Q.Find the foot of perpendicular from the point $(2, 3, -8)$ to the line $\dfrac{4-x}{2} = \dfrac{y}{6} = \dfrac{1-z}{3}$. Also, find the perpendicular distance from the given point to the line.

Concept: Distance From Point To Line (3D) — the foot of the perpendicular is the point on the line that minimises distance; the perpendicular distance is then the length of that segment.

Step 1: Rewrite the line in symmetric form.

The given line is $\dfrac{4-x}{2} = \dfrac{y}{6} = \dfrac{1-z}{3}$.

Rewrite as $\dfrac{x-4}{-2} = \dfrac{y}{6} = \dfrac{z-1}{-3}$.

So direction ratios are $(-2, 6, -3)$ and a point on the line is $A(4, 0, 1)$.

Step 2: Parameterise the line.

Let $\dfrac{x-4}{-2} = \dfrac{y}{6} = \dfrac{z-1}{-3} = t$.

Then any point $P$ on the line is:

$P = (4 - 2t,\; 6t,\; 1 - 3t)$.

Step 3: Condition for foot of perpendicular.

Let the given point be $Q(2, 3, -8)$.

Vector $\overrightarrow{PQ} = (2 - (4-2t),\; 3 - 6t,\; -8 - (1-3t)) = (-2 + 2t,\; 3 - 6t,\; -9 + 3t)$.

This must be perpendicular to the direction vector $(-2, 6, -3)$: …

Concept: foot of the perpendicular from a point to a line in 3D

The foot of the perpendicular is the unique point $P$ on the line at which the segment from the given point $A$ meets the line at a right angle. So take a general point $P(t)$ on the line, form $\overrightarrow{AP}$, and impose $\overrightarrow{AP}\cdot\vec d=0$ (perpendicular to the direction $\vec d$). Solving for $t$ locates $P$; the distance is $|\overrightarrow{AP}|$.

Step 1 - Write the line in standard form.

The line passes through $(4,0,1)$ with direction $\vec d=(-2,\,6,\,-3)$.

Step 2 - General point on the line. Let the common ratio be $t$:

Step 3 - Apply the perpendicularity condition. With $A=(2,3,-8)$,

Set $\overrightarrow{AP}\cdot\vec d=0$:

$-4+4t+36t-18-27+9t=0\ \Rightarrow\ 49t-49=0\ \Rightarrow\ t=1.$ …

Method: Foot of the perpendicular from a point to a line (and the distance)

Use this to find the point on a line closest to a given external point, and the shortest distance to it.

Steps

Step 1: Standardise the line.

Rewrite it as $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}=t$, reading off a point $(x_0,y_0,z_0)$ and direction $\vec d=(a,b,c)$. Watch signs: a numerator like $4-x$ hides a $-1$, so that direction ratio is negative.

Step 2: Take a general point on the line.

Write the foot as $F(t)=(x_0+at,\ y_0+bt,\ z_0+ct)$ — one unknown $t$.

Step 3: Impose perpendicularity.

The foot is where the join from the given point $A$ to $F(t)$ is perpendicular to the line: …

Common Mistakes

Mistake 1: Misreading the direction because of a reversed numerator.

Why it's wrong: $\dfrac{4-x}{2}$ is $\dfrac{x-4}{-2}$, so the $x$ direction ratio is $-2$, not $+2$; the same flips the $z$ term of $\dfrac{1-z}{3}$. Correct approach: rewrite every fraction as $\dfrac{x-x_0}{a}$ before reading $\vec d$.

Mistake 2: Setting the join perpendicular to a point on the line instead of its direction. …