Question

Perpendicular distance of the point $(3,4,5)$ from the $y$- axis is
A. $\sqrt {34} $
B. $\sqrt {41} $
C. $4$
D. $5$

Answer 1

Hint: We learnt how to calculate the distance between two points like $A$ and $B$ , in coordinate geometry. Assume that two points $A = ({x_1},{y_1},{z_1})$ and $B = ({x_2},{y_2},{z_2})$ have coordinates that are on the same line. The length between the two is equal to $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $ .

Formula used:
Distance Formula: $d = \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:
Consider the given point be $A(3,4,5)$ .
We have to find the perpendicular distance of the point $A$ from $y$- axis.
In coordinate geometry, we know that any point on the $y$- axis has coordinate $B(0,y,0)$.
Since we have to find perpendicular distance so we will consider y-coordinate according to the point
With respect to the given point $A(3,4,5)$ , we have $y$- axis coordinates as $B(0,4,0)$ .
Now, we need to find the distance between the points $A$ and $B$ using the formula
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $
Since we have
$({x_1},{y_1},{z_1}) = (3,4,5)$ and $({x_2},{y_2},{z_2}) = (0,4,0)$ .
Substituting these values in our distance formula, we get
$d = \sqrt {{{\left( {0 - 3} \right)}^2} + {{\left( {4 - 4} \right)}^2} + {{\left( {0 - 5} \right)}^2}} $
$d = \sqrt {{3^2} + {5^2}} $
Solving the root, we get the value of the perpendicular distance of point $A(3,4,5)$ from $y$- axis
$d = \sqrt {34} $ .

Hence, the correct option is option (A).

Note: We have $({y_2} - {y_1}) = 0$ in the problem mentioned above, as you can see. The points are therefore on the same horizontal line, as indicated by this. If we had $({x_2} - {x_1}) = 0$ , this implies that the points are therefore on the same vertical line. Similarly, goes for $({z_2} - {z_1}) = 0$. There can be no negative distance between two points.