3x+4y+2z=13x+4y+2z=1 touches the conicoid 3x2+2y2+z2=1.3x^2+2y^2+z^2=1.

True or false

Solution:

Suppose that the plane

3x+4y+2z=13x+4y+2z=1

touches the conicoid

3x2+2y2+z2=13x^2+2y^2+z^2=1

at some point

M(x0,y0,z0).M(x_0,y_0,z_0).

Then the plane equation can be written as follows:

FxM(xx0)+FyM(yy0)+FzM(zz0)=0.F_{x'}\mid_M(x-x_0)+F_{y'}\mid_M(y-y_0)+F_{z'}\mid_M(z-z_0)=0.

where

F(x,y,z)=3x2+2y2+z21,F(x,y,z)=3x^2+2y^2+z^2-1,

FxM=6x0, FyM=4y0, FzM=2z0F_{x'}\mid_M=6x_0,\space F_{y'}|_M=4y_0,\space F_{z'}|_M=2z_0

or

6x0(xx0)+4y0(yy0)+2z0(zz0=0),6x_0(x-x_0)+4y_0(y-y_0)+2z_0(z-z_0=0),

6x0x+4y0y+2z0z(6x02+4y02+2z02)=0.6x_0x+4y_0y+2z_0z-(6x_0^2+4y_0^2+2z_0^2)=0.

The coefficients of the last equation must be equal to the coefficients of the equation

3x+4y+2z1=0.3x+4y+2z-1=0.

Check it out:

6x0=3, 4y0=4, 2z0=2, 6x02+4y02+2z02=1.6x_0=3,\space 4y_0=4,\space 2z_0=2,\space 6x_0^2+4y_0^2+2z_0^2=1.

x0=12, y0=1, z0=1, 7.5=1.x_0=\frac{1}{2},\space y_0=1,\space z_0=1,\space 7.5=1.

The last equality is wrong,

Answer:

so the plane

3x+4y+2z=13x+4y+2z=1

does not touch the conicoid

3x2+2y2+z2=13x^2+2y^2+z^2=1

and there is no point of contact.


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