Lines and planes in 3D — vector equations

WHY do we even need vector equations?

In 2D you can describe a line with $y = mx + c$ . In 3D this breaks: a single equation $y=mx+c$ in 3D space carves out a plane, not a line. We need a language that:

works in any number of dimensions,
separates position (where we start) from direction (where we go),
handles vertical / weird-slope objects without dividing by zero.

Vectors do exactly this. A point gives us position; a direction vector gives us travel.

1. The Line

HOW we derive it (from scratch). Take any point $P$ (position $\mathbf{r}$ ) on the line and the fixed point $A$ (position $\mathbf{a}$ ). The vector $\overrightarrow{AP} = \mathbf{r}-\mathbf{a}$ lies along the line, so it must be parallel to the direction $\mathbf{d}$ . Two parallel vectors are scalar multiples: $\mathbf{r}-\mathbf{a} = t\,\mathbf{d}\;\Rightarrow\; \mathbf{r}=\mathbf{a}+t\mathbf{d}.$ Why this step? "Lies along the line" is the entire geometric meaning of a line — every displacement on it is a stretch/shrink of the same direction.

Line through two points

Through $A(\mathbf a)$ and $B(\mathbf b)$ : the direction is $\mathbf{d}=\mathbf{b}-\mathbf{a}$ , so $\mathbf{r}=\mathbf{a}+t(\mathbf{b}-\mathbf{a}) = (1-t)\mathbf{a}+t\mathbf{b}.$ At $t=0$ you're at $A$ , at $t=1$ at $B$ , $0\le t\le 1$ traces the segment.

2. The Plane

A plane needs two non-parallel directions, or — cleaner — one perpendicular direction (its normal).

WHY the normal form works. For any point $P$ in the plane, $\mathbf r-\mathbf a$ lies in the plane, hence is perpendicular to the normal $\mathbf n$ . Perpendicular ⇒ dot product zero. That single equation captures every in-plane point.

HOW to go from parametric to normal: the normal must be perpendicular to both spanning directions, so $\mathbf n=\mathbf u\times\mathbf v.$ Why this step? The cross product produces exactly the vector orthogonal to both inputs — the geometric definition of the plane's normal.

Figure — Lines and planes in 3D — vector equations

3. Distances (the 80/20 high-value formulas)

Worked Examples

Common mistake Steel-manned errors

(a) "Different parameter $t$ values ⇒ different lines." Why it feels right: changing numbers usually changes objects. Fix: a line is a set of points; many $(\mathbf a,\mathbf d)$ pairs describe the same set. Lines are equal iff directions are parallel and they share a point.

(b) "The Cartesian plane $ax+by+cz=d$ — the normal is $(a,b,c,d)$ ." Why it feels right: all four numbers look symmetric. Fix: the normal is only $(a,b,c)$ . The constant $d$ shifts the plane along $\mathbf n$ ; it isn't part of the direction.

(c) Using symmetric form when a denominator is 0. Fix: never write $\frac{z-z_0}{0}$ . If $c=0$ , state $z=z_0$ and use symmetric form only on the other coordinates.

(d) Confusing line "direction" with plane "normal." A line's vector points along it; a plane's normal points across it. Mixing them flips parallel ↔ perpendicular.

Recall Feynman: explain to a 12-year-old

Imagine you're standing in a giant gym. A line is "stand here, then keep walking in one arrow's direction" — you can only go forward or backward along that arrow. A plane is the whole floor: from where you stand you can slide in two arrow directions and reach anywhere flat. To describe a floor quickly, instead of giving two slide-directions, you give one arrow pointing straight up out of the floor (the normal). Any point is "on the floor" only if walking to it never goes up or down relative to that up-arrow — that's exactly the dot product being zero.

Connections

Vectors and scalar (dot) product — powers the normal-form & projections.
Cross product — builds the plane's normal from two directions.
Vector projection — engine behind both distance formulas.
Intersection of lines and planes — solve the parametric equations simultaneously.
Systems of linear equations — a Cartesian plane is one linear equation; intersections = solving systems.
Parametrisation and parameters — the role of $t,s$ .

Vector equation of a line through

\mathbf a

with direction

\mathbf d

\mathbf r=\mathbf a+t\mathbf d,\ t\in\mathbb R

Direction vector of the line through points

A,B

\mathbf b-\mathbf a

Parametric form of

\mathbf r=\mathbf a+t\mathbf d

with

\mathbf a=(x_0,y_0,z_0),\mathbf d=(a,b,c)

x=x_0+ta,\ y=y_0+tb,\ z=z_0+tc

Symmetric form of a line

\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}

(each denominator nonzero)

Parametric (two-direction) equation of a plane

\mathbf r=\mathbf a+s\mathbf u+t\mathbf v

Normal (scalar) form of a plane

(\mathbf r-\mathbf a)\cdot\mathbf n=0

, i.e.

\mathbf r\cdot\mathbf n=d

How to get a plane's normal from two spanning directions

\mathbf n=\mathbf u\times\mathbf v

What are the coefficients in

ax+by+cz=d

components of the normal vector

(a,b,c)

Distance from point

\mathbf p

to plane

\mathbf r\cdot\mathbf n=d

\dfrac{|\mathbf p\cdot\mathbf n-d|}{|\mathbf n|}

Distance from point

\mathbf p

to line

\mathbf r=\mathbf a+t\mathbf d

\dfrac{|(\mathbf p-\mathbf a)\times\mathbf d|}{|\mathbf d|}

Condition for two lines to be identical

directions parallel AND they share a common point

What does

t

represent in

\mathbf r=\mathbf a+t\mathbf d

a scalar parameter selecting which point on the line

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, 3D me ek line ko samajhne ka sabse easy tareeka: ek starting point lo ( $\mathbf a$ ), aur ek arrow (direction $\mathbf d$ ). Ab us arrow ke direction me aage-peeche jitna chaho chalo — har point line par aa jayega. Isi ko likhte hain $\mathbf r=\mathbf a+t\mathbf d$ , jahan $t$ ek number hai jo batata hai kitna chalna hai. $t=0$ pe tum point $A$ par ho, badhao toh aage, negative karo toh peeche.

Plane thoda alag hai. Yahan ek arrow se kaam nahi chalta, kyunki plane ek pura flat floor hai. Toh do options hain: ya toh do non-parallel arrows ( $\mathbf u,\mathbf v$ ) do aur unka combination $\mathbf r=\mathbf a+s\mathbf u+t\mathbf v$ banao; ya phir — aur ye zyada smart tareeka hai — ek normal vector $\mathbf n$ do jo floor ke bilkul perpendicular (straight upar) ho. Koi bhi point plane par tabhi hoga jab us point tak jaane wala displacement normal ke perpendicular ho, yaani $(\mathbf r-\mathbf a)\cdot\mathbf n=0$ . Dot product zero matlab perpendicular — yeh trick poore chapter ki jaan hai.

Ek important baat yaad rakhna: Cartesian form $ax+by+cz=d$ me jo coefficients $(a,b,c)$ hain, wahi normal vector hai — seedha utha lo! Aur $d$ sirf plane ko normal ke direction me shift karta hai, woh direction ka part nahi hai. Ye chhoti si baat exams me bahut marks bachati hai.

Distances ke liye do formula 80/20 wale hain: point se plane $\frac{|\mathbf p\cdot\mathbf n-d|}{|\mathbf n|}$ , aur point se line $\frac{|(\mathbf p-\mathbf a)\times\mathbf d|}{|\mathbf d|}$ . Dono projection ki idea se aate hain — plane ke liye normal pe project karo, line ke liye cross product se perpendicular height nikalo. Inhe rata maat, derive karke samjho, phir kabhi bhoologe nahi.

Lines and planes in 3D — vector equations

WHY do we even need vector equations?

1. The Line

Line through two points

2. The Plane

3. Distances (the 80/20 high-value formulas)

Worked Examples

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Linear Algebra (Full)

Connections