
The distance of the point (1,2,3) is $\sqrt {10} $ from which of the following:
A. Origin
B. x-axis
C. y-axis
D. z-axis
Answer
233.1k+ views
Hint: Use the distance formula to calculate the distance between points (1,2,3) and each of the four options. The distance between two points, $P({x_1},{y_1},{z_1})\,{\text{and }}Q({x_2},{y_2},{z_2})$ is given by the distance formula \[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \].
Formula Used:
If two points $P({x_1},{y_1},{z_1})\,{\text{and }}Q({x_2},{y_2},{z_2})$ then \[PQ = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}}\]
Complete step by step solution:
We know that the distance between two points, $P({x_1},{y_1},{z_1})\,{\text{and }}Q({x_2},{y_2},{z_2})$ is given by the formula
\[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]
The coordinate of the origin is (0,0,0). So, the distance between the origin and point (1,2,3) is \[\sqrt {{{(1-0)}^2} + {{(2-0)}^2} + {{(3-0)}^2}} = \sqrt {14} \] units
The coordinates of the point on the x-axis whose distance from the point (1,2,3) we need to calculate is (1,0,0). So, the distance between point (1,0,0) and point (1,2,3) is \[\sqrt {{{(1 - 1)}^2} + {{(2-0)}^2} + {{(3-0)}^2}} = \sqrt {13}\] units.
The coordinates of the point on the y-axis whose distance from the point (1,2,3) we need to calculate is (0,2,0). So, the distance between point (0,2,0) and point (1,2,3) is \[\sqrt {{{(1-0)}^2} + {{(2 - 2)}^2} + {{(3-0)}^2}} = \sqrt {10} \] units.
The coordinates of the point on the z-axis whose distance from the point (1,2,3) we need to calculate are (0,0,3). So, the distance between point (0,0,3) and point (1,2,3) is \[\sqrt {{{(1-0)}^2} + {{(2-0)}^2} + {{(3 - 3)}^2}} = \sqrt 5 \] units
Therefore, option (C) y-axis is the correct answer.
Note: In this question, for a given point A (p,q,r), to calculate the distance from A to the x-axis, we need to calculate the distance between A and (p,0,0) as that is the shortest distance between A and any point on the x-axis.
Formula Used:
If two points $P({x_1},{y_1},{z_1})\,{\text{and }}Q({x_2},{y_2},{z_2})$ then \[PQ = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}}\]
Complete step by step solution:
We know that the distance between two points, $P({x_1},{y_1},{z_1})\,{\text{and }}Q({x_2},{y_2},{z_2})$ is given by the formula
\[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]
The coordinate of the origin is (0,0,0). So, the distance between the origin and point (1,2,3) is \[\sqrt {{{(1-0)}^2} + {{(2-0)}^2} + {{(3-0)}^2}} = \sqrt {14} \] units
The coordinates of the point on the x-axis whose distance from the point (1,2,3) we need to calculate is (1,0,0). So, the distance between point (1,0,0) and point (1,2,3) is \[\sqrt {{{(1 - 1)}^2} + {{(2-0)}^2} + {{(3-0)}^2}} = \sqrt {13}\] units.
The coordinates of the point on the y-axis whose distance from the point (1,2,3) we need to calculate is (0,2,0). So, the distance between point (0,2,0) and point (1,2,3) is \[\sqrt {{{(1-0)}^2} + {{(2 - 2)}^2} + {{(3-0)}^2}} = \sqrt {10} \] units.
The coordinates of the point on the z-axis whose distance from the point (1,2,3) we need to calculate are (0,0,3). So, the distance between point (0,0,3) and point (1,2,3) is \[\sqrt {{{(1-0)}^2} + {{(2-0)}^2} + {{(3 - 3)}^2}} = \sqrt 5 \] units
Therefore, option (C) y-axis is the correct answer.
Note: In this question, for a given point A (p,q,r), to calculate the distance from A to the x-axis, we need to calculate the distance between A and (p,0,0) as that is the shortest distance between A and any point on the x-axis.
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