Q.What is the length of foot of perpendicular drawn from the point $P(3,4,5)$ on $y$-axis
(A) $\sqrt{41}$
(B) $\sqrt{34}$
(C) $5$
(D) none of these

The problem asks for the length of the perpendicular segment from a point to the $y$-axis, which is equivalent to finding the shortest distance from the point to the $y$-axis.

Concept: Distance From Point To Line (specifically, a coordinate axis).

1. Identify the given point as $P(x_0, y_0, z_0) = P(3,4,5)$.
2. The $y$-axis is the line where $x=0$ and $z=0$. The foot of the perpendicular from any point $(x_0, y_0, z_0)$ to the $y$-axis will have coordinates $(0, y_0, 0)$.
3. For $P(3,4,5)$, the foot of the perpendicular on the $y$-axis is $F(0,4,0)$.
4. Calculate the distance between $P(3,4,5)$ and $F(0,4,0)$ using the distance formula: …

When we talk about the "foot of the perpendicular" from a point to a line (or an axis in 3D), we are referring to the specific point on that line which is closest to the given point. Imagine dropping a plumb line from the point straight down to the axis; where it lands is the foot of the perpendicular. The "length of the foot of the perpendicular" is a common phrasing that actually means the distance from the original point to this foot.

For a point in 3D space, say $P(x_0, y_0, z_0)$, and an axis, the foot of the perpendicular has a very specific form:

• If the axis is the x-axis, the foot will be $(x_0, 0, 0)$.
• If the axis is the y-axis, the foot will be $(0, y_0, 0)$.
• If the axis is the z-axis, the foot will be $(0, 0, z_0)$.

This is because the perpendicular line segment from the point to the axis must be parallel to the plane formed by the other two axes. For instance, for the y-axis, the perpendicular segment is parallel to the xz-plane, meaning its x and z components change to zero, while its y-component remains fixed.

Let's apply this understanding to the given problem.

1. Identify the given point:

   We are given the point $P(3,4,5)$. Here, $x_0=3$, $y_0=4$, and $z_0=5$.

2. Identify the target axis:

   The perpendicular is drawn to the $y$-axis. The $y$-axis is the line where $x=0$ and $z=0$.

3. Determine the coordinates of the foot of the perpendicular:

   As discussed, for a point $(x_0, y_0, z_0)$ and the $y$-axis, the foot of the perpendicular will have coordinates $(0, y_0, 0)$.

   For our point $P(3,4,5)$, the foot of the perpendicular, let's call it $F$, will be $(0, 4, 0)$.

   ›Proof

   To be rigorous, let $P(x_0, y_0, z_0)$ be the given point. Any point on the y-axis can be represented as $Q(0, k, 0)$ for some scalar $k$.

   The vector $\vec{PQ}$ connecting $P$ to $Q$ is $(0-x_0, k-y_0, 0-z_0) = (-x_0, k-y_0, -z_0)$.

   The direction vector of the y-axis is $\vec{j} = (0, 1, 0)$.

   For $\vec{PQ}$ to be perpendicular to the y-axis, their dot product must be zero:

   $\vec{PQ} \cdot \vec{j} = (-x_0)(0) + (k-y_0)(1) + (-z_0)(0) = 0$

   This simplifies to $k-y_0 = 0$, which means $k = y_0$.

   Therefore, the foot of the perpendicular $F$ is indeed $(0, y_0, 0)$. …