
A point moves in such a way that the sum of its distance from \[xy\]-plane and \[yz\]-plane remains equal to its distance from \[xz\]-plane. Then find the locus of the point.
A. \[x - y + z = 2\]
B. \[x + y - z = 0\]
C. \[x - y + z = 0\]
D. \[x - y - z = 2\]
Answer
162.6k+ views
Hint: First, apply the distance formula and calculate the distance from any plane to that point. Then, substitute those values in the given condition. In the end, solve that 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: \[PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \]
Complete step by step solution: Given:
A point moves in a 3-D plane.
The sum of its distance from \[xy\]-plane and \[yz\]-plane remains equal to its distance from \[xz\]-plane.
Let’s calculate the distance between that point and the \[xy\]-plane.
Let consider, the coordinates of the point are \[P\left( {x,y,z} \right)\].
We know that the coordinates of any point in xy-plane is \[Q\left( {x,y,0} \right)\].
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,
\[PQ = \sqrt {{{\left( {x - x} \right)}^2} + {{\left( {y - y} \right)}^2} + {{\left( {z - 0} \right)}^2}} \]
\[ \Rightarrow PQ = \sqrt {{z^2}} \]
\[ \Rightarrow PQ = \left| z \right|\]
Similarly, we get
The distance between that point and the \[yz\]-plane: \[\left| x \right|\]
The distance between that point and the \[xz\]-plane: \[\left| y \right|\]
It is given that the sum of its distance from \[xy\]-plane and \[yz\]-plane remains equal to its distance from \xz\]-plane.
From the above information, we get
\[\left| z \right| + \left| x \right| = \left| y \right|\]
We know that the distance is always positive.
So,
\[x + z = y\]
\[ \Rightarrow x - y + z = 0\]
Thus, the locus of the point is \[x - y + z = 0\].
Thus, Option (C) is correct.
Note: A locus is a curve or shape that is formed by the collection of points whose position is represented by certain conditions.
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: \[PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \]
Complete step by step solution: Given:
A point moves in a 3-D plane.
The sum of its distance from \[xy\]-plane and \[yz\]-plane remains equal to its distance from \[xz\]-plane.
Let’s calculate the distance between that point and the \[xy\]-plane.
Let consider, the coordinates of the point are \[P\left( {x,y,z} \right)\].
We know that the coordinates of any point in xy-plane is \[Q\left( {x,y,0} \right)\].
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,
\[PQ = \sqrt {{{\left( {x - x} \right)}^2} + {{\left( {y - y} \right)}^2} + {{\left( {z - 0} \right)}^2}} \]
\[ \Rightarrow PQ = \sqrt {{z^2}} \]
\[ \Rightarrow PQ = \left| z \right|\]
Similarly, we get
The distance between that point and the \[yz\]-plane: \[\left| x \right|\]
The distance between that point and the \[xz\]-plane: \[\left| y \right|\]
It is given that the sum of its distance from \[xy\]-plane and \[yz\]-plane remains equal to its distance from \xz\]-plane.
From the above information, we get
\[\left| z \right| + \left| x \right| = \left| y \right|\]
We know that the distance is always positive.
So,
\[x + z = y\]
\[ \Rightarrow x - y + z = 0\]
Thus, the locus of the point is \[x - y + z = 0\].
Thus, Option (C) is correct.
Note: A locus is a curve or shape that is formed by the collection of points whose position is represented by certain conditions.
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