Question

If A(1,-3,2) find the coordinates of two points which are equidistant from A and have the magnitude of 5?

Answer 1

`color(blue)((1,-3,7) and (1,-3,-3)`

Explanation:

Let one point be `(x_1,y_1,z_1)`

Distance between `(1,-3,2)` and `(x_1,y_1,z_1)`

`5=sqrt((x_1-1)^2+(y_1-(-3))^2+(z_1-2)^2)`

`(x_1-1)^2+(y_1-(-3))^2+(z_1-2)^2=25`

Substitute arbitrary values for `x_1` and `y_1`:

We can do this, since we are not restricted to any line or plane.

`x_1=1`

`y_1=-3`

`(1-1)^2+(-3-(-3))^2+(z_1-2)^2=25`

`(z_1-2)^2=25`

`z_1=7, z_1=-3`

Let `(1,-3,2)` be the co-ordinates of the midpoint.

and the other point be `(x_2,y_2,z_2)`

Then:

`((1+x_2)/2,(-3+y_2)/2,(7+z_2)/2)=(1,-3,2)`

`:.`

`(1+x_2)/2=1=>x_2=1`

`(-3+y_2)/2=-3=>y_2=-3`

`(7+z_2)/2=2=>z_2=-3`

Co-ordinates of the 2 points are:

`(1,-3,7)` and `(1,-3,-3)`

Notice that we had `z_1=7` and `z_1=-3` it didn't matter what one we chose. This can be seen in the points we found. i.e. we could switch the `7` and `-3` without changing the results. Also there are an infinite number of possible points that meet the conditions

PLOT:

Answer 2

See below.

Explanation:

Another way of seeing this problem.

All points equidistant from some given point in 3 dimensional space lie on the surface of a sphere.

The general equation of a sphere is given as:

`(x-h)^2+(y-k)^2+(z-l)^2=r^2`

Where:

`bbh`, `bbk` and `bbl` are the `bbx`, `bby` and `bbz` co-ordinates of the centre respectively.

If `A=(1,-3,2)` be the centre of the sphere with radius `5`, then all points that satisfy the equation:

`(x-1)^2+(y+3)^2+(z-2)^2=25`

will lie on the surface of the sphere and be 5 units from the centre.

So assigning arbitrary values to two of the variables and calculating the third will be solutions.

Let.

`x=1` and `y=-3`

`(1-1)^2+(-3+3)^2+(z-2)^2=25=>z=7 and z=-3`

Two points are:

`(1,-3,7)` and `(1,-3,-3)`

We could write a general solution as:

`(x,y,2+sqrt(25-(x-1)^2+(y+3)^2))`

With `x` and `y` arbitrary.

PLOT: