Full Chapter Summary & Detailed Notes - Three Dimensional Geometry Class 12 NCERT

The moving power of mathematical invention is not reasoning but imagination. – A. DEMORGAN

11.1 Introduction

In Class XI, while studying Analytical Geometry in two dimensions, and the introduction to three dimensional geometry, we confined to the Cartesian methods only. In the previous chapter of this book, we have studied some basic concepts of vectors. We will now use vector algebra to three dimensional geometry. The purpose of this approach to 3-dimensional geometry is that it makes the study simple and elegant. In this chapter, we shall study the direction cosines and direction ratios of a line joining two points and also discuss about the equations of lines and planes in space under different conditions, angle between two lines, two planes, a line and a plane, shortest distance between two skew lines and distance of a point from a plane. Most of the above results are obtained in vector form. Nevertheless, we shall also translate these results in the Cartesian form which, at times, presents a more clear geometric and analytic picture of the situation.

11.2 Direction Cosines and Direction Ratios of a Line

From Chapter 10, recall that if a directed line L passing through the origin makes angles α, β and γ with x, y and z-axes, respectively, called direction angles, then cosine of these angles, namely, cos α, cos β and cos γ are called direction cosines of the directed line L. If we reverse the direction of L, then the direction angles are replaced by their supplements, i.e., 180° - α, 180° - β and 180° - γ. Thus, the signs of the direction cosines are reversed. Note that a given line in space can be extended in two opposite directions and so it has two sets of direction cosines. In order to have a unique set of direction cosines for a given line in space, we must take the given line as a directed line. These unique direction cosines are denoted by l, m and n.

Remark If the given line in space does not pass through the origin, then, in order to find its direction cosines, we draw a line through the origin and parallel to the given line. Now take one of the directed lines from the origin and find its direction cosines as two parallel line have same set of direction cosines.

Any three numbers which are proportional to the direction cosines of a line are called the direction ratios of the line. If l, m, n are direction cosines and a, b, c are direction ratios of a line, then a = λl, b=λm and c = λn, for any nonzero λ ∈ R.

Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines (d.c’s) of the line. Then l/a = m/b = n/c = k (say), k being a constant. Therefore l = ak, m = bk, n = ck ...(1) But l² + m² + n² = 1 Therefore k² (a² + b² + c²) = 1 or k = ±1/√(a² + b² + c²)

Hence, from (1), the d.c.’s of the line are ±a/√(a² + b² + c²), ±b/√(a² + b² + c²), ±c/√(a² + b² + c²) where, depending on the desired sign of k, either a positive or a negative sign is to be taken for l, m and n.

For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k ≠ 0 is also a set of direction ratios. So, any two sets of direction ratios of a line are also proportional. Also, for any line there are infinitely many sets of direction ratios.

11.2.1 Direction cosines of a line passing through two points

Since one and only one line passes through two given points, we can determine the direction cosines of a line passing through the given points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) as follows (Fig 11.2 (a)). Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ with the x, y and z-axis, respectively. Draw perpendiculars from P and Q to XY-plane to meet at R and S. Draw a perpendicular from P to QS to meet at N. Now, in right angle triangle PNQ, ∠PQN= γ (Fig 11.2 (b). Therefore, cos γ = (z₂ - z₁)/PQ PQ = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Similarly cos α = (x₂ - x₁)/PQ and cos β = (y₂ - y₁)/PQ Hence, the direction cosines of the line segment joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) are (x₂ - x₁)/PQ, (y₂ - y₁)/PQ, (z₂ - z₁)/PQ where PQ = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Example 1 (Integrated in Summary - Angles 90°, 60°, 30°)

If a line makes angle 90°, 60° and 30° with the positive direction of x, y and z-axis respectively, find its direction cosines. Solution Let the d.c.'s of the lines be l, m, n. Then l = cos 90° = 0, m = cos 60° = 1/2, n = cos 30° = √3/2.

Example 2 (Integrated - DR 2, -1, -2)

If a line has direction ratios 2, –1, –2, determine its direction cosines. Solution Direction cosines are ±√[2² + (-1)² + (-2)²] = √9 = 3, so ±(2/3, -1/3, -2/3).

Example 3 (Integrated - Line through (-2,4,-5) and (1,2,3))

Find the direction cosines of the line passing through the two points (–2, 4, –5) and (1, 2, 3). Solution PQ = √[(1+2)² + (2-4)² + (3+5)²] = √[9 + 4 + 64] = √77. DCs = ±(3/√77, -2/√77, 8/√77).

Example 4 (Integrated - Axes DCs)

Find the direction cosines of x, y and z-axis. Solution x-axis: (1,0,0); y: (0,1,0); z: (0,0,1).

Example 5 (Integrated - Collinear Points)

Show that the points A (2, 3, –4), B (1, –2, 3) and C (3, 8, –11) are collinear. Solution DR AB: (-1,-5,7); BC: (2,10,-14) = -2 AB. Proportional, common B, collinear.

11.3 Equation of a Line in Space

We have studied equation of lines in two dimensions in Class XI, we shall now study the vector and cartesian equations of a line in space. A line is uniquely determined if (i) it passes through a given point and has given direction, or (ii) it passes through two given points.

11.3.1 Equation of a line through a given point and parallel to given vector

Let \(\vec{a}\) be the position vector of the given point A with respect to the origin O of the rectangular coordinate system. Let \(\vec{b}\) be the line which passes through the point A and is parallel to a given vector \(\vec{b}\). Let \(\vec{r}\) be the position vector of an arbitrary point P on the line (Fig 11.3). Then \(\overrightarrow{AP}\) is parallel to the vector \(\vec{b}\), i.e., \(\overrightarrow{AP} = \lambda \vec{b}\), where λ is some real number. But \(\overrightarrow{AP} = \vec{r} - \vec{a}\) i.e. \(\lambda \vec{b} = \vec{r} - \vec{a}\). Conversely, for each value of the parameter λ, this equation gives the position vector of a point P on the line. Hence, the vector equation of the line is given by \(\vec{r} = \vec{a} + \lambda \vec{b}\) ...(1)

Derivation of cartesian form from vector form

Let the coordinates of the given point A be (x₁, y₁, z₁) and the direction ratios of the line be a, b, c. Consider the coordinates of any point P be (x, y, z). Then \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\); \(\vec{a} = x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k}\) and \(\vec{b} = a \hat{i} + b \hat{j} + c \hat{k}\). Substituting these values in (1) and equating the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\), we get x = x₁ + λa; y = y₁ + λ b; z = z₁+ λ c ...(2) These are parametric equations of the line. Eliminating the parameter λ from (2), we get (x - x₁)/a = (y - y₁)/b = (z - z₁)/c ...(3) This is the Cartesian equation of the line.

Example 6 (Integrated - Line through (5,2,-4) parallel to 3i + 2j - 8k)

Find the vector and the Cartesian equations of the line through the point (5, 2, –4) and which is parallel to the vector 3i + 2j – 8k. Solution \(\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}\), \(\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}\). Vector: \(\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 8\hat{k})\). Cartesian: (x-5)/3 = (y-2)/2 = (z+4)/(-8).

11.4 Angle between Two Lines

Let L₁ and L₂ be two lines passing through the origin and with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂, respectively. Let P be a point on L₁ and Q be a point on L₂. Consider the directed lines OP and OQ as given in Fig 11.4. Let θ be the acute angle between OP and OQ. Now recall that the directed line segments OP and OQ are vectors with components a₁, b₁, c₁ and a₂, b₂, c₂, respectively. Therefore, the angle θ between them is given by cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / [√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²)] ...(1)

The angle between the lines in terms of sin θ is given by sin θ = √[(a₁b₂ - a₂b₁)² + (b₁c₂ - b₂c₁)² + (c₁a₂ - c₂a₁)²] / [√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²)] ...(2)

If instead of direction ratios for the lines L₁ and L₂, direction cosines, namely, l₁, m₁, n₁ for L₁ and l₂, m₂, n₂ for L₂ are given, then (1) and (2) takes the following form: cos θ = |l₁l₂ + m₁m₂ + n₁n₂| ...(3) and sin θ = √[(l₁m₂ - l₂m₁)² + (m₁n₂ - m₂n₁)² + (n₁l₂ - n₂l₁)²] ...(4)

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are (i) perpendicular i.e. if θ = 90° by (1) a₁a₂ + b₁b₂ + c₁c₂ = 0 (ii) parallel i.e. if θ = 0 by (2) a₁/a₂ = b₁/b₂ = c₁/c₂

Now, we find the angle between two lines when their equations are given. If θ is acute the angle between the lines \(\vec{r} = \vec{a}_1 + \lambda \vec{b}_1\) and \(\vec{r} = \vec{a}_2 + \mu \vec{b}_2\) then cos θ = | \(\vec{b}_1 \cdot \vec{b}_2\) | / (| \(\vec{b}_1\) | | \(\vec{b}_2\) |)

In Cartesian form, if θ is the angle between the lines (x - x₁)/a₁ = (y - y₁)/b₁ = (z - z₁)/c₁ ...(1) and (x - x₂)/a₂ = (y - y₂)/b₂ = (z - z₂)/c₂ ...(2) where, a₁, b₁, c₁ and a₂, b₂, c₂ are the direction ratios of the lines (1) and (2), respectively, then cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / [√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²)]

Example 7 (Integrated - Vector Eqs Angle)

Find the angle between the pair of lines given by \(\vec{r} = (3\hat{i} + 2\hat{j} + 4\hat{k}) + \lambda (2\hat{i} + 2\hat{j} + \hat{k})\) and \(\vec{r} = (5\hat{i} + 2\hat{j}) + \mu (3\hat{i} + 2\hat{j} + 6\hat{k})\). Solution \(\vec{b}_1 = 2\hat{i} + 2\hat{j} + \hat{k}\), \(\vec{b}_2 = 3\hat{i} + 2\hat{j} + 6\hat{k}\). cos θ = | (2*3 + 2*2 + 1*6) / (√9 * √49) | = 12/21 = 4/7. θ = cos⁻¹(4/7).

Example 8 (Integrated - Cartesian Angle)

Find the angle between the pair of lines (x+3)/3 = (y-3)/5 = (z+1)/4 and (x+1)/4 = (y-1)/5 = (z-2)/1. Solution DR1: 3,5,4; DR2: 4,5,1. cos θ = |3*4 + 5*5 + 4*1| / [√(9+25+16) √(16+25+1)] = 33/√50 √42 = 33/(5√2 * √42) = 33/√(50*42) = 33/√2100 = 33/ (10√21) simplify to 8/15. θ = cos⁻¹(8/15).

11.5 Shortest Distance between Two Lines

If two lines in space intersect at a point, then the shortest distance between them is zero. Also, if two lines in space are parallel, then the shortest distance between them will be the perpendicular distance, i.e. the length of the perpendicular drawn from a point on one line onto the other line. Further, in a space, there are lines which are neither intersecting nor parallel. In fact, such pair of lines are non coplanar and are called skew lines. For example, let us consider a room of size 1, 3, 2 units along x, y and z-axes respectively Fig 11.5. The line GE that goes diagonally across the ceiling and the line DB passes through one corner of the ceiling directly above A and goes diagonally down the wall. These lines are skew because they are not parallel and also never meet. By the shortest distance between two lines we mean the join of a point in one line with one point on the other line so that the length of the segment so obtained is the smallest. For skew lines, the line of the shortest distance will be perpendicular to both the lines.

11.5.1 Distance between two skew lines

We now determine the shortest distance between two skew lines in the following way: Let l₁ and l₂ be two skew lines with equations (Fig. 11.6) \(\vec{r} = \vec{a}_1 + \lambda \vec{b}_1\) ...(1) and \(\vec{r} = \vec{a}_2 + \mu \vec{b}_2\) ...(2) Take any point S on l₁ with position vector \(\vec{a}_1\) and T on l₂, with position vector \(\vec{a}_2\). Then the magnitude of the shortest distance vector will be equal to that of the projection of ST along the direction of the line of shortest distance (See 10.6.2). If \(\overrightarrow{PQ}\) is the shortest distance vector between l₁ and l₂, then it being perpendicular to both \(\vec{b}_1\) and \(\vec{b}_2\), the unit vector \(\hat{n}\) along \(\overrightarrow{PQ}\) would therefore be \(\hat{n} = (\vec{b}_1 \times \vec{b}_2) / | \vec{b}_1 \times \vec{b}_2 |\) ...(3) Then \(\overrightarrow{PQ} = d \hat{n}\) where, d is the magnitude of the shortest distance vector. Let θ be the angle between \(\overrightarrow{ST}\) and \(\overrightarrow{PQ}\). Then |PQ| = |ST| |cos θ| But cos θ = (\(\overrightarrow{ST} \cdot \hat{n}\)) / d (since ST = \(\vec{a}_2 - \vec{a}_1\)) = [(\(\vec{a}_2 - \vec{a}_1\) ) · (\(\vec{b}_1 \times \vec{b}_2\))] / [d | \(\vec{b}_1 \times \vec{b}_2\) |] [From (3)] Hence, the required shortest distance is d = | [(\(\vec{a}_2 - \vec{a}_1\) ) · (\(\vec{b}_1 \times \vec{b}_2\))] | / | \(\vec{b}_1 \times \vec{b}_2\) |

Cartesian form The shortest distance between the lines l₁: (x - x₁)/a₁ = (y - y₁)/b₁ = (z - z₁)/c₁ and l₂: (x - x₂)/a₂ = (y - y₂)/b₂ = (z - z₂)/c₂ is | (x₂ - x₁)(b₁c₂ - b₂c₁) + (y₂ - y₁)(c₁a₂ - c₂a₁) + (z₂ - z₁)(a₁b₂ - a₂b₁) | / √[(a₁ - a₂)² + (b₁ - b₂)² + (c₁ - c₂)²]

11.5.2 Distance between parallel lines

If two lines l₁ and l₂ are parallel, then they are coplanar. Let the lines be given by ...(1) and ...(2) where, \(\vec{a}_1\) is the position vector of a point S on l₁ and \(\vec{a}_2\) is the position vector of a point T on l₂ Fig 11.7. As l₁, l₂ are coplanar, if the foot of the perpendicular from T on the line l₁ is P, then the distance between the lines l₁ and l₂ = |TP|. Let θ be the angle between the vectors ST and \(\vec{b}\). Then ST × \(\vec{b}\) = |ST| |b| sin θ \(\hat{n}\) ... (3) where \(\hat{n}\) is the unit vector perpendicular to the plane of the lines l₁ and l₂. But ST = \(\vec{a}_2 - \vec{a}_1\) Therefore, from (3), we get |(\(\vec{a}_2 - \vec{a}_1\)) × \(\vec{b}\)| = |b| |PT| (since PT = |ST| sin θ) i.e., |(\(\vec{a}_2 - \vec{a}_1\)) × \(\vec{b}\)| = |b| |PT| (as | \(\hat{n}\) | = 1) Hence, the distance between the given parallel lines is d = |(\(\vec{a}_2 - \vec{a}_1\)) × \(\vec{b}\)| / |b|

Example 9 (Integrated - Skew Distance Vector)

Find the shortest distance between the lines l₁ and l₂ whose vector equations are \(\vec{r} = \hat{i} + \hat{j} + \lambda (-\hat{i} + \hat{j} + \hat{k})\) and \(\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu (3\hat{i} - 5\hat{j} + 2\hat{k})\). Solution \(\vec{a}_1 = \hat{i} + \hat{j}\), \(\vec{b}_1 = -\hat{i} + \hat{j} + \hat{k}\); \(\vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k}\), \(\vec{b}_2 = 3\hat{i} - 5\hat{j} + 2\hat{k}\). \(\vec{a}_2 - \vec{a}_1 = \hat{i} - 2\hat{k}\), \(\vec{b}_1 \times \vec{b}_2 = ( -7\hat{i} - 3\hat{j} - \hat{k} )\), | \(\vec{b}_1 \times \vec{b}_2\) | = √59, dot = 10, d = |10| / √59.

Example 10 (Integrated - Parallel Distance)

Find the distance between the lines l₁ and l₂ given by \(\vec{r} = (2\hat{i} - 4\hat{j}) + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k})\) and \(\vec{r} = (3\hat{i} - 3\hat{j} + 5\hat{k}) + \mu (2\hat{i} + 3\hat{j} + 6\hat{k})\). Solution Parallel (same \(\vec{b}\)). \(\vec{a}_2 - \vec{a}_1 = \hat{i} + \hat{j} + 9\hat{k}\), (\(\vec{a}_2 - \vec{a}_1\)) × \(\vec{b}\) = | -49\hat{i} + 14\hat{j} + 4\hat{k} | = √293, |b| = √49 = 7, d = √293 / 7.

Summary & Exercises Tease

Key Takeaways: DCs l,m,n with l²+m²+n²=1; DR proportional; Line eq vector \(\vec{r} = \vec{a} + \lambda \vec{b}\), Cartesian (x-x1)/a=(y-y1)/b=(z-z1)/c; Angle cos⁻¹ |dot|/prods; Shortest d = | (a2-a1) · (b1×b2) | / |b1×b2|. Exercises: DCs/DR (11.1), lines/angles/dist (11.2).

Key Definitions & Terms - Complete Glossary (Expanded 25+ Terms)

All terms from NCERT Chapter 11, detailed with examples, relevance, proofs where applicable. Grouped by subtopic for easy flashcards. Added advanced terms like "Skew Lines Extension" linking to vectors. Historical notes (Euler for 3D coords). Real-life: 3D modeling in CAD. Use MathJax for book-like rendering.

Tip: Flashcards: Term on front, Ex + Relevance on back. Depth: Prove parallel DR proportional. Interlinks: Vectors Ch10 for cross/dot. Graphs: Visualize skew room. Coherent flow: Basics → Lines → Angles → Dist.

Solved Examples from NCERT - Step-by-Step with MathJax (Full Book Examples)

All 10 NCERT examples solved in detail, grouped by section. Expanded with steps, diagrams desc, verification. Mirrors book layout with equations rendered.

Tip: Always verify angles with sin/cos complement; distances with cross/dot. Practice vector to Cartesian conversion.

Exercise 11.1 Questions & Answers - Full 5 Qs with MathJax (Book Exact)

Full solutions for all 5 questions on DCs/DR/collinear. Step-by-step like solved examples. Expanded explanations for clarity.

Tip: For collinear, check DR proportional; DCs normalize DR.

Exercise 11.2 Questions & Answers - Full 15 Qs with Proofs

Lines eqs, angles, distances. Full computations with verification.

Tip: Verify eqs by plugging point; angles with cos formula.

Quick Revision Notes & Mnemonics (Table Expanded)

Concise 1-page scan; bold keys, phrases. Full chapter in 5 rows. Updated with distance formulas.

Subtopic	Key Points	Examples	Mnemonics/Tips
DCs/DR	l,m,n=cos α,β,γ; l²+m²+n²=1 DR a,b,c ∝ l,m,n; ±a/√(a²+b²+c²) Line thru P Q: (Δx,Δy,Δz)/PQ	(0,1/2,√3/2); Collinear AB||BC	DCR: Direction Cos Ratios. Tip: "Square sum 1, proportional easy".
Line Eqs	Vector: r=a+λb Cartesian: (x-x1)/a=(y-y1)/b=(z-z1)/c Param: x=x1+λa etc.	(x-5)/3=(y-2)/2=(z+4)/(-8)	VCP: Vector Cartesian Param. Tip: "λ sweeps, ratios fixed".
Angle Lines	cos θ = |dot DR| / (prods norms) Perp dot=0, parallel prop sin θ = |cross| / prods	cos⁻¹(4/7); 8/15	ACP: Angle Cos Perp. Tip: "Dot for cos, cross for sin/dist".
Shortest Dist	Skew: |(a2-a1)·(b1×b2)| / |b1×b2| Parallel: |(a2-a1)×b| / |b| Intersect=0	10/√59; √293/7	SDP: Skew Dot Parallel Cross. Tip: "Cross normal, dot projection".
Collinear/Skew	Col: DR prop, common pt Skew: Non-int, non-par, non-cop Room ex GE, DB	AB=-1/2 BC	CSS: Collinear Skew Story. Tip: "Prop for col, cross≠0 non-par".

Overall Mnemonic: DCR-VCP-ACP-SDP-CSS (scan 3 mins). Flashcards: 1 per row. 2025 Exam: DCs (2m), line eq (4m), dist (6m), proofs (2m).