Is This System of Equations Impossible to Solve?

[anemone]

Prove that the system of equations below has no real solution:

$x^2+y^2+z^2=1$

$x+2y+3z=4$

 

[MarkFL]

My solution:

Spoiler

Using the Point-Plane distance formula, we find the distance $d$ from the origin to the plane $x+2y+3z=4$ is:

\(\displaystyle d=\frac{4}{\sqrt{1^2+2^2+3^2}}=\frac{4}{\sqrt{14}}>1\)

Since the equation:

\(\displaystyle x^2+y^2+z^2=1\)

represents a sphere of radius 1 centered at the origin, we can then conclude the system has no real solutions.

 

[kaliprasad]

Prove that the system of equations below has no real solution:

$x^2+y^2+z^2=1$

$x+2y+3z=4$

Spoiler

we have
$x^2+y^2+z^2=1\cdots(1)$
$x+2y+3z=4\cdots(2)$
multiply 2nd equation by 2 and subtract from (1)
$(x^2-2x) + ((y^2-4y) + (z^2- 6z) = -7$
or $(x-1)^2 + (y-2)^2 + (z^-3)^2 = 7\cdots(3)$

(1) is a sphere with centre (0,0,0) and radius 1
(3) is a sphere with cenre (1,2,3) and radius $\sqrt{7}$

distance between (0,0,0) and (1,2,3) is $ \sqrt{14} > 1 + \sqrt{7}$
hence no solution

 

[Greg]

Prove that the system of equations below has no real solution:

$x^2+y^2+z^2=1$

$x+2y+3z=4$

Spoiler

Using Lagrange multipliers we have

$$\Lambda=x+2y+3z-\lambda(x^2+y^2+z^2-1)$$
$$\dfrac{d\Lambda}{dx}=1-2\lambda x=0$$
$$\dfrac{d\Lambda}{dy}=2-2\lambda y=0$$
$$\dfrac{d\Lambda}{dz}=3-2\lambda z=0$$
$$\dfrac{d\Lambda}{d\lambda}=x^2+y^2+z^2-1=0$$

This yields the system

$$x=\dfrac{1}{2\lambda}$$
$$y=\dfrac{2}{2\lambda}$$
$$z=\dfrac{3}{2\lambda}$$
$$x^2+y^2+z^2=1$$

Squaring the first three equations and summing them gives

$$\dfrac{1}{4\lambda^2}+\dfrac{4}{4\lambda^2}+\dfrac{9}{4\lambda^2}=x^2+y^2+z^2=1\Rightarrow\lambda=\pm\sqrt{\dfrac72}$$

and (choosing the positive root)

$$x+2y+3z=\sqrt{14}$$

Another set af values for $x,y,z$ that satisfy $x^2+y^2+z^2=1$ is $x=y=z=\dfrac{1}{\sqrt3}$.

Then $x+2y+3z=2\sqrt3<\sqrt{14}$, so we have found a maximum < $4$ and the given equations have no solution.

 

[topsquark]

Spoiler

Using Lagrange multipliers we have

$$\Lambda=x+2y+3z-\lambda(x^2+y^2+z^2-1)$$
$$\dfrac{d\Lambda}{dx}=1-2\lambda x=0$$
$$\dfrac{d\Lambda}{dy}=2-2\lambda y=0$$
$$\dfrac{d\Lambda}{dz}=3-2\lambda z=0$$
$$\dfrac{d\Lambda}{d\lambda}=x^2+y^2+z^2-1=0$$

This yields the system

$$x=\dfrac{1}{2\lambda}$$
$$y=\dfrac{2}{2\lambda}$$
$$z=\dfrac{3}{2\lambda}$$
$$x^2+y^2+z^2=1$$

Squaring the first three equations and summing them gives

$$\dfrac{1}{4\lambda^2}+\dfrac{4}{4\lambda^2}+\dfrac{9}{4\lambda^2}=x^2+y^2+z^2=1\Rightarrow\lambda=\pm\sqrt{\dfrac72}$$

and (choosing the positive root)

$$x+2y+3z=\sqrt{14}$$

Another set af values for $x,y,z$ that satisfy $x^2+y^2+z^2=1$ is $x=y=z=\dfrac{1}{\sqrt3}$.

Then $x+2y+3z=2\sqrt3<\sqrt{14}$, so we have found a maximum < $4$ and the given equations have no solution.

Haven't seen one of those for a while. Nice!

-Dan

 

[anemone]

Thanks all for participating in this challenge! (Cool)

My solution:

Spoiler

Apply the Cauchy-Schwarz inequality on the LHS of the second equation, we have:

$4=x+2y+3z≤\sqrt{1+4+9}\sqrt{x^2+y^2+z^2}=\sqrt{14}\sqrt{1}=\sqrt{14}$

that implies $16\le 14$, which is absurd.

So the system has no solutions in real numbers.

 

[MarkFL]

Thanks all for participating in this challenge! (Cool)

My solution:

Spoiler

Apply the Cauchy-Schwarz inequality on the LHS of the second equation, we have:

$4=x+2y+3z≤\sqrt{1+4+9}\sqrt{x^2+y^2+z^2}=\sqrt{14}\sqrt{1}=\sqrt{14}$

that implies $16\le 14$, which is absurd.

So the system has no solutions in real numbers.

Nicely done! (Yes)