
The distances of the point P (1,2,3) from the coordinate axes are:
A. $\sqrt {13} ,\sqrt {10} ,\sqrt 5 $
B. $\sqrt {11} ,\sqrt {10} ,\sqrt 5 $
C. $\sqrt {13} ,\sqrt {20} ,\sqrt {15} $
D. $\sqrt {23} ,\sqrt {10} ,\sqrt 5 $
Answer
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Hint: Find the points on the coordinate axes from which we need to find the distance till P. Then use the distance formula to find all three distances. 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 \[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \].
Formula used: \[{\text{Distance}}\, = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]
Complete step by step solution:
The point on the x-axis from which we need to calculate the distance to point P is (1,0,0) as this the perpendicular from P to the x-axis.
The point on the y-axis from which we need to calculate the distance to point P is (0,2,0) as this the perpendicular from P to the y-axis.
The point on the z-axis from which we need to calculate the distance to point P is (0,0,3) as this the perpendicular from P to the z-axis.
We know that given two points $A({x_1},{y_1},{z_1})\,{\text{and B}}({x_2},{y_2},{z_2})$ the distance AB is found using the following formula:
\[{\text{distance}}\, = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]. This is called the distance formula.
Distance from x-axis to point P:
$\sqrt {{{(1 - 1)}^2} + {{(2 - 0)}^2} + {{(3 - 0)}^2}} = \sqrt {4 + 9} = \sqrt {13} \,units$
Distance from y-axis to point P:
$\sqrt {{{(1 - 0)}^2} + {{(2 - 2)}^2} + {{(3 - 0)}^2}} = \sqrt {1 + 9} = \sqrt {10} \,units$
Distance from z-axis to point P:
$\sqrt {{{(1 - 0)}^2} + {{(2 - 0)}^2} + {{(3 - 3)}^2}} = \sqrt {1 + 4} = \sqrt 5 \,units$
Therefore, the distances of the point P (1,2,3) from the coordinate axes are $\sqrt {13} ,\sqrt {10} ,\sqrt 5 $.
The correct option is option A. $\sqrt {13} ,\sqrt {10} ,\sqrt 5 $
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: \[{\text{Distance}}\, = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]
Complete step by step solution:
The point on the x-axis from which we need to calculate the distance to point P is (1,0,0) as this the perpendicular from P to the x-axis.
The point on the y-axis from which we need to calculate the distance to point P is (0,2,0) as this the perpendicular from P to the y-axis.
The point on the z-axis from which we need to calculate the distance to point P is (0,0,3) as this the perpendicular from P to the z-axis.
We know that given two points $A({x_1},{y_1},{z_1})\,{\text{and B}}({x_2},{y_2},{z_2})$ the distance AB is found using the following formula:
\[{\text{distance}}\, = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \]. This is called the distance formula.
Distance from x-axis to point P:
$\sqrt {{{(1 - 1)}^2} + {{(2 - 0)}^2} + {{(3 - 0)}^2}} = \sqrt {4 + 9} = \sqrt {13} \,units$
Distance from y-axis to point P:
$\sqrt {{{(1 - 0)}^2} + {{(2 - 2)}^2} + {{(3 - 0)}^2}} = \sqrt {1 + 9} = \sqrt {10} \,units$
Distance from z-axis to point P:
$\sqrt {{{(1 - 0)}^2} + {{(2 - 0)}^2} + {{(3 - 3)}^2}} = \sqrt {1 + 4} = \sqrt 5 \,units$
Therefore, the distances of the point P (1,2,3) from the coordinate axes are $\sqrt {13} ,\sqrt {10} ,\sqrt 5 $.
The correct option is option A. $\sqrt {13} ,\sqrt {10} ,\sqrt 5 $
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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