Q.L is the foot of the perpendicular drawn from a point $(3,4,5)$ on $x$-axis. The coordinates of L are
(A) $(3,0,0)$
(B) $(0,4,0)$
(C) $(0,0,5)$
(D) none of these

The key idea is that any point on the $x$-axis has coordinates $(x,0,0)$ — the $y$ and $z$ coordinates are zero. The foot of the perpendicular from a given point to the $x$-axis is found by dropping the $y$ and $z$ coordinates to zero while keeping the $x$-coordinate unchanged.

Step 1: The given point is $P(3,4,5)$. …

The key idea here is understanding what the $x$-axis actually is in 3D space. The $x$-axis consists of all points where the $y$ and $z$ coordinates are zero — only $x$ can vary. So any point on the $x$-axis has the form $(a, 0, 0)$.

When we drop a perpendicular from a point to the $x$-axis, we are essentially finding the closest point on that axis. Because the $x$-axis runs parallel to the $x$-direction, the perpendicular from a point will land directly "below" or "above" it in the $y$ and $z$ directions. That means the $x$-coordinate of the foot is the same as the original point's $x$-coordinate, and the $y$ and $z$ coordinates become zero.

Let's work through it step by step.

1. Identify the target line. The $x$-axis is the set of points $\{(x,0,0) \mid x \in \mathbb{R}\}$. So any foot of the perpendicular must have $y=0$ and $z=0$.

2. Find the perpendicular condition. The line from the given point $P(3,4,5)$ to the foot $L$ must be perpendicular to the $x$-axis. The $x$-axis direction vector is $\hat{i} = (1,0,0)$. The vector $\overrightarrow{PL}$ is $(x-3, 0-4, 0-5) = (x-3, -4, -5)$. For perpendicularity, the dot product with $(1,0,0)$ must be zero:

So $x = 3$. …