Hint: For solving this question, we need to find the coordinates of the foot of the perpendicular from the given point on the y-axis. Since on the y axis, the x and the z coordinates are equal to zero, the x and the z coordinates of the foot of the perpendicular will be equal to zero. Also, since the perpendicular line is normal to the y-axis, the y-coordinate of the foot will be equal to that of the given point. Finally, using the distance formula given by $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$, we can determine the required length of the perpendicular.
Complete step by step solution:
For finding out the length of perpendicular from the given point $P\left( 3,4,5 \right)$ on the y-axis, we first need to determine the coordinates of the foot of the perpendicular from the given point on the y-axis.
We know that on the y-axis, the x and the y coordinates are equal to zero. Since the foot of the perpendicular from the given point will lie on the y-axis, both the x and the y coordinates of the foot of the perpendicular will be equal to zero.
Now, since the perpendicular line is normal to the y-axis, it will be parallel to the x-z plane. Therefore, the y-coordinate has to remain constant. Therefore, the y coordinates of all the points on the perpendicular will be equal to the y coordinate of the given point P, which is equal to P. Now, since the foot of the perpendicular lies on the perpendicular, its y coordinate will be equal to $4$.
Therefore, the coordinates of the foot of the perpendicular will be equal to $F\left( 0,4,0 \right)$.
Now, the length of the perpendicular is equal to the distance of the point P from the foot of the perpendicular F, which can be calculated by using the distance formula which is given by
$\Rightarrow d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$
Therefore, the distance PF will be given by
\[\begin{align}
& \Rightarrow PF=\sqrt{{{\left( 0-3 \right)}^{2}}+{{\left( 4-4 \right)}^{2}}+{{\left( 0-5 \right)}^{2}}} \\
& \Rightarrow PF=\sqrt{9+0+25} \\
& \Rightarrow PF=\sqrt{34} \\
\end{align}\]
Hence, the required length of the perpendicular of the point $P\left( 3,4,5 \right)$ on the y-axis is equal to \[\sqrt{34}\] units.
Note: Do not forget the z component in the formula for the distance between two points given as $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$. This is because it is the distance in three dimensions. For the two dimensional geometry, the distance formula is given by $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$, which involves no z component.
