# Thread: Operations with Algebraic Vectors

1. ## Operations with Algebraic Vectors

The question that I am having trouble with is:
a) Find the point on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3).
b) Find a point not on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3).

I'm unsure of where to even start, so if anyone can give me any hints or suggestions that would be much appreciated! Please and thank you!

2. ## Re: Operations with Algebraic Vectors

Originally Posted by misiaizeska
The question that I am having trouble with is:
a) Find the point on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3).
b) Find a point not on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3).
For part a):
find the y-intercept of the plane through (1,0,2) with normal <2,-2,-3>.

For part b), find any other point on that same plane.

3. ## Re: Operations with Algebraic Vectors

Hello, misiaizeska!

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

The word "equidistant" should suggest the Distance Formula . . .

Let point $P$ be $(0,y,0)$

We have: . $\begin{Bmatrix}\overline{PA} &=& \sqrt{(0-2)^2 + (y+1)^2 + (0-1)^2} &=& \sqrt{y^2 + 2y - 6} \\ \overline{PB} &=& \sqrt{(0-0)^2 + (y-1)^2 + (0-30^2} &=& \sqrt{y^2-2y+10} \end{Bmatrix}$

Then: . $\sqrt{y^2+2y+6} \:=\:\sqrt{y^2-2y+10}$

. . . . . . . $y^2+2y + 6 \:=\:y^2 - 2y + 10$

. . . . . . . . $4y \:=\:4 \quad\Rightarrow\quad y \:=\:1$

$\text{Therefore: }\:P(0,1,0)$