Question

What is the distance of point $(4,3,5)$ from the y-axis?
A. $\sqrt {34} $
B.$5$
C. $\sqrt {41} $
D. $\sqrt {15} $

Answer 1

Hint: The distance of a point from an axis is the square root of the sum of squares of the two coordinates other than the one included in the axis itself. For example, the distance of point $(x,y,z)$ from the x-axis is $\sqrt {{y^2} + {z^2}} $ units respectively.

Complete step by step Solution:
The given point is: $(4,3,5)$
As the distance of a point say, $(x,y,z)$ from the x-axis is calculated as:
$\sqrt {{y^2} + {z^2}} $ units
Similarly,
The distance of the same point from the y-axis is calculated as:
$\sqrt {{x^2} + {z^2}} $ units
Therefore, the distance of the point $(4,3,5)$ from the y-axis is calculated as:
$\sqrt {{4^2} + {5^2}} $
On simplifying further, we get $\sqrt {41} $ units.
Thus, the distance of the point $(4,3,5)$ from y-axis is $ \sqrt {41} $ units respectively.

Hence, the correct option is (C).

Note: The distance of a point $P,(x,y,z)$ in the vector form of $\mathop {OP}\limits^ \to $ from a line $\vec l$ , with direction vectors as $\vec d$ , in space is calculate by $\dfrac{{\left| {\mathop {OP \times \vec d}\limits^ \to } \right|}}{{\left| {\vec d} \right|}}$ . Now, the direction cosines of y-axis are $(0,1,0)$ making its direction vector as $\hat j$ . This formula then reduces to $\sqrt {{x^2} + {z^2}} $ which was used to solve the above question.