Question

A point moves so that its distances from the points \[\left( {3,4, - 2} \right)\] and \[\left( {2,3, - 3} \right)\] remains equal. Then find the locus of the point.
A. A line
B. A plane whose normal is equally inclined to axes
C. A plane which passes through the origin
D. A sphere

Answer 1

Hint: First, apply the distance formula and calculate the distance from the given points to that point. Then, equate both distances and solve the equation. After that, check the behaviour of the equation and get the required answer.

Formula used: The distance formula:
The distance between the two points \[P\left( {{x_1},{y_1},{z_1}} \right)\] and \[Q\left( {{x_2},{y_2},{z_2}} \right)\] is: \[PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \]
\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]

Complete step by step solution: Given:
A point moves so that its distances from the points \[\left( {3,4, - 2} \right)\] and \[\left( {2,3, - 3} \right)\] remains equal.

Let consider, the coordinates of the points are \[P\left( {x,y,z} \right)\].
Now calculate the distance between the points \[Q\left( {3,4, - 2} \right)\] and \[P\left( {x,y,z} \right)\].
Apply the distance formula \[PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \].
We get,
\[PQ = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 4} \right)}^2} + {{\left( {z + 2} \right)}^2}} \] \[.....\left( 1 \right)\]

Similarly, calculate the distance between the points \[R\left( {2,3, - 3} \right)\] and \[P\left( {x,y,z} \right)\].
We get,
\[PR = \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 3} \right)}^2} + {{\left( {z + 3} \right)}^2}} \] \[.....\left( 2 \right)\]

It is given that the distances from the points \[\left( {3,4, - 2} \right)\] and \[\left( {2,3, - 3} \right)\].
So, equate the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[\sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 4} \right)}^2} + {{\left( {z + 2} \right)}^2}} = \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 3} \right)}^2} + {{\left( {z + 3} \right)}^2}} \]
\[ \Rightarrow {\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} + {\left( {z + 2} \right)^2} = {\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {z + 3} \right)^2}\]
\[ \Rightarrow {x^2} + 9 - 6x + {y^2} + 16 - 8y + {z^2} + 4 + 4z = {x^2} + 4 - 4x + {y^2} + 9 - 6y + {z^2} + 9 + 6z\]
\[ \Rightarrow - 6x - 8y + 4z + 29 = - 4x - 6y + 6z + 22\]
\[ \Rightarrow - 2x - 2y - 2z + 7 = 0\]
\[ \Rightarrow 2x + 2y + 2z = 7\]
From observing the above equation, we get
The coefficients of all variables are the same. So, they are equally inclined to all axes.

Thus, Option (B) is correct.

Note: Students often get confused with the formula of the square of the differences of the numbers. Sometimes they consider it as \[{\left( {a - b} \right)^2} = {a^2} - 2ab - {b^2}\], because of the negative sign of \[b\]. But the correct formula is \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\].