Question

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

• A. $(0,1,3)$
• B. $(1,0,3)$
• C. $(0,2,3)$
• D. $(0,3,1)$

Answer 1

Answer: (A)

Since $x$-coordinate of every point in $yz$-plane is zero. Let $P(0,y,z)$ be a point on the $yz$-plane such that $PA=PB=PC$.
Now $P{A}^2=P{B}^2$
$\Rightarrow {\left(0-2\right)}^2+{\left(y-0\right)}^2+{\left(z-3\right)}^2$
$={\left(0-0\right)}^2+{\left(y-3\right)}^2+{\left(z-2\right)}^2$
i.e., $z-3y=0...\left(1\right)$
and $P{B}^2=P{C}^2$
$\Rightarrow y^2+9-6y+z^2+4-4z$
$=y^2+z^2+1-2z$
i.e., $3y+z=6...\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get $y=1$, $z=3$
Hence, the coordinates of the point $P$ are $\left(0,1,3\right)$.