Question

Let $P$ be an arbitrary point having the sum of the squares of the distances from the planes $x+y+z=0$, $lx-nz=0$ and $x-2y+z=0$, equal to $9$. If the locus of the point $P$ is $x^2+y^2+z^2=9$, then the value of $l-n$ is equal to______.

Answer 1

Solving for the value $l-n$:

Step 1: Finding the locus of $P$

Let the point $P$ has coordinates $(\alpha ,\beta ,\gamma )$.

The point $P$ have the sum of the squares of the distances from the planes $x+y+z=0$, $lx-nz=0$ and $x-2y+z=0$, equal to $9$.

$$\begin{gathered} \Rightarrow {\left(\frac{\left[\alpha +\beta +\gamma \right]}{\sqrt{1+1+1}}\right)}^2+{\left(\frac{\left[l\alpha -n\gamma \right]}{\sqrt{l^2+n^2}}\right)}^2+{\left(\frac{\left[\alpha -2\beta +\gamma \right]}{\sqrt{1+4+1}}\right)}^2=9 \\ \Rightarrow {\left(\frac{\left[\alpha +\beta +\gamma \right]}{\sqrt{3}}\right)}^2+{\left(\frac{\left[l\alpha -n\gamma \right]}{\sqrt{l^2+n^2}}\right)}^2+{\left(\frac{\left[\alpha -2\beta +\gamma \right]}{\sqrt{6}}\right)}^2=9 \\ \Rightarrow \frac{{\left[\alpha +\beta +\gamma \right]}^2}{3}+\frac{{\left[l\alpha -n\gamma \right]}^2}{l^2+n^2}+\frac{{\left[\alpha -2\beta +\gamma \right]}^2}{6}=9 \end{gathered}$$

So the locus is,
$\frac{{\left[x+y+z\right]}^2}{3}+\frac{{\left[lx-nz\right]}^2}{l^2+n^2}+\frac{{\left[x-2y+z\right]}^2}{6}=9$

Step 2: Expressing the above locus in the form of $x^2+y^2+z^2=9$

It is given that the locus of the point is $x^2+y^2+z^2=9$.

So, equating the two equations

$\frac{{\left[x+y+z\right]}^2}{3}+\frac{{\left[lx-nz\right]}^2}{l^2+n^2}+\frac{{\left[x-2y+z\right]}^2}{6}=9$

$$\begin{gathered} \therefore \frac{{\left[x+y+z\right]}^2}{3}+\frac{{\left[lx-nz\right]}^2}{l^2+n^2}+\frac{{\left[x-2y+z\right]}^2}{6}=9 \\ \Rightarrow \frac{\left[x^2+y^2+z^2+2xy+2yz+2xz-xyz\right]}{3}+\frac{\left[l^2{x}^2+n^2{z}^2-2lxnz\right]}{l^2+n^2}+\frac{\left[x^2+4y^2+z^2-4xy+2xz-4yz+2xyz\right]}{6}=9 \\ \Rightarrow \frac{9+2xy+2yz+2xz-xyz}{3}+\frac{9+3y^2-4xy+2xz-4yz+2xyz}{6}+\frac{\left[l^2{x}^2+n^2{z}^2-2lxnz\right]}{l^2+n^2}=9 \\ \Rightarrow \frac{18+2xy+2yz+2xz-xyz+9+3y^2-4xy+2xz-4yz+2xyz}{6}+\frac{\left[l^2{x}^2+n^2{z}^2-2\ln xz\right]}{l^2+n^2}=9 \\ \Rightarrow \frac{27-2xy-2yz+4xz+xyz+3y^2}{6}+\frac{\left[l^2{x}^2+n^2{z}^2-2lxnz\right]}{l^2+n^2}=9 \\ \Rightarrow \frac{(l^2+n^2)(27-2xy-2yz+4xz+xyz+3y^2)+6l^2{x}^2+6n^2{z}^2-12\ln xz}{6(l^2+n^2)}=9 \\ \Rightarrow \frac{27(l^2+n^2)-2(l^2+n^2)xy-2(l^2+n^2)yz+\left(4\right(l^2+n^2)-12\ln )xz+(l^2+n^2)xyz+3(l^2+n^2)y^2+6l^2{x}^2+6n^2{z}^2}{6(l^2+n^2)}=9 \\ \Rightarrow \left(\frac{2l^2}{(l^2+n^2)}\right)x^2+y^2+\left(\frac{2n^2}{(l^2+n^2)}\right)z^2+\frac{54(l^2+n^2)-4(l^2+n^2)xy-4(l^2+n^2)yz+\left(8\right(l^2+n^2)-24\ln )xz}{6(l^2+n^2)}=18 \\ \Rightarrow \left(\frac{2l^2}{(l^2+n^2)}\right)x^2+y^2+\left(\frac{2n^2}{(l^2+n^2)}\right)z^2+9+\frac{-4(l^2+n^2)xy-4(l^2+n^2)yz+\left(8\right(l^2+n^2)-24\ln )xz}{6(l^2+n^2)}=18 \\ \Rightarrow \left(\frac{2l^2}{(l^2+n^2)}\right)x^2+y^2+\left(\frac{2n^2}{(l^2+n^2)}\right)z^2+\frac{-4(l^2+n^2)xy-4(l^2+n^2)yz+\left(8\right(l^2+n^2)-24\ln )xz}{6(l^2+n^2)}=9.....\left(1\right) \end{gathered}$$

Step 3: Equating the given locus and calculated locus

Comparing equation $\left(1\right)$ this with $x^2+y^2+z^2=9$, we get

$\frac{2l^2}{l^2+n^2}=1...\left(2\right)$ and

$\frac{2n^2}{l^2+n^2}=1.....\left(3\right)$

Step 4: Subtracting equation $\left(1\right)-\left(2\right)$

$$\begin{gathered} \Rightarrow \frac{2{\mathrm{l}}^2}{{\mathrm{l}}^2+{\mathrm{n}}^2}-\frac{2{\mathrm{n}}^2}{{\mathrm{l}}^2+{\mathrm{n}}^2}=0 \\ \Rightarrow \frac{2({\mathrm{l}}^2-{\mathrm{n}}^2)}{{\mathrm{l}}^2+{\mathrm{n}}^2}=0 \\ \Rightarrow {\mathrm{l}}^2-{\mathrm{n}}^2=0 \\ \Rightarrow (\mathrm{l}+\mathrm{n})(\mathrm{l}-\mathrm{n})=0 \end{gathered}$$

Here, $\mathrm{l}+\mathrm{n}$cannot be zero as both $\mathrm{l},\mathrm{n}$ are natural numbers.

Hence, $l-n=0$

Therefore, the value of $l-n$ is equal to $0$.