Question

Determine the point in the YZ-plane which is equidistant from three points $A(2,0,3),B(0,3,2)$ and $C(0,0,1)$.

Answer 1

STEP 1 : Given

Three points $A(2,0,3),B(0,3,2)$ and $C(0,0,1)$ are equidistant from a point $P$ on YZ-plane

Since $x$ coordinate of every point in the YZ-plane is zero, therefore

Let $P(0,y,z)$ be a point on the YZ-plane such that $PA=PB=PC$.

STEP 2 : Forming an equations to find $y$ and $z$ coordinates

We know that $PA=PB$

$\Rightarrow \sqrt{{\left(0-2\right)}^2+{\left(y-0\right)}^2+{\left(z-3\right)}^2}=\sqrt{{\left(0-0\right)}^2+{\left(y-3\right)}^2+{\left(z-2\right)}^2}$

Squaring both sides we get,

$\Rightarrow {\left(0-2\right)}^2+{\left(y-0\right)}^2+{\left(z-3\right)}^2={\left(0-0\right)}^2+{\left(y-3\right)}^2+{\left(z-2\right)}^2$

$\Rightarrow {\left(2\right)}^2+{\left(y\right)}^2+{\left(z-3\right)}^2={\left(0\right)}^2+{\left(y-3\right)}^2+{\left(z-2\right)}^2$

$\Rightarrow 4+y^2+(z^2+3^2-6z)=\left(y^2+3^2-6y\right)+\left(z^2+2^2-4z\right)$

$\Rightarrow 4+y^2+z^2+9-6z=y^2+9-6y+z^2+4-4z$

$\Rightarrow y^2+z^2-6z+13=y^2-6y+z^2-4z+13$

Cancelling the same terms from both the sides of the equation we get,

$\Rightarrow -6z=-6y-4z$

$\Rightarrow 6y-6z+4z=0$

$\Rightarrow 6y-2z=0$

$\Rightarrow 3y-z=0$ $\dots \left(1\right)$

STEP 3 : Forming another equations to find $y$ and $z$ coordinates

We also know that $PB=PC$

$\Rightarrow \sqrt{{\left(0-0\right)}^2+{\left(y-3\right)}^2+{\left(z-2\right)}^2}=\sqrt{{\left(0-0\right)}^2+{\left(y-0\right)}^2+{\left(z-1\right)}^2}$

Squaring both sides we get,

$\Rightarrow {\left(0-0\right)}^2+{\left(y-3\right)}^2+{\left(z-2\right)}^2={\left(0-0\right)}^2+{\left(y-0\right)}^2+{\left(z-1\right)}^2$

$\Rightarrow {\left(0\right)}^2+\left(y^2+3^2-6y\right)+\left(z^2+2^2-4z\right)={\left(0\right)}^2+{\left(y\right)}^2+\left(z^2+1^2-2z\right)$

$\Rightarrow y^2+9-6y+z^2+4-4z=y^2+z^2+1-2z$

Cancelling the same terms from both the sides of the equation we get,

$\Rightarrow 13-6y-4z=1-2z$

$\Rightarrow 12=6y+2z$

$\Rightarrow 6=3y+z$

$\Rightarrow 3y+z=6$ $\dots \left(2\right)$

STEP 4 : Solving the above $2$ equations

Adding equation $1$ and $2$ we get,

$3y+3y-z+z=0+6$

$6y=6$

$y=1$

Substitute in equation $2$ we get,

$3\times 1+z=6$

$z=6-3=3$

Therefore the coordinate of the point $P$ is $(0,1,3)$.

Hence, the point in the YZ-plane which is equidistant from three points $A(2,0,3),B(0,3,2)$ and $C(0,0,1)$ is $P(0,1,3)$