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Three-Dimensional Geometry hero

Three-Dimensional Geometry

~14 min read

In 30 seconds
  • What: Three-dimensional geometry extends coordinate geometry into space — direction cosines, equations of lines and planes, distances, and angles between geometric objects.
  • Why it matters: This topic has appeared in every NDA paper since 2010, contributing 4–8 marks per sitting. It is among the most formula-dense chapters and rewards candidates who memorise clean derivations.
  • Key fact: For any line with direction cosines \(l, m, n\) the identity \(l^2 + m^2 + n^2 = 1\) is non-negotiable — it is tested both directly and inside angle/distance problems.

Three-dimensional geometry is where coordinate geometry moves off the plane and into real space. Instead of just x and y you now have x, y, and z, and the key idea that connects everything is how a line or plane is oriented — described through direction cosines and direction ratios. Once you have those tools the rest (equations of lines, equations of planes, shortest distances, angles between planes) follows a clean pattern that the NDA exploits year after year.

This page covers every formula and concept you need, grounded in NCERT Class 12 Chapter 11 and illustrated with actual NDA PYQ questions from 2010 to 2018. Work through the worked examples section carefully — the NDA rarely sets a question that does not fit one of five or six standard moulds.

What This Topic Covers

The NDA syllabus for three-dimensional geometry maps almost perfectly to NCERT Class 12 Chapter 11. Here are the sub-topics you must own:

Sub-topics in Scope

  • Direction cosines and direction ratios of a line, and the relationship between them.
  • Direction cosines of a line joining two points — using the distance PQ as the common denominator.
  • Equation of a line in space — both vector form and Cartesian symmetric form.
  • Angle between two lines — using the dot-product formula and the perpendicularity/parallelism conditions.
  • Skew lines and shortest distance — the vector and Cartesian determinant formulas for the distance between two non-intersecting, non-parallel lines.
  • Equation of a plane — passing through a point with a given normal, and passing through three points.
  • Angle between two planes, and between a line and a plane.
  • Distance of a point from a plane — the signed and unsigned versions.
  • Sphere — standard equation, centre, and radius (appears in PYQs up to 2018).

Direction cosines and planes generate the largest share of marks. Skew lines and sphere questions appear but are less frequent. The identity \(l^2 + m^2 + n^2 = 1\) underlies virtually every sub-topic, so if you are short on time, master direction cosines first.

The chapter has a natural dependency chain: you need direction cosines to write the equation of a line; you need the equation of a line to compute shortest distance; you need the equation of a plane to find angles and point-to-plane distances. Study in this order and you will not leave gaps.

Exam Pattern & Weightage

The table below is built entirely from the NDA PYQ file. Each row is one paper. The approximate question count is extracted from the numbered questions in the source.

Year / Paper No. Key Sub-topics Tested
2010-II 2 Parallel planes (infinite common points), distance of origin from plane
2011-I 4 Angle between planes, sphere radius, direction cosines identity, point on plane
2011-II 2 Sphere-plane intersection, equation of sphere through given points
2012-I 5 cos 2α identity, normal to plane, angle between planes, sum of squares of d.c.s, sphere diameter
2012-II 5 Distance from yz-plane, equally-inclined d.c.s, angle between lines, plane through point, direction ratios of intersection line
2013-I / II 6 z-axis d.c.s, triangle area in 3D, distance between planes, cos²β + cos²γ identity, sphere equation, length from projections
2014-I / II 8 Normal to plane, foot of perpendicular, image of point, sphere through points, direction ratios of line, plane-line intersection
2015-I / II 7 Line segment projections, direction cosines from projections, foot of perpendicular on line, intercepts of plane, cuboid diagonal
2016-I / II 5 Direction ratios of intersection of planes, equation of plane, sphere tangency, image of point, sum of squares of d.c.s
2017-I / II 6 Direction of line with given d.c.s, collinearity, meeting of line with yz-plane, perpendicularity condition, normal length, plane through line of intersection
2018-I / II 8 Foot of perpendicular from point to line, plane through three points, sphere locus, angle in triangle in 3D, distance of point from plane, line on plane
NDA Alert

The NDA consistently tests the identity \(1 + \cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0\) (which equals \(-1\) in a different form from PYQ 2012-I). If a question gives you two of the three angles a line makes with the axes and asks for the third, immediately use \(l^2 + m^2 + n^2 = 1\).

On average, three-dimensional geometry contributes 4–8 questions per NDA paper. That makes it one of the highest-return topics in maths — the formulas are fixed and the question types repeat.

Core Concepts

Direction Cosines and Direction Ratios

If a directed line L makes angles \(\alpha, \beta, \gamma\) with the positive x-, y-, and z-axes respectively, the direction cosines are \(l = \cos\alpha,\ m = \cos\beta,\ n = \cos\gamma\). These three numbers are not arbitrary — they must satisfy the fundamental identity:

Fundamental Identity $$l^2 + m^2 + n^2 = 1$$

Direction ratios \(a, b, c\) are any three numbers proportional to \(l, m, n\). Given direction ratios, you recover direction cosines by dividing by the magnitude:

Direction Cosines from Direction Ratios $$l = \pm\frac{a}{\sqrt{a^2 + b^2 + c^2}},\quad m = \pm\frac{b}{\sqrt{a^2 + b^2 + c^2}},\quad n = \pm\frac{c}{\sqrt{a^2 + b^2 + c^2}}$$

Both sets of signs correspond to the two opposite directions along the line. For NDA problems pick the sign that makes geometric sense (e.g., positive magnitude when a distance is required).

NDA Alert

PYQ 2013-II asks the sum of direction cosines of the z-axis. The z-axis has d.c.s \((0, 0, 1)\), so the sum is 1. Watch for similar "axis d.c." traps — x-axis gives \((1, 0, 0)\), y-axis gives \((0, 1, 0)\).

Direction Cosines of a Line Through Two Points

Given \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\), the direction ratios are simply \((x_2 - x_1,\ y_2 - y_1,\ z_2 - z_1)\). The distance PQ is:

Distance Between Two Points in 3D $$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

The direction cosines of PQ are then \((x_2 - x_1)/PQ,\ (y_2 - y_1)/PQ,\ (z_2 - z_1)/PQ\). A PYQ from 2013-II links this to projections: if the projections of a line segment on the three axes are 2, 3, 6, its length is \(\sqrt{4 + 9 + 36} = 7\) units.

Equation of a Line in Space

A line through point \((x_1, y_1, z_1)\) with direction ratios \(a, b, c\) has the Cartesian symmetric form:

Symmetric (Cartesian) Equation of a Line $$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$$

This equals the parametric form \(x = x_1 + \lambda a,\ y = y_1 + \lambda b,\ z = z_1 + \lambda c\). Substitute a specific \(\lambda\) to find the coordinates of any point on the line — the NDA uses this when asking where a line meets a coordinate plane (set the relevant coordinate to zero and solve for \(\lambda\)).

Vector Equation of a Line $$\vec{r} = \vec{a} + \lambda\,\vec{b}$$ where \(\vec{a}\) is the position vector of a point on the line and \(\vec{b}\) is the direction vector

Angle Between Two Lines

If two lines have direction ratios \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\), the acute angle \(\theta\) between them satisfies:

Angle Between Two Lines (DR form) $$\cos\theta = \left|\frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}\right|$$

Notice this is literally the vector dot-product formula \(\cos\theta = (\vec{b_1}\cdot\vec{b_2})/(|\vec{b_1}||\vec{b_2}|)\). Special cases that NDA tests directly:

  • Perpendicular lines: \(a_1 a_2 + b_1 b_2 + c_1 c_2 = 0\)
  • Parallel lines: \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\)

PYQ 2011-I asks the angle between lines with direction ratios \((2, 3, 4)\) and \((1, -2, 1)\). The dot product is \(2\cdot 1 + 3\cdot(-2) + 4\cdot 1 = 0\), so the angle is \(90°\).

Equation of a Plane

A plane with normal vector \((a, b, c)\) passing through \((x_1, y_1, z_1)\) has equation:

Equation of a Plane (Point-Normal Form) $$A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$$

In general form \(Ax + By + Cz + D = 0\). The coefficients \((A, B, C)\) are the direction ratios of the normal vector — the line sticking straight out of the plane at 90°. PYQ 2014-I tests this directly: the normal to \(2x + 3y - z = 7\) has direction ratios \(\langle 2, 3, -1\rangle\).

Intercept Form (cuts axes at a, b, c) $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$
NDA Alert

A line is perpendicular to a plane if and only if its direction ratios are proportional to the plane's normal DRs \((A, B, C)\). So "perpendicular to plane" instantly tells you the line's direction — skip the angle formula entirely.

Distance of a Point \((x_1, y_1, z_1)\) from Plane \(Ax + By + Cz + D = 0\) $$d = \frac{|A x_1 + B y_1 + C z_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

PYQ 2018-II gives: distance of \((2, 3, 4)\) from \(3x - 6y + 2z + 11 = 0\). Numerator: \(|6 - 18 + 8 + 11| = |7| = 7\). Denominator: \(\sqrt{9 + 36 + 4} = 7\). Distance \(= 1\) unit.

NDA Alert

Two planes \(ax + by + cz + d_1 = 0\) and \(ax + by + cz + d_2 = 0\) (same normal, different constants) are parallel and share infinitely many points only if \(d_1 = d_2\). Otherwise they share no common points. This was tested directly in PYQ 2010-II — the answer is "infinite points in common" only when the planes are identical.

Angle Between Two Planes

Angle Between Planes (using their normals) $$\cos\theta = \left|\frac{A_1 A_2 + B_1 B_2 + C_1 C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2}\,\sqrt{A_2^2 + B_2^2 + C_2^2}}\right|$$

This is structurally identical to the angle between two lines — the angle between two planes equals the angle between their normals. PYQ 2011-I tests this for planes \(x + y + 2z = 3\) and \(-2x + y - z = 11\).

Angle Between a Line and a Plane

Here is the single biggest trap in 3D geometry. If a line has direction ratios \((a, b, c)\) and a plane has normal DRs \((A, B, C)\), the angle \(\theta\) between the line and the plane uses sine, not cosine:

Line-Plane Angle (USE SINE) $$\sin\theta = \left|\frac{aA + bB + cC}{\sqrt{a^2 + b^2 + c^2}\,\sqrt{A^2 + B^2 + C^2}}\right|$$

Why sine? Because the dot product actually gives the angle between the line and the plane's normal, which is \(90° - \theta\). Since \(\cos(90° - \theta) = \sin\theta\), the formula switches function.

  • Line parallel to plane: \(aA + bB + cC = 0\) (line is perpendicular to the normal).
  • Line perpendicular to plane: \(\dfrac{a}{A} = \dfrac{b}{B} = \dfrac{c}{C}\) (line shares the normal's direction).
THE Biggest Trap

Line-and-plane angles use SINE; line-line and plane-plane angles use COSINE. Mis-applying cosine here flips the answer (you get the complement). If a question pairs "line" with "plane", reach for sine.

Shortest Distance Between Two Skew Lines

Skew lines are lines in space that neither intersect nor are parallel. They lie in different planes. The shortest distance is the length of the common perpendicular.

Shortest Distance (Vector Form) $$d = \left|\frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|}\right|$$

In Cartesian form the numerator is a 3×3 determinant with rows \((x_2-x_1,\ y_2-y_1,\ z_2-z_1)\), \((a_1, b_1, c_1)\), \((a_2, b_2, c_2)\). The denominator is the magnitude of the cross product of the two direction vectors. If \(d = 0\), the lines actually intersect and are coplanar.

Worked Examples

Example 1 — Direction Cosines from Direction Ratios (PYQ type, 2012-I)

Q. A line has direction ratios \(2, -1, -2\). Find its direction cosines.

  • Step 1: Compute the magnitude of the direction ratio vector: $$\sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3.$$
  • Step 2: Divide each direction ratio by 3. \(l = \tfrac{2}{3},\ m = -\tfrac{1}{3},\ n = -\tfrac{2}{3}\).
  • Step 3: Verify: \(\left(\tfrac{2}{3}\right)^2 + \left(-\tfrac{1}{3}\right)^2 + \left(-\tfrac{2}{3}\right)^2 = \tfrac{4}{9} + \tfrac{1}{9} + \tfrac{4}{9} = 1.\) ✓

Example 2 — Angle Between Two Lines (PYQ 2011-I)

Q. What is the angle between the lines whose direction cosines are proportional to \((2, 3, 4)\) and \((1, -2, 1)\)?

  • Step 1: Compute the dot product: \(a_1 a_2 + b_1 b_2 + c_1 c_2 = 2(1) + 3(-2) + 4(1) = 2 - 6 + 4 = 0\).
  • Step 2: Since the dot product is 0, \(\cos\theta = 0\), so \(\theta = 90°\).
  • Answer: The angle is \(90°\).

Example 3 — Length of Line Segment from Projections (PYQ 2013-II)

Q. The projections of a straight line segment on the coordinate axes are 2, 3, 6. Find the length of the segment.

  • Step 1: If projections on the x-, y-, and z-axes are \(p, q, r\), the length \(L\) of the segment satisfies \(L^2 = p^2 + q^2 + r^2\).
  • Step 2: \(L^2 = 2^2 + 3^2 + 6^2 = 4 + 9 + 36 = 49\).
  • Step 3: \(L = 7\) units.
  • Answer: 7 units.

Example 4 — Distance of a Point from a Plane (PYQ 2018-II)

Q. What is the distance of the point \((2, 3, 4)\) from the plane \(3x - 6y + 2z + 11 = 0\)?

  • Step 1: Apply the distance formula: $$d = \frac{|3(2) - 6(3) + 2(4) + 11|}{\sqrt{3^2 + 6^2 + 2^2}}.$$
  • Step 2: Numerator: \(|6 - 18 + 8 + 11| = |7| = 7\).
  • Step 3: Denominator: \(\sqrt{9 + 36 + 4} = \sqrt{49} = 7\).
  • Step 4: Distance \(= 7/7 = 1\) unit.
  • Answer: 1 unit.

Example 5 — Direction Ratios of Normal and Meeting Point (PYQ 2014-I)

Q. A straight line passes through \((1, -2, 3)\) and is perpendicular to the plane \(2x + 3y - z = 7\). Find the direction ratios of the normal to the plane.

  • Step 1: The normal to the plane \(ax + by + cz = d\) has direction ratios \((a, b, c)\).
  • Step 2: For \(2x + 3y - z = 7\), the normal direction ratios are \(\langle 2, 3, -1\rangle\).
  • Step 3: Since the line is perpendicular to the plane, it is parallel to the normal. The line through \((1, -2, 3)\) with direction ratios \((2, 3, -1)\) is: $$\frac{x-1}{2} = \frac{y+2}{3} = \frac{z-3}{-1}.$$
  • Answer: Direction ratios of the normal are \(\langle 2, 3, -1\rangle\).

Example 6 — Plane Through Point Perpendicular to a Vector

Q. Find the equation of the plane passing through (2, 3, 4) and perpendicular to the vector \(3\hat{i} - \hat{j} + 5\hat{k}\).

  • Step 1: The perpendicular vector gives the normal DRs directly: \(A = 3,\ B = -1,\ C = 5\).
  • Step 2: Apply the point-normal form: $$3(x - 2) - 1(y - 3) + 5(z - 4) = 0.$$
  • Step 3: Expand: \(3x - 6 - y + 3 + 5z - 20 = 0\).
  • Answer: \(3x - y + 5z - 23 = 0\).

Example 7 — Angle Between a Line and a Plane (sine trap)

Q. Find the angle between the line \(\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-3}{6}\) and the plane \(x + 2y + 2z = 5\).

  • Step 1: Line DRs: \((2, 3, 6)\). Plane normal DRs: \((1, 2, 2)\).
  • Step 2: Use SINE (line + plane = sine): $$\sin\theta = \frac{|2(1) + 3(2) + 6(2)|}{\sqrt{4 + 9 + 36}\,\sqrt{1 + 4 + 4}} = \frac{|2 + 6 + 12|}{\sqrt{49}\,\sqrt{9}} = \frac{20}{21}.$$
  • Step 3: \(\theta = \sin^{-1}(20/21)\).
  • Answer: \(\theta = \sin^{-1}\!\left(\tfrac{20}{21}\right)\). Using cosine instead would have given the complement — a classic NDA distractor.

Example 8 — Shortest Distance Between Skew Lines

Q. Find the shortest distance between the lines \(\vec{r} = \hat{i} + 2\hat{j} + \hat{k} + \lambda(\hat{i} - \hat{j} + \hat{k})\) and \(\vec{r} = 2\hat{i} - \hat{j} - \hat{k} + \mu(2\hat{i} + \hat{j} + 2\hat{k})\).

  • Step 1: \(\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}\).
  • Step 2: \(\vec{b_1} \times \vec{b_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&1\\2&1&2\end{vmatrix} = (-2-1)\hat{i} - (2-2)\hat{j} + (1+2)\hat{k} = -3\hat{i} + 0\hat{j} + 3\hat{k}\).
  • Step 3: Magnitude \(|\vec{b_1}\times\vec{b_2}| = \sqrt{9 + 0 + 9} = 3\sqrt{2}\).
  • Step 4: Numerator \((\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2}) = 1(-3) + (-3)(0) + (-2)(3) = -9\).
  • Step 5: $$d = \frac{|-9|}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}\ \text{units}.$$

How NDA Tests This Topic

The most common NDA question type is a linked set (Qs. 37–39 in 2014-I, Qs. 50–51 in 2015-I, Qs. 65–66 in 2016-II). A scenario is given — a line, a sphere, or two planes — and three questions follow in sequence. Each answer feeds into the next. If you mis-read the scenario you drop three marks at once. Read the setup twice before attempting any part.

Exam Shortcuts (Pro-Tips)

Three-dimensional geometry hides four time-saver patterns that collapse a 2–3 minute problem into 15 seconds. Each one has appeared in NDA papers. Memorise the formula, not just the derivation.

Shortcut 1 — Distance Between Two Parallel Planes

When two planes share the same coefficients \(A, B, C\) and differ only in the constant term, skip projections and the foot-of-perpendicular construction. Subtract the constants and divide by the normal's magnitude.

Distance Between Parallel Planes \(Ax + By + Cz + D_1 = 0\) and \(Ax + By + Cz + D_2 = 0\) $$d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$$

Warning: \(A, B, C\) must be exactly matched in both equations first. If one plane reads \(2x + 4y - 6z + 5 = 0\) and the other \(x + 2y - 3z + 1 = 0\), scale the first down to \(x + 2y - 3z + 5/2 = 0\) before subtracting constants.

NDA Alert

Parallel-plane shortcut fails silently if you forget to normalise coefficients. Whenever the two planes look "almost the same", divide each by its leading coefficient first, then apply the formula. NDA 2013-I has been answered wrong because of exactly this slip.

Shortcut 2 — Coplanarity of Four Points (Scalar Triple Product)

To test whether four points \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\), \((x_3, y_3, z_3)\), \((x_4, y_4, z_4)\) lie on the same plane, form three vectors from the first point and set their scalar triple product equal to zero.

Coplanarity Condition $$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \\ x_4 - x_1 & y_4 - y_1 & z_4 - z_1 \end{vmatrix} = 0$$

If the determinant is zero, the four points are coplanar. This is faster than first finding the plane through three of them and then checking whether the fourth satisfies it.

Shortcut 3 — Family-of-Planes Through an Intersection Line

Any plane containing the line of intersection of two given planes \(P_1 = 0\) and \(P_2 = 0\) can be written as a single one-parameter family:

Family of Planes Through Intersection $$P_1 + \lambda P_2 = 0$$

Plug the additional condition (a given point, perpendicularity to another plane, etc.) into this combined equation to solve for the scalar λ, then substitute back. You never have to find the intersection line in parametric form — a huge saver when the line is messy.

Shortcut 4 — Line-Plane Angle Uses Sine

The single biggest trap in this chapter. The dot product of the line's direction vector with the plane's normal gives \(\cos(90° - \theta) = \sin\theta\), not \(\cos\theta\). Always use sine for line-to-plane angles, cosine for line-to-line and plane-to-plane.

Quick Reference Line + Line → cos ·  Plane + Plane → cos ·  Line + Plane → sin

A 5-second sanity check before submitting: if the question pairs a line with a plane and your formula contains cosine, stop and switch to sine.

Common Question Patterns

After reviewing PYQ data from 2010 to 2018, three-dimensional geometry questions cluster into six recurring moulds. Recognising the mould saves you 30–60 seconds per question.

Mould 1 — Identity on Direction Cosines

The question gives two angles and asks for a derived quantity. Immediate move: use \(l^2 + m^2 + n^2 = 1\) and the double-angle formula \(\cos 2\alpha = 2\cos^2\alpha - 1\). The derived identity is \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1\), or equivalently \(1 + \cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0\) (tested 2012-I). PYQ 2013-I gives: if a line makes \(30°\) with x-axis, find \(\cos^2\beta + \cos^2\gamma\). Answer: \(1 - \cos^2 30° = 1 - \tfrac{3}{4} = \tfrac{3}{4}\).

Mould 2 — Angle Between Two Objects

Can be two lines, two planes, or a line and a plane. In every case you compute a dot product and divide by the product of magnitudes. Check: if the dot product comes out zero, the objects are perpendicular. If the direction vectors are proportional, they are parallel.

Mould 3 — Plane Through Given Conditions

The question states three points, or a point and a normal direction, or a line of intersection of two planes. Write the general plane equation, plug in the conditions, solve for the constants. PYQ 2012-II: plane through \((1, 2, 3)\) parallel to \(3x + 4y - 5z = 0\). Since parallel planes share the same normal, the equation is \(3x + 4y - 5z = k\), and substituting \((1, 2, 3)\) gives \(k = 3 + 8 - 15 = -4\), so the plane is \(3x + 4y - 5z + 4 = 0\).

Mould 4 — Distance Calculation

Either point-to-plane (use the formula directly) or distance of origin from a plane (set \(x_0 = y_0 = z_0 = 0\)). PYQ 2010-II: distance of origin from \(2x + 6y - 3z + 7 = 0\) is \(|7| / \sqrt{4 + 36 + 9} = 7/7 = 1\) unit. PYQ 2013-II confirms: distance of plane \(2x + y + 2z = 3\) from origin is \(|-3| / \sqrt{4 + 1 + 4} = 3/3 = 1\) unit.

Mould 5 — Line Meeting a Coordinate Plane

Write the line in parametric form. Set the coordinate for the target plane equal to zero. Solve for \(\lambda\), then substitute back. PYQ 2017-I: line through \((1, 2, -1)\) and \((3, -1, 2)\) meets the yz-plane. Set \(x = 1 + 2\lambda = 0 \implies \lambda = -1/2\). Then \(y = 2 - 3/2 = 1/2\) and \(z = -1 - 3/2 = -5/2\). Meeting point: \((0,\ 1/2,\ -5/2)\).

Mould 6 — Sphere Problems

Complete the square to convert the general sphere equation \(x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0\) into standard form \((x + u)^2 + (y + v)^2 + (z + w)^2 = u^2 + v^2 + w^2 - d\). Centre is \((-u, -v, -w)\) and radius is \(\sqrt{u^2 + v^2 + w^2 - d}\). PYQ 2012-I: \(x^2 + y^2 + z^2 - 4x + 6y - 8z - 7 = 0 \implies\) centre \((2, -3, 4)\), radius \(= \sqrt{4 + 9 + 16 + 7} = \sqrt{36} = 6\) units, so diameter \(= 12\) units.

Preparation Strategy

Week 1 — Build the Formula Sheet

Write every formula on this page by hand, once. Do not copy-paste. The act of writing embeds the formula structure. Pay special attention to the three distance formulas: point-to-point, point-to-plane, and shortest distance between skew lines. These look similar but are structurally different — confusing them is the single most common error in this topic.

Week 2 — Categorise by Mould

Go through PYQs from 2010 to 2018 and label each question with its mould number (1–6 from the patterns section). You will quickly see that roughly 70% of questions are Moulds 2 and 3. Prioritise those. Once you can solve any angle-between and plane-through question in under 90 seconds, move to the distance and skew-line types.

Week 3 — Timed Sets and Error Log

NDA gives roughly 1.5 minutes per question. Three-dimensional geometry questions should take you 60–90 seconds each — the computation is short once the formula is identified. If you are taking longer, the bottleneck is formula recall, not arithmetic. Go back to Week 1.

Keep an error log with two columns: the mould that caught you, and the exact mistake. Review it before each mock. The NDA recycles question scenarios — a question from 2013 about projections will reappear in 2018 with different numbers but the same logic.

High-Priority Items for Revision

  • \(l^2 + m^2 + n^2 = 1\) and its derived identities (\(\cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1\)).
  • Symmetric equation of a line and the parametric form — especially for meeting-with-plane questions.
  • Distance of a point from a plane — numerator is the signed substitution, denominator is the norm of the normal vector.
  • Perpendicularity condition for lines: \(a_1 a_2 + b_1 b_2 + c_1 c_2 = 0\).
  • Completing the square for sphere centre and radius.
  • Direction ratios of the normal to a plane \(ax + by + cz + d = 0\) are \((a, b, c)\).

Three-dimensional geometry rewards systematic revision more than almost any other NDA chapter. The formulas are finite, the question types are finite, and the numbers change but the logic does not. If you put in three weeks of structured work you can expect near-perfect accuracy on this topic in the exam.

Link your study of this chapter to Vector Algebra — the vector form of line and plane equations is the same framework, and many NDA questions can be solved faster using vectors once you know both chapters. Also revisit Straight Lines and the Cartesian System if the 2D analogues feel shaky — the 3D extensions follow the exact same logic.

Test your understanding on our full-length NDA mock tests which include three-dimensional geometry in the correct proportion and difficulty.

Test Your 3D Geometry Under Exam Conditions

Our NDA mock tests include three-dimensional geometry questions drawn from the same patterns as PYQs — timed, scored, and with full solutions. Find out exactly where your formula recall breaks down before the real exam does.

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Frequently Asked Questions

What is the difference between direction cosines and direction ratios?

Direction cosines \((l, m, n)\) are the cosines of the angles a line makes with the positive x-, y-, and z-axes. They always satisfy \(l^2 + m^2 + n^2 = 1\). Direction ratios \((a, b, c)\) are any three numbers proportional to the direction cosines — they do not need to satisfy the unit-length condition. Given direction ratios you recover direction cosines by dividing by \(\sqrt{a^2 + b^2 + c^2}\).

How many questions from three-dimensional geometry appear in NDA Maths?

Based on PYQ data from 2010 to 2018, three-dimensional geometry contributes approximately 4 to 8 questions per NDA paper. The count varies by year but the chapter is always represented. In linked-set questions (3 parts per scenario) you can gain or lose 3 marks on a single reading error, so accuracy matters more than speed here.

What is the equation of a line parallel to the z-axis through a point (a, b, c)?

The z-axis has direction ratios \((0, 0, 1)\). A line parallel to the z-axis through \((a, b, c)\) is given by \(\dfrac{x - a}{0} = \dfrac{y - b}{0} = \dfrac{z - c}{1}\). The conventional way to write this is \(x = a,\ y = b\) (the first two coordinates are fixed and \(z\) varies freely). This was tested in PYQ 2012-II.

How do I find where a line meets a coordinate plane?

Write the line in parametric form: \(x = x_1 + \lambda a,\ y = y_1 + \lambda b,\ z = z_1 + \lambda c\). To find where it meets the xy-plane, set \(z = 0\) and solve for \(\lambda\). Substitute this \(\lambda\) back into the \(x\) and \(y\) equations. For the yz-plane set \(x = 0\); for the xz-plane set \(y = 0\). This technique solves PYQ 2017-I directly.

What is the identity relating cos 2α, cos 2β, cos 2γ for a line's direction angles?

Start from \(l^2 + m^2 + n^2 = 1\), where \(l = \cos\alpha\) etc. Use \(\cos 2\alpha = 2\cos^2\alpha - 1\). Then: \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 2(l^2 + m^2 + n^2) - 3 = 2(1) - 3 = -1\). Equivalently, \(1 + \cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0\). This identity was tested in PYQ 2012-I.

How do you find the distance between two parallel planes?

Two parallel planes can be written as \(ax + by + cz + d_1 = 0\) and \(ax + by + cz + d_2 = 0\) (same \(a, b, c\)). The distance between them is \(\dfrac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}\). PYQ 2013-I tests this with planes \(x - 2y + z - 1 = 0\) and \(-3x + 6y - 3z + 2 = 0\). Rewrite the second as \(x - 2y + z - 2/3 = 0\). Distance \(= |-1 - (-2/3)| / \sqrt{1 + 4 + 1} = (1/3)/\sqrt{6}\), which the source records as 0 (the planes are actually the same family — check the ratio to confirm before applying the formula).

What does it mean for two planes to have "infinitely many points in common"?

Two planes that are identical (same equation up to a scalar multiple) coincide — every point on one plane is on the other, giving infinitely many common points. Two planes with the same normal vector but different constants are parallel and share no common points. Two planes with different normal vectors meet in a line — infinitely many points in common but along a line, not the whole plane. PYQ 2010-II tests the first case.