Question

Find the total number of points, whose perpendicular distances from yz, zx and xy-planes are in A.P and whose distances from x, y and z axes are $\sqrt{13}$, $\sqrt{10}$ and $\sqrt{5}$ respectively AP is increasing in nature.

Answer 1

8

Answer 2

Let the coordinates of a point be $P(x, y, z)$.

The perpendicular distance of the point $P(x, y, z)$ from the yz-plane is $|x|$.
The perpendicular distance of the point $P(x, y, z)$ from the zx-plane is $|y|$.
The perpendicular distance of the point $P(x, y, z)$ from the xy-plane is $|z|$.

These perpendicular distances from yz, zx, and xy-planes, in that order, are in an increasing A.P.
So, the sequence $(|x|, |y|, |z|)$ is an increasing A.P.
This means $|y| - |x| = |z| - |y| = d$ for some common difference $d > 0$.
Also, $|x| < |y| < |z|$.

The distance of the point $P(x, y, z)$ from the x-axis is $\sqrt{y^2 + z^2}$.
Given that this distance is $\sqrt{13}$. So, $y^2 + z^2 = 13$.

The distance of the point $P(x, y, z)$ from the y-axis is $\sqrt{x^2 + z^2}$.
Given that this distance is $\sqrt{10}$. So, $x^2 + z^2 = 10$.

The distance of the point $P(x, y, z)$ from the z-axis is $\sqrt{x^2 + y^2}$.
Given that this distance is $\sqrt{5}$. So, $x^2 + y^2 = 5$.

We have the following system of equations:

1. $y^2 + z^2 = 13$
2. $x^2 + z^2 = 10$
3. $x^2 + y^2 = 5$

Adding all three equations gives:
$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 13 + 10 + 5$
$2x^2 + 2y^2 + 2z^2 = 28$
$x^2 + y^2 + z^2 = 14$

Now, subtract each original equation from $x^2 + y^2 + z^2 = 14$:
$(x^2 + y^2 + z^2) - (y^2 + z^2) = 14 - 13 \implies x^2 = 1$
$(x^2 + y^2 + z^2) - (x^2 + z^2) = 14 - 10 \implies y^2 = 4$
$(x^2 + y^2 + z^2) - (x^2 + y^2) = 14 - 5 \implies z^2 = 9$

From these results, we find the absolute values of the coordinates:
$|x| = \sqrt{x^2} = \sqrt{1} = 1$
$|y| = \sqrt{y^2} = \sqrt{4} = 2$
$|z| = \sqrt{z^2} = \sqrt{9} = 3$

Now we check if the sequence of perpendicular distances $(|x|, |y|, |z|)$ is an increasing A.P.
The sequence is $(1, 2, 3)$.
The differences between consecutive terms are $2 - 1 = 1$ and $3 - 2 = 1$. The common difference is $d=1$.
Since the common difference $d=1 > 0$, the A.P. is increasing.
Also, the condition $|x| < |y| < |z|$ is satisfied since $1 < 2 < 3$.
Thus, the values $|x|=1, |y|=2, |z|=3$ satisfy the condition that the perpendicular distances from yz, zx, and xy-planes are in an increasing A.P.

The possible values for $x, y, z$ are:
$x^2 = 1 \implies x = \pm 1$
$y^2 = 4 \implies y = \pm 2$
$z^2 = 9 \implies z = \pm 3$

Any combination of these values for $x, y, z$ will result in $|x|=1, |y|=2, |z|=3$, which satisfies the A.P. condition.
The number of possible values for $x$ is 2.
The number of possible values for $y$ is 2.
The number of possible values for $z$ is 2.

The total number of points $(x, y, z)$ that satisfy these conditions is the product of the number of possibilities for each coordinate.
Total number of points = (Number of choices for x) $\times$ (Number of choices for y) $\times$ (Number of choices for z)
Total number of points = $2 \times 2 \times 2 = 8$.

These 8 points are:
(1, 2, 3), (1, 2, -3), (1, -2, 3), (1, -2, -3), (-1, 2, 3), (-1, 2, -3), (-1, -2, 3), (-1, -2, -3).
All these points have $|x|=1, |y|=2, |z|=3$, which forms the increasing A.P. (1, 2, 3).
Also, for any of these points, $x^2=1, y^2=4, z^2=9$, which satisfies the distance conditions from the axes:
$\sqrt{y^2+z^2} = \sqrt{4+9} = \sqrt{13}$
$\sqrt{x^2+z^2} = \sqrt{1+9} = \sqrt{10}$
$\sqrt{x^2+y^2} = \sqrt{1+4} = \sqrt{5}$

Thus, there are 8 such points.