# 3d distance problem

• Jan 29th 2009, 08:43 PM
collegestudent123
3d distance problem
The question is what is the equation that includes the points T so that the distance from T to O = (4,0,1) is twice the distance from T to C = (1, -2, 4)?
I set the equation up like this \$\displaystyle sqrt((x-4)^2 + (y-0)^2 + (z-4)^2) = 2sqrt((x-1)^2 + (y+2)^2 + (z-4)^2)\$

I get it down to this:
\$\displaystyle x^2 + (y + 16/6)^2 + (z - 5)^2 = -137/9\$
The problem is the equation doesn't work because the second side is negative
• Jan 30th 2009, 03:13 AM
Plato
Quote:

Originally Posted by collegestudent123
The question is what is the equation that includes the points T so that the distance from T to O = (4,0,1) is twice the distance from T to C = (1, -2, 4)?
I set the equation up like this \$\displaystyle sqrt((x-4)^2 + (y-0)^2 + (z-{\color{red}4})^2) = 2sqrt((x-1)^2 + (y+2)^2 + (z-4)^2)\$

I get it down to this:
\$\displaystyle x^2 + (y + 16/6)^2 + (z - 5)^2 = -137/9\$
The problem is the equation doesn't work because the second side is negative

See your error in red above.
• Jan 30th 2009, 06:52 AM
collegestudent123
Sorry the 4 was actually a typo, in the work I did it was a 1
I tried it again and still come up with that answer

Edit: I made a algebra error, I figured it out