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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