NDA Maths · 3D Geometry

The Straight Line in Space

A line is a point plus a direction; everything — points on it, where it pierces a plane, whether it's parallel to one — follows from its parametric form.

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Why this matters

Eleven PYQs, almost all EASY or MODERATE — the most reliably scorable subtopic in the chapter once you commit to parametrising. The bank repeats four moves: write the line, generate a point on it, find where it meets a coordinate plane or a given plane, and test whether it is parallel to (or lies in) a plane. Master the parametric form and the rest are substitutions.

Concept 1 of 5

Equation of a line — symmetric and two-point forms

Intuition

A line is pinned down by one point on it and one direction. Symmetric form stacks the three coordinate equations into one chain; two-point form builds the direction from the difference of the points. A line in space is also the intersection of two planes — which is why a single symmetric line can be written as a pair of plane equations.

Definition

Through point $(x_0,y_0,z_0)$ with direction ratios $\langle a,b,c\rangle$ , the symmetric form is $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ . Through two points $A, B$ , the direction ratios are $\langle x_2-x_1, y_2-y_1, z_2-z_1\rangle$ . Setting the chain equal to a parameter $t$ gives the parametric form $(x_0+at,\ y_0+bt,\ z_0+ct)$ . Splitting the chain into two equalities expresses the line as the intersection of two planes. Conversely, when a line is GIVEN as the intersection of two planes $P_1, P_2$ , its direction ratios are $\vec{n_1} \times \vec{n_2}$ — the cross product of the two normals (the line lies in both planes, so it is perpendicular to both normals).

Symmetric form of a line

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

Worked example

Write the line through

A(1, 2, -1)

and

B(3, -1, 2)

in symmetric form.

Direction ratios $= B - A = \langle 3-1,\ -1-2,\ 2-(-1)\rangle = \langle 2, -3, 3\rangle$ .
Use point $A(1,2,-1)$ and these ratios.
Symmetric form: $\frac{x-1}{2} = \frac{y-2}{-3} = \frac{z+1}{3}$ .

Answer:

\dfrac{x-1}{2} = \dfrac{y-2}{-3} = \dfrac{z+1}{3}

Practice this conceptself-check · 5 quick reps

From the bank · past-year question

Example 13D GeometryHARD

The line

\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}

is given by

[Q58 · Sep · 2018]

Drill 3 more on equation of a line — symmetric and two-point forms

Concept 2 of 5

Points on a line

Intuition

Every point on a line is the base point pushed along the direction by some amount

t

. Plug a value of

t

and you get a point; or set the parametric coordinates equal to a candidate point and check they share a single

t

Definition

Points on $(x_0+at,\ y_0+bt,\ z_0+ct)$ are generated by choosing $t$ . To test whether $(p,q,r)$ lies on the line, solve $p = x_0+at$ for $t$ and confirm the SAME $t$ reproduces $q$ and $r$ .

Worked example

Find a point on the line

\frac{x-2}{3} = \frac{y+1}{1} = \frac{z-4}{2}

other than

(2,-1,4)

Set the chain equal to $t$ : point $= (2+3t,\ -1+t,\ 4+2t)$ .
Take $t = 1$ : $(5, 0, 6)$ .
Check: $\frac{5-2}{3} = 1,\ \frac{0+1}{1} = 1,\ \frac{6-4}{2} = 1$ — all equal. ✓

Answer:

(5, 0, 6)

(any

t \neq 0

works).

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 23D GeometryEASY

A point on the line

\dfrac{x-1}{1} = \dfrac{y-3}{2} = \dfrac{z+2}{7}

has coordinates

[Q57 · Sep · 2019]

Concept 3 of 5

Where a line meets a coordinate plane

Intuition

To find where a line crosses the XY-plane, ask: at what

t

z = 0

? Solve for that

t

, substitute back, and read off the point. Same idea for the YZ-plane (

x=0

) and ZX-plane (

y=0

Definition

A line meets the XY-plane where its $z$ -coordinate is 0, the YZ-plane where $x=0$ , and the ZX-plane where $y=0$ . Parametrise the line, set the relevant coordinate to 0, solve for $t$ , and back-substitute.

Diagram · line piercing a plane (drag to rotate)

Substitute the line's point (x₀+at, y₀+bt, z₀+ct) into the plane equation → one equation in t → solve → back-substitute to get the pierce point P.

Worked example

Find where the line joining

(2, -1, 3)

and

(4, 3, -1)

meets the XY-plane.

Direction $= \langle 2, 4, -4\rangle$ . Point $= (2+2t,\ -1+4t,\ 3-4t)$ .
XY-plane: $z = 0 \Rightarrow 3 - 4t = 0 \Rightarrow t = \tfrac{3}{4}$ .
Then $x = 2 + 2(\tfrac34) = \tfrac{7}{2}$ and $y = -1 + 4(\tfrac34) = 2$ .

Answer:

\left( \tfrac{7}{2}, 2, 0 \right)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 33D GeometryEASY

The point of intersection of the line joining the points

(-3, 4, -8)

and

(5, -6, 4)

with the

XY

-plane is

[Q52 · Sep · 2017]

Drill 1 more on where a line meets a coordinate plane

Concept 4 of 5

Intersection of a line and a plane

Intuition

A line generally pierces a plane at one point. Plug the line's parametric coordinates into the plane equation — that collapses to one equation in

t

. Solve it, then feed

t

back to get the pierce point.

Definition

For a line $(x_0+at,\ y_0+bt,\ z_0+ct)$ and plane $Ax+By+Cz = D$ , substitute to get a linear equation in $t$ . One solution → a single pierce point. No solution (the $t$ terms cancel but the constants disagree) → the line is parallel and misses. Identity (all $t$ work) → the line lies in the plane.

Diagram · line piercing a plane (drag to rotate)

Substitute the line's point (x₀+at, y₀+bt, z₀+ct) into the plane equation → one equation in t → solve → back-substitute to get the pierce point P.

Worked example

A line through

(1, 0, -1)

with direction ratios

\langle 1, 2, 1\rangle

meets the plane

x + y + z = 8

. Find the point of intersection.

Parametrise: $(1+t,\ 2t,\ -1+t)$ .
Substitute into the plane: $(1+t) + 2t + (-1+t) = 8$ .
$4t = 8 \Rightarrow t = 2$ .
Back-substitute $t = 2$ : $(3, 4, 1)$ .

Answer:

(3, 4, 1)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 43D GeometryEASY

A line through

(1,-1,2)

with direction ratios

\langle3,2,2\rangle

meets the plane

x+2y+3z=18

. What is the point of intersection?

[Q55 · Apr · 2024]

Drill 1 more on intersection of a line and a plane

Concept 5 of 5

Line parallel to, or lying in, a plane

Intuition

A line is parallel to a plane when its direction is perpendicular to the plane's normal — their dot product is zero, so the line never tilts toward the plane. If, in addition, one point of the line satisfies the plane equation, the whole line LIES in the plane.

Definition

For a line with direction $\langle a,b,c\rangle$ and a plane with normal $\langle A,B,C\rangle$ :

Parallel iff $Aa + Bb + Cc = 0$ (direction ⟂ normal) AND a point of the line is NOT on the plane.
Lies in the plane iff $Aa+Bb+Cc = 0$ AND a point of the line satisfies the plane equation.

Parallel/contained condition (direction ⟂ normal)

A a + B b + C c = 0

Worked example

If the line

\frac{x-1}{2} = \frac{y-m}{1} = \frac{z-3}{1}

lies in the plane

2x - 3y - z = 2

, find

m

Direction $\langle 2,1,1\rangle$ , normal $\langle 2,-3,-1\rangle$ : dot $= 4 - 3 - 1 = 0$ ✓ (so it can lie in the plane).
Now the point $(1, m, 3)$ must satisfy the plane: $2(1) - 3(m) - 3 = 2$ .
$-3m - 1 = 2 \Rightarrow m = -1$ .

Answer:

m = -1

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 53D GeometryMODERATE

If the line

\dfrac{x-4}{1} = \dfrac{y-2}{1} = \dfrac{z-k}{2}

lies on the plane

2x - 4y + z = 7

, then what is the value of

k

[Q58 · Sep · 2019]

Parallel vs lying-in needs the second check

Direction ⟂ normal (dot = 0) only means the line doesn't tilt toward the plane — it could be strictly parallel (misses) OR contained (lies in). You MUST then test a point: on the plane → contained; off → parallel.

direction · normal = 0 means the line is PARALLEL to the plane, not perpendicular

A line is PERPENDICULAR to a plane when its direction is PARALLEL to the normal — the direction ratios are proportional to

\langle A,B,C\rangle

. When the direction-normal dot product is

0

, the line is perpendicular to the NORMAL, hence PARALLEL to (or lying in) the plane. Don't read "dot = 0" as "line ⟂ plane" — it's the exact opposite.

Drill 1 more on line parallel to, or lying in, a plane

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Equation of a line — symmetric and two-point forms
Symmetric form of a line
$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = t$
Line parallel to, or lying in, a plane
Parallel/contained condition (direction ⟂ normal)
$A a + B b + C c = 0$

Watch out for (2)

Parallel vs lying-in needs the second check
→ Line parallel to, or lying in, a plane
direction · normal = 0 means the line is PARALLEL to the plane, not perpendicular
→ Line parallel to, or lying in, a plane

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