vectors - find equidistant point

**theowne** · May 19th 2008, 07:39 PM

Using vectors:

Find the point on the y axis that is equidistant from the points (2,-1,1) and (0,1,3)

**flyingsquirrel** · May 20th 2008, 12:20 PM

Hi

The point $P$ we're looking for is on the $y$ axis hence $P(0,\,y,\,0)$

The distance between $P(0,\,y,\,0)$ and $A(2,\,-1,\,1)$ is $\|\vec{AP}\|=\sqrt{(2-0)^2+(-1-y)^2+(1-0)^2}=\sqrt{5+(1+y)^2}$

The distance between $P(0,\,y,\,0)$ and $B(0,\,1,\,3)$ is $\|\vec{BP}\|=\sqrt{(0-0)^2+(1-y)^2+(3-0)^2}=\sqrt{9+(1-y)^2}$

We want these two distances to be equal thus $\|\vec{AP}\|=\|\vec{BP}\| \Leftrightarrow \sqrt{9+(1-y)^2}=\sqrt{5+(1+y)^2}$

Squaring both sides gives $9+(1-y)^2=5+(1+y)^2$ which you can solve for $y$ .

Does it help ?

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