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Question: The points equidistant from the point \(O\left( {0,0,0} \right),A\left( {a,0,0} \right),B\left( {0,b...

The points equidistant from the point O(0,0,0),A(a,0,0),B(0,b,0)O\left( {0,0,0} \right),A\left( {a,0,0} \right),B\left( {0,b,0} \right) and C(0,0,c)C\left( {0,0,c} \right) has the coordinates.

  1. (a,b,c)\left( {a,b,c} \right)
  2. (a2,b2,c2)\left( {\dfrac{a}{2},\dfrac{b}{2},\dfrac{c}{2}} \right)
  3. (a3,b3,c3)\left( {\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}} \right)
  4. (a4,b4,c4)\left( {\dfrac{a}{4},\dfrac{b}{4},\dfrac{c}{4}} \right)
Explanation

Solution

Hint: First of all we will let a point P(x,y,z)P\left( {x,y,z} \right) and then find the distance from it to each of the given points using the distance formula, (x1−x2)2+(y1−y2)2+(z1−z2)2\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2} + {{\left( {{z_1} - {z_2}} \right)}^2}} . Since, each point is equidistant from the point P(x,y,z)P\left( {x,y,z} \right). Equate the distance POPO to PAPA and calculate the value of xx.
Similarly, we will equate POPO to PBPB and calculate the value of yy and POPO to PCPC and calculate the value of zz. Also, we will write the final answer in coordinate form.

Complete step by step answer:

In this type of questions, where we have to find a point, we first let that point.
So, let P(x,y,z)P\left( {x,y,z} \right) be a point such that the point is equidistant from the point O(0,0,0),A(a,0,0),B(0,b,0)O\left( {0,0,0} \right),A\left( {a,0,0} \right),B\left( {0,b,0} \right) and C(0,0,c)C\left( {0,0,c} \right).
Next, let us find the distances of all the points from PP using the formula, (x1−x2)2+(y1−y2)2+(z1−z2)2\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2} + {{\left( {{z_1} - {z_2}} \right)}^2}} .
We begin by finding the distance POPO where coordinates of PP are (x,y,z)\left( {x,y,z} \right) and coordinates of OO are (0,0,0)\left( {0,0,0} \right).
PO=(x−0)2+(y−0)2+(z−0)2 PO=x2+y2+z2  PO = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2} + {{\left( {z - 0} \right)}^2}} \\\ PO = \sqrt {{x^2} + {y^2} + {z^2}} \\\
Similarly, find the distance PAPA where coordinates of PPare (x,y,z)\left( {x,y,z} \right) and coordinates of AA are (a,0,0)\left( {a,0,0} \right).
PA=(a−x)2+(0−y)2+(0−z)2 PA=(a−x)2+y2+z2  PA = \sqrt {{{\left( {a - x} \right)}^2} + {{\left( {0 - y} \right)}^2} + {{\left( {0 - z} \right)}^2}} \\\ PA = \sqrt {{{\left( {a - x} \right)}^2} + {y^2} + {z^2}} \\\
Next, find the distance PBPB where coordinates of PPare (x,y,z)\left( {x,y,z} \right) and coordinates of BB are (0,b,0)\left( {0,b,0} \right).
PB=(0−x)2+(b−y)2+(0−z)2 PB=x2+(b−y)2+z2  PB = \sqrt {{{\left( {0 - x} \right)}^2} + {{\left( {b - y} \right)}^2} + {{\left( {0 - z} \right)}^2}} \\\ PB = \sqrt {{x^2} + {{\left( {b - y} \right)}^2} + {z^2}} \\\
And lastly, find the distance PCPC where coordinates of PPare (x,y,z)\left( {x,y,z} \right) and coordinates of CC are (0,0,c)\left( {0,0,c} \right).
PC=(0−x)2+(0−y)2+(c−z)2 PC=x2+y2+(c−z)2  PC = \sqrt {{{\left( {0 - x} \right)}^2} + {{\left( {0 - y} \right)}^2} + {{\left( {c - z} \right)}^2}} \\\ PC = \sqrt {{x^2} + {y^2} + {{\left( {c - z} \right)}^2}} \\\
Since, each point is equidistant, let PO=PAPO = PA
x2+y2+z2=(a−x)2+y2+z2\sqrt {{x^2} + {y^2} + {z^2}} = \sqrt {{{\left( {a - x} \right)}^2} + {y^2} + {z^2}}
Squaring both sides, we get,
x2+y2+z2=(a−x)2+y2+z2{x^2} + {y^2} + {z^2} = {\left( {a - x} \right)^2} + {y^2} + {z^2}
Terms y2{y^2} and z2{z^2} gets cancelled and we get,
x2=(a−x)2 x=±(a−x)  {x^2} = {\left( {a - x} \right)^2} \\\ x = \pm \left( {a - x} \right) \\\
If we take, (a−x)\left( {a - x} \right), then,
x=a−x 2x=a x=a2  x = a - x \\\ 2x = a \\\ x = \dfrac{a}{2} \\\
If we take, (a−x)\left( {a - x} \right), then,
x=−a+x \-a=0  x = - a + x \\\ \- a = 0 \\\
This does not give us the value of xx.

Similarly, let PO=PBPO = PB
x2+y2+z2=x2+(b−y)2+z2\sqrt {{x^2} + {y^2} + {z^2}} = \sqrt {{x^2} + {{\left( {b - y} \right)}^2} + {z^2}}
Squaring both sides, we get,
x2+y2+z2=x2+(b−y)2+z2{x^2} + {y^2} + {z^2} = {x^2} + {\left( {b - y} \right)^2} + {z^2}
Terms x2{x^2} and z2{z^2} gets cancelled and we get,
y2=(b−y)2 y=±(b−y)  {y^2} = {\left( {b - y} \right)^2} \\\ y = \pm \left( {b - y} \right) \\\
If y=b−yy = b - y, then, we get,
y=b−y 2y=b y=b2  y = b - y \\\ 2y = b \\\ y = \dfrac{b}{2} \\\
If we take y=−(b−y)y = - \left( {b - y} \right), then,
y=−(b−y) y=−b+y b=0  y = - \left( {b - y} \right) \\\ y = - b + y \\\ b = 0 \\\
This case does not give any value of yy.
Also, PO=PCPO = PC
x2+y2+z2=x2+y2+(z−c)2\sqrt {{x^2} + {y^2} + {z^2}} = \sqrt {{x^2} + {y^2} + {{\left( {z - c} \right)}^2}}
Squaring both sides, we get,
x2+y2+z2=x2+y2+(z−c)2{x^2} + {y^2} + {z^2} = {x^2} + {y^2} + {\left( {z - c} \right)^2}
Terms y2{y^2} and x2{x^2} gets cancelled and we get,
z2=(c−z)2 z=±(c−z)  {z^2} = {\left( {c - z} \right)^2} \\\ z = \pm \left( {c - z} \right) \\\
If we take z=(c−z)z = \left( {c - z} \right), then we get,
z=c−z 2z=c z=c2  z = c - z \\\ 2z = c \\\ z = \dfrac{c}{2} \\\
If we take z=−(c−z)z = - \left( {c - z} \right), then we get,
z=−(c−z) z=−c+z c=0  z = - \left( {c - z} \right) \\\ z = - c + z \\\ c = 0 \\\
This case will not give the value of zz.
Write the answer in coordinates form.
Therefore, point P(x,y,z)P\left( {x,y,z} \right) is (a2,b2,c2)\left( {\dfrac{a}{2},\dfrac{b}{2},\dfrac{c}{2}} \right).
Hence, option B is the correct option.

Note: It is important to square both sides, after equating the distances to avoid mistakes in calculation. Also, while solving the variables, there are two answers possible when square-root is done. Take the condition that holds true, that if we take x=−(a−x)x = - \left( {a - x} \right), then we will get, x=−a+x⇒a=0x = - a + x \Rightarrow a = 0, which is incorrect.