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

2 Answers
Jun 23, 2018

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:

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Jun 23, 2018

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:

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