MathsISC Class 12 · Chapter 11

Three-dimensional Geometry ISC Class 12: Lines, Planes and Solved Questions

Updated 14 October 2026 · Fact-checked

Three-dimensional Geometry is Chapter 11 of ISC Class 12 Mathematics. It covers direction cosines and ratios, vector and Cartesian equations of lines and planes, coplanar and skew lines with the shortest distance between them, angles between lines and planes, intersection of a line and a plane, and the distance of a point from a line and from a plane.

What Three-dimensional Geometry is about

This chapter applies vectors to lines and planes in space. It is a compulsory 6-mark unit in ISC 2027, examined through one-mark items, 2-mark plane questions and longer problems set in real contexts (a tower on a hillside, flight paths of insects or aircraft).

Direction cosines and ratios: a line making angles α, β, γ with the axes has direction cosines l = cos α, m = cos β, n = cos γ, with $l^{2} + m^{2} + n^{2} = 1$ ; the line joining $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ has direction ratios $x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1}$ . The angle between two lines is given by $cos θ = \frac{∣ a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2} ∣}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$ ; lines are perpendicular if $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ and parallel if the ratios are proportional.

Lines: vector form $r = a + λ b$ and Cartesian form $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$ , through one point with a direction or through two points. Two lines may intersect, be parallel, or be skew (neither). The shortest distance between skew lines is $\frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$ ; it is zero exactly when the lines are coplanar. You also need the distance of a point from a line.

Planes: normal form $r \cdot \overset{n}{^} = d$ (or ax + by + cz = d with normal direction ratios a, b, c), one-point form $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$ , intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ , and the plane through three points. Then angles between two planes and between a line and a plane, the intersection of a line and a plane, and the distance of a point from a plane, $\frac{∣ a x _{1} + b y _{1} + c z _{1} - d ∣}{a ^{2} + b ^{2} + c ^{2}}$ .

Weightage: Three-dimensional Geometry is part of the unit Three-dimensional Geometry, which carries 6 of the 80 marks in the ISC Mathematics. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Direction cosines, direction ratios and the angle between lines: If a line makes angles α, β, γ with the positive x-, y-, z-axes, its direction cosines are l = cos α, m = cos β, n = cos γ, and $l^{2} + m^{2} + n^{2} = 1$ (so $sin^{2} α + sin^{2} β + sin^{2} γ = 2$ ). Direction ratios a, b, c are any numbers proportional to l, m, n; divide by $a^{2} + b^{2} + c^{2}$ to get the cosines. The line through two points has direction ratios equal to the differences of coordinates. Angle between lines with direction ratios $(a_{1}, b_{1}, c_{1})$ and $(a_{2}, b_{2}, c_{2})$ : $cos θ = \frac{∣ a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2} ∣}{\sum a _{1}^{2} \sum a _{2}^{2}}$ .; Worked example
Find the direction cosines of the line through A(1, 2, 3) and B(3, 5, 9), and the angle between it and a line with direction ratios 1, 2, 2.
Try it first, then show the solutionSolution
Direction ratios 2, 3, 6; magnitude 7; direction cosines 2/7, 3/7, 6/7.
$cos θ = \frac{∣2 + 6 + 12∣}{7 \times 3} = \frac{20}{21}$
Answer: (2/7, 3/7, 6/7); θ = $cos^{- 1} \frac{20}{21}$
Equations of a line: Through the point with position vector $a$ parallel to $b$ : $r = a + λ b$ . In Cartesian form, through $(x_{1}, y_{1}, z_{1})$ with direction ratios a, b, c: $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$ . Through two points: $r = a + λ (b - a)$ , or $\frac{x - x _{1}}{x _{2} - x _{1}} = \dots$ . Any point on the line can be written in terms of the parameter: $(x_{1} + aλ, y_{1} + bλ, z_{1} + c λ)$ ; this is the key to intersections and feet of perpendiculars. Axes: the x-axis is y = 0, z = 0; lines parallel to it have direction ratios 1, 0, 0.; Worked example
Write the vector and Cartesian equations of the line through (1, 2, 3) and (3, 5, 9).
Try it first, then show the solutionSolution
Direction $b = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ .
Vector: $r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ ; Cartesian: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ .
Answer: As above
Intersecting, parallel and skew lines; shortest distance: Two lines in space are parallel (direction ratios proportional), intersecting (they share a point), or skew (neither, so they are not coplanar). To test intersection, write general points on both lines and try to solve; consistent values of both parameters give the common point. Shortest distance between $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ : $d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$ . If d = 0 the lines are coplanar and (if not parallel) intersect. For parallel lines ( $b_{1} = b_{2} = b$ ): $d = \frac{∣ b \times ( a _{2} - a _{1} ) ∣}{∣ b ∣}$ .; Worked example
Find the shortest distance between $r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - 3 \hat{j} + 2 \hat{k})$ and $r = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$ .
Try it first, then show the solutionSolution
$b_{1} \times b_{2} = - 9 \hat{i} + 3 \hat{j} + 9 \hat{k}$ , magnitude $171 = 319$ .
$a_{2} - a_{1} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k}$ ; dot product = −27 + 9 + 27 = 9.
$d = \frac{9}{3 19} = \frac{3}{19} \approx 0.69$
Answer: $\frac{3}{19}$ units (the lines are skew)
Distance of a point from a line: Write the general point Q of the line in terms of λ. The foot of the perpendicular from P is the Q for which $P Q \cdot b = 0$ ; solve for λ, then the distance is |PQ|. Alternatively $d = \frac{∣ b \times ( p - a ) ∣}{∣ b ∣}$ .; Worked example
Find the foot of the perpendicular from P(3, 0, 0) to the line $r = λ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ and the distance of P from the line.
Try it first, then show the solutionSolution
General point Q(λ, 2λ, 2λ); $P Q = (λ - 3, 2 λ, 2 λ)$ .
$P Q \cdot (1, 2, 2) = λ - 3 + 4 λ + 4 λ = 9 λ - 3 = 0 \Rightarrow λ = \frac{1}{3}$ : foot $(\frac{1}{3}, \frac{2}{3}, \frac{2}{3})$ .
$P Q = (\frac{8}{3})^{2} + (\frac{2}{3})^{2} + (\frac{2}{3})^{2} = \frac{72}{3} = 22$
Answer: Foot (1/3, 2/3, 2/3); distance 2√2
Equations of a plane: Normal form: $r \cdot \overset{n}{^} = d$ , where $\overset{n}{^}$ is the unit normal and d the distance from the origin; Cartesian $l x + m y + n z = d$ . In general ax + by + cz = d has normal direction ratios a, b, c. One-point form: $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$ . Intercept form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ . Through three points A, B, C: normal $n = A B \times A C$ , then use the one-point form. A plane parallel to an axis has its normal perpendicular to that axis. Coordinate planes: xy-plane z = 0, yz-plane x = 0, zx-plane y = 0; the distance of (a, b, c) from the xy-plane is |c|.; Worked example
Find the equation of the plane through A(1, 1, 1), B(2, 3, 1) and C(3, 1, 2).
Try it first, then show the solutionSolution
$A B = (1, 2, 0)$ , $A C = (2, 0, 1)$ ; $n = A B \times A C = (2, - 1, - 4)$ .
$2 (x - 1) - (y - 1) - 4 (z - 1) = 0 \Rightarrow 2 x - y - 4 z + 3 = 0$ .
Check C: 6 − 1 − 8 + 3 = 0 ✓.
Answer: 2x − y − 4z + 3 = 0
Angles: between planes and between a line and a plane: The angle between two planes is the angle between their normals: $cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}$ ; planes are perpendicular if $n_{1} \cdot n_{2} = 0$ , parallel if the normals are proportional. The angle φ between a line (direction $b$ ) and a plane (normal $n$ ) is the complement of the angle between $b$ and $n$ : $sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$ . The line is parallel to the plane if $b \cdot n = 0$ , perpendicular if $b ∥ n$ .; Worked example
Find the angle between the planes x + 2y + 2z = 5 and 2x − y + 2z = 7, and the angle between the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ and the plane x + y + z = 17.
Try it first, then show the solutionSolution
Planes: $cos θ = \frac{∣2 - 2 + 4∣}{3 \times 3} = \frac{4}{9}$ .
Line and plane: $sin ϕ = \frac{∣2 + 3 + 6∣}{7 3} = \frac{11}{7 3} \approx 0.907$ , so φ ≈ 65.1°.
Answer: $cos^{- 1} \frac{4}{9}$ ; about 65.1°
Intersection of a line and a plane; distance of a point from a plane: Intersection: substitute the general point of the line $(x_{1} + aλ, y_{1} + bλ, z_{1} + c λ)$ into the plane's equation, solve for λ, and substitute back. **Distance of $(x_{1}, y_{1}, z_{1})$ from ax + by + cz = d:** $D = \frac{∣ a x _{1} + b y _{1} + c z _{1} - d ∣}{a ^{2} + b ^{2} + c ^{2}}$ ; in vector form $\frac{∣ a \cdot n - d ∣}{∣ n ∣}$ . The foot of the perpendicular is found by moving from the point along the normal: $(x_{1}, y_{1}, z_{1}) + t (a, b, c)$ with t chosen to satisfy the plane.; Worked example
Find where the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ meets the plane x + y + z = 17, and the distance of (1, 2, 3) from the plane 2x − y − 4z + 3 = 0.
Try it first, then show the solutionSolution
General point (1 + 2λ, 2 + 3λ, 3 + 6λ): 6 + 11λ = 17 ⇒ λ = 1 ⇒ (3, 5, 9).
$D = \frac{∣2 - 2 - 12 + 3∣}{4 + 1 + 16} = \frac{9}{21} \approx 1.96$
Answer: (3, 5, 9); $\frac{9}{21}$ units

Maths in real life: Three-dimensional Geometry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Distance of a point from a plane

Suppose a flagpole is fixed perpendicular to a sloping hillside modelled by the plane x + 2y + 2z = 9, and its top is at P(2, 3, 5), in metres.

Find the length of the pole (the perpendicular distance from P to the hillside).

Try it first, then show the solutionSolution

$D = \frac{∣2 + 6 + 10 - 9∣}{1 + 4 + 4}$
$= \frac{9}{3} = 3$

Answer: 3 m

Memory hook: Plug the point in, take the size, divide by the normal's length.

Shortest distance between skew lines

Imagine two aircraft flying straight, level-crossing paths at different heights: path 1 through (1, 2, 3) with direction (1, −3, 2), path 2 through (4, 5, 6) with direction (2, 3, 1), in kilometres.

What is the closest the two paths ever come?

Try it first, then show the solutionSolution

$b_{1} \times b_{2} = (- 9, 3, 9)$ , magnitude $319$
$d = \frac{∣ ( 3 , 3 , 3 ) \cdot ( - 9 , 3 , 9 ) ∣}{3 19} = \frac{9}{3 19} \approx 0.69$

Answer: About 0.69 km (the paths themselves never meet; whether the aircraft come that close also depends on timing)

Memory hook: Skew lines: project the joining vector onto the common perpendicular.

Intersection of a line and a plane

Suppose a laser pointer beam travels along $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ towards a wall in the plane x + y + z = 17.

Where does the spot appear on the wall?

Try it first, then show the solutionSolution

General point (1 + 2λ, 2 + 3λ, 3 + 6λ); substitute: 6 + 11λ = 17.
λ = 1, point (3, 5, 9).

Answer: At (3, 5, 9)

Memory hook: Parametrise the line, substitute into the plane, solve for the parameter.

Topics in this chapter

Three-dimensional Geometry formula sheet

Direction cosines: $l^{2} + m^{2} + n^{2} = 1; l = \frac{a}{a ^{2} + b ^{2} + c ^{2}}, etc.$
Angle between lines: $cos θ = \frac{∣ a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2} ∣}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$; Perpendicular: numerator 0; parallel: ratios proportional.
Line: $r = a + λ b; \frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$; Through two points A and B, use the direction $O B - O A$ .
Shortest distance (skew lines): $d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$; d = 0: coplanar.
Distance between parallel lines: $d = \frac{∣ b \times ( a _{2} - a _{1} ) ∣}{∣ b ∣}$
Plane forms: $r \cdot \overset{n}{^} = d; a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0; \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$; Normal, one-point and intercept forms.
Plane through three points: $n = A B \times A C$; Then one-point form.
Angle between planes: $cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}$
Angle between line and plane: $sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$; Sine, not cosine.
Distance of a point from a plane: $D = \frac{∣ a x _{1} + b y _{1} + c z _{1} - d ∣}{a ^{2} + b ^{2} + c ^{2}}$; Plane ax + by + cz = d.

How the board asks Three-dimensional Geometry

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Direction cosines and coordinates (Section A)	Sum of sin² or cos² of direction angles; distances of a point from coordinate planes; equation of an axis.	Use l² + m² + n² = 1; distance from the xy-plane is \|z\|.
Lines (2-3 marks)	Vector ↔ Cartesian forms; line through two points; parallel line through a given point; angle between lines; do lines intersect?	Read point and direction from the symmetric form; general point with a parameter.
Planes (2-3 marks)	Plane through three points; through two points and parallel to an axis; intercept or normal form.	Find a normal by a cross product; parallel to an axis means normal ⊥ that axis.
Distances and angles	Shortest distance between skew lines; distance of a point from a plane or line; angles between planes or a line and a plane.	Use the formulas carefully with absolute values; sine for line-plane angles.
Real-life case (5 marks)	Tower on a sloping plane, bees' or aircraft paths, laser beam hitting a wall.	Translate into plane/line equations; distance from a point to a plane gives heights.

Tricks to remember Three-dimensional Geometry

Short memory aids. Say each one aloud once, then use it on a question the same day.

Cos squares add to one
cos²α + cos²β + cos²γ = 1, so sin²α + sin²β + sin²γ = 2.
Use it when: Direction-angle MCQs.
Plane: normal is the coefficients
In ax + by + cz = d, (a, b, c) points perpendicular to the plane.
Use it when: Angles between planes and line-plane problems.
Line-plane uses sine
Between a line and a plane, use sin φ (the line meets the normal at 90° − φ).
Use it when: Avoiding the cos/sin swap.
Three points: cross then plug
Normal = AB × AC, then a(x − x₁) + b(y − y₁) + c(z − z₁) = 0.
Use it when: Plane through three points, avoiding error-prone cross-multiplication.
Parallel to an axis
A plane parallel to the x-axis has no x term in its normal (a = 0).
Use it when: Planes parallel to axes.

Three-dimensional Geometry: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Line equally inclined to the axes · 1 mark · MCQBoard level

The direction cosines of a line equally inclined to the three coordinate axes are:
(a) 1, 1, 1
(b) $\pm \frac{1}{3}, \pm \frac{1}{3}, \pm \frac{1}{3}$ (same sign)
(c) $\frac{1}{3}, \frac{1}{3}, \frac{1}{3}$
(d) $\frac{1}{2}, \frac{1}{2}, 0$

Try it first, then show the solutionSolution

l = m = n and 3l² = 1 ⇒ l = ±1/√3.

Answer: (b)

Example 2: Perpendicular planes · 1 mark · Assertion-reasonBoard level

Assertion: The planes 2x − y + z = 3 and x + 3y + z = 5 are perpendicular.
Reason: Two planes are perpendicular if the scalar product of their normals is zero.
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

Normals (2, −1, 1) and (1, 3, 1): 2 − 3 + 1 = 0, so the planes are perpendicular by exactly the stated test.

Answer: (a)

Example 3: Line through two points · 2 marks · Short answerHarder board level

Find the vector and Cartesian equations of the line through (1, 2, 3) and (3, 5, 9). [2]

Try it first, then show the solutionSolution

Direction ratios 2, 3, 6.
$r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ ; $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ .

Answer: As shown

Example 4: Distance of a point from a plane · 3 marks · Short answerHarder board level

Find the distance of the point (1, 2, 3) from the plane 2x − y − 4z + 3 = 0, and the foot of the perpendicular. [3]

Try it first, then show the solutionSolution

$D = \frac{∣2 - 2 - 12 + 3∣}{21} = \frac{9}{21}$ .
Foot: (1 + 2t, 2 − t, 3 − 4t) on the plane: 2 + 4t − 2 + t − 12 + 16t + 3 = 0 ⇒ 21t = 9 ⇒ t = 3/7.
Foot $= (\frac{13}{7}, \frac{11}{7}, \frac{9}{7})$ .

Answer: $\frac{9}{21}$ ; foot (13/7, 11/7, 9/7)

Example 5: Plane through three points · 3 marks · Short answerHarder board level

Find the equation of the plane through A(1, 1, 1), B(2, 3, 1) and C(3, 1, 2). [3]

Try it first, then show the solutionSolution

$A B = (1, 2, 0)$ , $A C = (2, 0, 1)$ .
$n = A B \times A C = (2, - 1, - 4)$ .
2(x − 1) − (y − 1) − 4(z − 1) = 0 ⇒ 2x − y − 4z + 3 = 0.

Answer: 2x − y − 4z + 3 = 0

Example 6: Skew lines: shortest distance and angle · 5 marks · Long answerHarder board level

Lines $L_{1} : r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - 3 \hat{j} + 2 \hat{k})$ and $L_{2} : r = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$ .
(a) Show that they are not parallel and find the angle between them. [2]
(b) Find the shortest distance between them and state whether they intersect. [3]

Try it first, then show the solutionSolution

(a) Directions (1, −3, 2) and (2, 3, 1) are not proportional, so not parallel.
$cos θ = \frac{∣2 - 9 + 2∣}{14 14} = \frac{5}{14}$ , θ = $cos^{- 1} \frac{5}{14}$ .
(b) $b_{1} \times b_{2} = - 9 \hat{i} + 3 \hat{j} + 9 \hat{k}$ , $∣ b_{1} \times b_{2} ∣ = 171$ .
$(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = (3, 3, 3) \cdot (- 9, 3, 9) = 9$ .
$d = \frac{9}{171} = \frac{3}{19} \neq = 0$ : the lines are skew and do not intersect.

Answer: θ = cos⁻¹(5/14); shortest distance 3/√19; they do not intersect

Example 7: Plane parallel to an axis · 3 marks · Short answerHOTS / competency

Find the equation of the plane passing through (1, 2, 3) and (3, −1, 2) and parallel to the y-axis. [3]

Try it first, then show the solutionSolution

The normal is perpendicular to both $\hat{j}$ and the direction (2, −3, −1) of the joining line.
$n = (2, - 3, - 1) \times (0, 1, 0) = (1, 0, 2)$ .
1(x − 1) + 0 + 2(z − 3) = 0 ⇒ x + 2z = 7 (check (3, −1, 2): 3 + 4 = 7 ✓).

Answer: x + 2z = 7

Example 8: Distance of a point from a line · 2 marks · Short answerHOTS / competency

Find the distance of P(3, 0, 0) from the line $r = λ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ . [2]

Try it first, then show the solutionSolution

Foot Q(λ, 2λ, 2λ) with $P Q \cdot (1, 2, 2) = 0$ : 9λ − 3 = 0, λ = 1/3.
$P Q = (8/3)^{2} + (2/3)^{2} + (2/3)^{2} = 22$ .

Answer: 2√2 units

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

A mast on a sloping roof

Suppose a wind-sensor mast is fixed to a large sloping roof whose flat surface passes through A(1, 1, 1), B(2, 3, 1) and C(3, 1, 2). The mast is perpendicular to the roof and its tip is at P(4, −2, 9) (units in metres).

(i) · 1 mark
Find the equation of the plane of the roof.
Answer
Answer: 2x − y − 4z + 3 = 0
Normal $A B \times A C = (2, - 1, - 4)$ through A.
(ii) · 1 mark
Write the equation of the line of the mast through P.
Answer
Answer: $\frac{x - 4}{2} = \frac{y + 2}{- 1} = \frac{z - 9}{- 4}$
The mast is along the normal to the roof.
(iii) · 2 marks
Find the length of the mast.
Answer
Answer: $\frac{23}{21} \approx 5.02$ m
$\frac{∣8 + 2 - 36 + 3∣}{21} = \frac{23}{21}$
(iv) · 1 mark
What is the angle between the mast and the vertical z-axis?
Answer
Answer: $cos^{- 1} \frac{4}{21} \approx 29. 2^{\circ}$
Direction (2, −1, −4) with (0, 0, 1): $\frac{∣ - 4∣}{21}$ .

Two aircraft paths

Suppose two aircraft fly along straight paths $L_{1} : r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - 3 \hat{j} + 2 \hat{k})$ and $L_{2} : r = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$ (in km).

(i) · 1 mark
Write $L_{1}$ in Cartesian form.
Answer
Answer: $\frac{x - 1}{1} = \frac{y - 2}{- 3} = \frac{z - 3}{2}$
Point (1, 2, 3), direction (1, −3, 2).
(ii) · 1 mark
Write the equation of a third path parallel to $L_{2}$ through (0, 0, 1).
Answer
Answer: $r = \hat{k} + t (2 \hat{i} + 3 \hat{j} + \hat{k})$
Same direction vector.
(iii) · 2 marks
Find the closest distance between $L_{1}$ and $L_{2}$ .
Answer
Answer: $\frac{3}{19} \approx 0.69$ km
Skew-line formula: 9/√171.
(iv) · 1 mark
Are the paths at right angles?
Answer
Answer: No.
(1)(2) + (−3)(3) + (2)(1) = −5 ≠ 0.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the equation of the plane with normal $2 \hat{i} - \hat{j} + 2 \hat{k}$ at a distance 3 from the origin. (2 marks)

Step	Marks
Unit normal $\frac{1}{3} (2 \hat{i} - \hat{j} + 2 \hat{k})$ ; $r \cdot \overset{n}{^} = 3$	1
Cartesian 2x − y + 2z = 9	1

Find the equation of the plane through (1, 1, 1), (2, 3, 1) and (3, 1, 2). (3 marks)

Step	Marks
Two vectors in the plane: (1, 2, 0) and (2, 0, 1)	1
Normal (2, −1, −4) by cross product	1
2x − y − 4z + 3 = 0	1

Find the shortest distance between $r = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k})$ and $r = (2 \hat{i} + \hat{j} - \hat{k}) + μ (3 \hat{i} - 5 \hat{j} + 2 \hat{k})$ . (5 marks)

Step	Marks
Identifies $a_{1}, b_{1}, a_{2}, b_{2}$	1
$b_{1} \times b_{2} = 3 \hat{i} - \hat{j} - 7 \hat{k}$	1
$∣ b_{1} \times b_{2} ∣ = 59$	1
$(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = (1, 0, - 1) \cdot (3, - 1, - 7) = 10$	1
Shortest distance $\frac{10}{59}$	1

Where students lose marks in Three-dimensional Geometry

The slips examiners see most often, and what to write instead.

Mistake: Not knowing the direction-angle identity
CISCE examiners' comments, 2024: candidates lacked knowledge of direction angles and cosines when asked for sin²α + sin²β + sin²γ.
Wrong working
sin²α + sin²β + sin²γ = 1
Right working
cos²α + cos²β + cos²γ = 1, so the sum of sines squared is 3 − 1 = 2.
Do this: Start every direction-angle question from l² + m² + n² = 1.
Usually costs: 1 mark.
Mistake: Mixing up distances from coordinate planes
CISCE examiners' comments, 2024: most candidates confused which coordinate gives the distance from which plane.
Wrong working
Distance of (a, b, c) from the xy-plane is a.
Right working
From the xy-plane: |c|; from the yz-plane: |a|; from the zx-plane: |b|.
Do this: The missing letter in the plane's name gives the coordinate.
Usually costs: 1 mark.
Mistake: Errors in cross-multiplication for a plane
CISCE examiners' comments, 2024 and 2025: finding the plane through three points, candidates solved the homogeneous equations in a, b, c by cross-multiplication inaccurately; others did not see that a plane parallel to the x-axis has its normal perpendicular to the x-axis.
Wrong working
Arithmetic slips in solving for a : b : c.
Right working
Use $n = A B \times A C$ and check all three points in the final equation; for 'parallel to the x-axis' include $\hat{i}$ in the cross product.
Do this: Always verify the plane passes through every given point.
Usually costs: 1-2 marks.
Mistake: Testing collinearity with position vectors
CISCE examiners' comments, 2025: to show three points collinear, candidates used the position vectors instead of the vectors PQ, QR and PR.
Wrong working
Shows $O P$ and $O Q$ are not parallel.
Right working
Show $P R = k P Q$ (displacement vectors from a common point).
Do this: Collinearity is about displacement vectors.
Usually costs: 1 mark.
Mistake: Using cosine for the line-plane angle
A frequent slip: the line-plane angle is found with the cosine formula, giving its complement.
Wrong working
$cos ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$
Right working
$sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$
Do this: The normal makes 90° with the plane; that switches cos to sin.
Usually costs: 1 mark.

Practice questions on Three-dimensional Geometry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The equations of the x-axis in space are:
(a) x = 0, y = 0
(b) y = 0, z = 0
(c) x = 0, z = 0
(d) x = y = z
Hint
On the x-axis y and z vanish.
Answer
(b)
Q2 · 1 mark · MCQBoard level
The perpendicular distance of the point (a, b, c) from the xy-plane is:
(a) |a|
(b) |b|
(c) |c|
(d) $a^{2} + b^{2}$
Hint
Height above the floor.
Answer
(c)
Q3 · 1 mark · MCQBoard level
The angle between the planes x + 2y + 2z = 5 and 2x − y + 2z = 7 is:
(a) $cos^{- 1} \frac{4}{9}$
(b) $cos^{- 1} \frac{2}{9}$
(c) $\frac{π}{2}$
(d) $\frac{π}{3}$
Hint
Angle between normals.
Answer
(a)
Q4 · 1 mark · Very short answerBoard level
Write the direction cosines of the line joining (1, 2, 3) and (3, 5, 9).
Hint
Divide (2, 3, 6) by 7.
Answer
2/7, 3/7, 6/7
Q5 · 2 marks · Short answerBoard level
Write the Cartesian form of $r = (\hat{i} - \hat{j} + 2 \hat{k}) + λ (3 \hat{i} + 2 \hat{j} - \hat{k})$ and the vector form of $\frac{x - 2}{1} = \frac{y}{4} = \frac{z + 1}{- 2}$ .
Hint
Point and direction.
Answer
$\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{- 1}$ ; $r = (2 \hat{i} - \hat{k}) + μ (\hat{i} + 4 \hat{j} - 2 \hat{k})$
Q6 · 2 marks · Short answerHarder board level
Find the distance of (1, 2, 3) from the plane 2x − y − 4z + 3 = 0.
Hint
|ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²).
Answer
9/√21
Q7 · 2 marks · Short answerHarder board level
Find the point where $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ meets the plane x + y + z = 17.
Hint
General point.
Answer
(3, 5, 9)
Q8 · 2 marks · Short answerHOTS / competency
Show that the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1}$ intersect, and find the point of intersection.
Hint
Equate general points.
Answer
(1 + 2s, 2 + 3s, 3 + 4s) = (4 + 5t, 1 + 2t, t) gives s = t = −1 consistently, so they meet at (−1, −1, −1).
Q9 · 3 marks · Short answerHarder board level
Find the angle between the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ and the plane x + y + z = 17.
Hint
Use sine.
Answer
$sin ϕ = \frac{11}{7 3}$ , φ ≈ 65.1°
Q10 · 3 marks · Short answerHarder board level
A plane cuts intercepts 2, 3 and 4 on the axes. Write its equation and find its distance from the origin.
Hint
Intercept form.
Answer
$\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ , i.e. 6x + 4y + 3z = 12; distance $\frac{12}{61} \approx 1.54$
Q11 · 3 marks · Short answerHarder board level
Find the foot of the perpendicular from (1, 2, 3) to the plane 2x − y − 4z + 3 = 0.
Hint
Move along the normal.
Answer
(13/7, 11/7, 9/7)
Q12 · 5 marks · Long answerHarder board level
Find the shortest distance between $r = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k})$ and $r = (2 \hat{i} + \hat{j} - \hat{k}) + μ (3 \hat{i} - 5 \hat{j} + 2 \hat{k})$ .
Hint
Cross product of directions.
Answer
$b_{1} \times b_{2} = 3 \hat{i} - \hat{j} - 7 \hat{k}$ , $∣ \cdot ∣ = 59$ ; $(a_{2} - a_{1}) \cdot (3, - 1, - 7) = 10$ ; distance $\frac{10}{59} \approx 1.30$ .
Q13 · 2 marks · Short answerHarder board level
Find the equation of the plane with normal $2 \hat{i} - \hat{j} + 2 \hat{k}$ at distance 3 from the origin.
Hint
Unit normal.
Answer
2x − y + 2z = 9

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Relation between direction cosines?Show answer
l² + m² + n² = 1.
2.Cartesian equation of a line through $(x_{1}, y_{1}, z_{1})$ with direction ratios a, b, c?Show answer
$\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$ .
3.What are skew lines?Show answer
Lines that are neither parallel nor intersecting (not coplanar).
4.Shortest distance between skew lines?Show answer
$\frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$ .
5.Intercept form of a plane?Show answer
x/a + y/b + z/c = 1.
6.Normal to the plane ax + by + cz = d?Show answer
Direction ratios a, b, c.
7.Distance of a point from a plane?Show answer
$\frac{∣ a x _{1} + b y _{1} + c z _{1} - d ∣}{a ^{2} + b ^{2} + c ^{2}}$ .
8.Angle between a line and a plane?Show answer
$sin ϕ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$ .

Best order to study it

Revise direction cosines and ratios and the angle between lines.
Learn line equations in both forms; practise conversion.
Practise intersection tests, skew lines and the shortest distance.
Learn plane forms; practise the plane through three points and planes parallel to an axis.
Practise angles between planes and between a line and a plane.
Finish with intersections and distances of a point from a line and a plane.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find direction cosines of a line joining two points?
Can you write a line in vector and Cartesian forms?
Can you test whether two lines intersect?
Can you find the shortest distance between skew lines?
Can you find a plane through three points using a cross product?
Can you find angles between planes and between a line and a plane?
Can you find where a line meets a plane?
Can you find the distance and foot of the perpendicular from a point to a plane?

Three-dimensional Geometry: frequently asked questions

Is Three-dimensional Geometry compulsory in ISC 2027?

Yes. It is Unit 5, worth 6 of the 80 marks; it was earlier in the optional Section B.

Is the distance of a point from a line in the syllabus?

Yes, the syllabus lists distance of a point from a line along with the shortest distance between two lines.

Which forms of the plane are required?

Cartesian and vector equations, the one-point form, normal form and intercept form, the direction ratios of the normal, and the plane through given points.

How do I know if two lines are coplanar?

They are coplanar if the shortest distance formula gives 0, i.e. (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0; non-parallel coplanar lines intersect.

Why do we use sin for the angle between a line and a plane?

The formula with the dot product gives the angle between the line and the plane's normal; the angle with the plane itself is its complement, so cos becomes sin.

Is the equation of a plane through the intersection of two planes required?

It is not named in the ISC 2027 list; the listed topics are the plane forms, angles, distance of a point from a plane and the intersection of a line and a plane.

Based on: CISCE ISC 2027 Mathematics syllabus; CISCE ISC 2027 Mathematics specimen question paper and answer key; CISCE Analysis of Pupil Performance, ISC Mathematics, 2025; CISCE Analysis of Pupil Performance, ISC Mathematics, 2024