Determine the points yz-plane equidistant from the points A(1, -1… - Vidyadip

Question

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

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Answer

Let $Q(0, y, z)$ be the required point.
So
$(A Q)^2=(B Q)^2 \Rightarrow(0-1)^2+(y+1)^2-(z-0)^2=(0-2)+(y+1)^2+(z-2)^2$
$\Rightarrow 1+y^2+1+2 y+z^2=4+y^2+1-2 y+z^2+4-42$
$\Rightarrow 4 y+4 z=7 \ldots(\text { i })$
$(B Q)^2=(C Q)^2 \Rightarrow(0-z)^2+(y-1)^2+(z-2)^2=(0-3)^2+(y-2)^2(2+1)^2$
$\Rightarrow 4+y^2+1-2 y+z^2+4-4 z-9+y^2+4-4 y+z^2+1+2 z$
$\Rightarrow 2 y-6 z=5 \ldots(\text { (ii) }$
$(\mathrm{AQ})^2=(C Q)^2 \Rightarrow(0-1)^2+(y+1)^2+(z-0)^2=(0-3)^2+(y-2)^2(z+1)^2$
$\Rightarrow 1+y^2+2 y+1+z^2=9+y^2-4 y+4+z^2+1+2 z$
$\Rightarrow 6 y-2 z=12 \ldots \text { (iii) }$
Solving equation (i) and (ii) we get
$\text{z}=\frac{-3}{16}$ and $\text{y} = \frac{31}{16}$
Put the value of y and z in equation (iii)
$6y - 2z = 12 = 12$
$6\Big(\frac{31}{16}\Big)-2\Big(\frac{-3}{16}\Big) = 12$
$\frac{186}{16}+\frac{6}{16}=12$
$\frac{192}{16}=12$
$12=12$
LHS = RHS.
so,
$\text{y}=\frac{31}{16},\ \text{z}=\frac{13}{16}$
Required point $=\Big(0,\ \frac{31}{16},\ \frac{-3}{16}\Big)$

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