Two walls and a ceiling of a room meet at right angles at a point P. A fly is in the air 1 m from one wall, 8 m from the other wall and 9 m from the point P. How many meters is the fly from the ceiling?

The correct answer is Option 4.

Let us choose a three-dimensional Cartesian coordinate system where the point P (where the two walls and the ceiling meet at right angles) is at the origin (0, 0, 0).

Let the two walls lie along the coordinate planes $x=0$ and $y=0$, and let the ceiling lie along the coordinate plane $z=0$.

Let the coordinates of the fly be $(x,y,z)$.

From the given information:

1. The distance of the fly from the first wall ($x=0$) is 1 m, so:

$\left|x\right|=1$

2. The distance of the fly from the second wall ($y=0$) is 8 m, so:

$\left|y\right|=8$

3. Let the distance of the fly from the ceiling ($z=0$) be $h$ meters, so:

$\left|z\right|=h$

4. The distance of the fly from the point P (the origin) is 9 m. Using the 3D distance formula:

$\sqrt{x^2+y^2+z^2}=9$

Squaring both sides of the equation gives:

$x^2+y^2+z^2=9^2$

$x^2+y^2+z^2=81$

Substitute the known values $x^2=1^2=1$ and $y^2=8^2=64$ into the equation:

$1+64+h^2=81$

$65+h^2=81$

Subtract 65 from both sides to solve for $h^2$:

$h^2=81-65$

$h^2=16$

Taking the positive square root for distance:

$h=\sqrt{16}=4$

Therefore, the fly is 4 meters from the ceiling.