calc III problem

[renegade05]

Question: Consider the points P such that the distance from P to A(1,-2,3) is three times the distance from P to B(0,1,-2). Describe this set of points, giving appropriate information.

So I'm not really sure what the heck this question is asking. Am I to take P(x,y,z) and find the distance from these points to A and B ? and then say the magnitude of the vector PA = 3d and PB = d, where d is the distance, and PA & PB are vectors... Im kinda lost. Any help would be lovely.

 

[burakaltr]

Question: Consider the points P such that the distance from P to A(1,-2,3) is three times the distance from P to B(0,1,-2). Describe this set of points, giving appropriate information.

So I'm not really sure what the heck this question is asking. Am I to take P(x,y,z) and find the distance from these points to A and B ? and then say the magnitude of the vector PA = 3d and PB = d, where d is the distance, and PA & PB are vectors... Im kinda lost. Any help would be lovely.

Call P=<x,y,z>

Form Vector AB --> Subtract coordinate of A from coordiantes of B ===> B-A

then

form PA ===>A-P

set that equal to

3*PB ==> BP = P-B
length(PA)=3*length(PB)

 

[renegade05]

Hmm i think i got it. I'm not really sure what you were saying but i figured it out.

The points are the surface of a sphere of radius $\displaystyle \frac{\sqrt{315}}{8}$ centered at $\displaystyle \left(\frac{-1}{8},\frac{11}{8},\frac{-21}{8}\right)$

Maybe a confirmation would be nice though, if someone disagrees.

 

[soroban]

Hello, renegade05!

I got the same answer!

Consider the points $\displaystyle P$ such that the distance from $\displaystyle P$ to $\displaystyle A(1,\text{-}2,3)$
. . is three times the distance from $\displaystyle P$ to $\displaystyle B(0,1,\text{-}2).$

Describe this set of points, giving appropriate information.

Let $\displaystyle (x,y,z)$ represent point $\displaystyle P.$


Then: .$\displaystyle \overline{PA} \:=\:\sqrt{(x-1)^2 + (y+2)^2 + (z-3)^2)}$

. and: .$\displaystyle \overline{PB} \:=\:\sqrt{x^2 + (y-1)^2 + (z+2)^2}$


Since $\displaystyle \overline{PA} \:=\:3\cdot\overline{PB}\!:$

. . . . . . . . $\displaystyle \sqrt{(x-1)^2 + (y+2)^2 + (z-3)^2} \;=\;3\sqrt{x^2 + (y-1)^2 + (z+2)^2}$

. . $\displaystyle x^2-2x+1 + y^2 + 4y + 4 + z^2 - 6z + 9 \;=\;$ .$\displaystyle 9(x^2 + y^2 - 2y + 1 + z^2 + 4z + 4)$

. . . . .
. . .$\displaystyle x^2 - 2x + y^2 + 4y + z^2 - 6z + 14 \;=\;9x^2 + 9y^2 - 18y + 9 + 9z^2 + 36z + 36$

. . . . . . . $\displaystyle 8x^2 + 2x + 8y^2 - 22y + 8z^2 + 42z \;=\; \text{-}31$


Divide by 8: .$\displaystyle x^2 + \frac{1}{4}x + y^2 - \frac{11}{4}y + z^2 + \frac{21}{4}z \;=\;\text{-}\frac{31}{8}$



Complete the square:

.$\displaystyle \left(x^2 + \frac{1}{4}x + \frac{1}{64}\right) + \left(y^2 - \frac{11}{4}y + \frac{121}{64}\right) + \left(z^2 + \frac{21}{4}z + \frac{441}{64}\right) \;=\;\text{-}\frac{31}{8} + \frac{1}{64} + \frac{121}{16} + \frac{441}{64}$

. . . . . . . . . . . . . . . . $\displaystyle \left(x + \frac{1}{8}\right)^2 + \left(y - \frac{11}{2}\right)^2 + \left(z + \frac{21}{8}\right)^2 \:=\:\frac{315}{64}$



The set of points is a sphere with center $\displaystyle \left(\text{-}\frac{1}{8},\:\frac{11}{2},\:\text{-}\frac{21}{8}\right)$ and radius $\displaystyle \frac{3\sqrt{35}}{8}$