5. Three-Dimensional Geometry

Angle between two lines

The angle between two lines in three-dimensional space is determined by the acute angle formed by their directional vectors or direction cosines.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Angle Between Two Lines in 3D Space via Direction Vectors

Let two lines $L_1$ and $L_2$ have direction vectors $\vec{u} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$ and $\vec{v} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$ respectively. The acute angle $\theta$ between these lines is given by $$\cos \theta = \frac{|\vec{u} \cdot \vec{v}|}{|\vec{u}| |\vec{v}|} = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}$$.

📐 Geometric Interpretation

Geometrically, the lines may intersect or be skew. The angle between them is measured by sliding their direction vectors to a common origin, capturing the inclination between their trajectories independent of their spatial positions.

⚙️ Algebraic & Structural Meaning

Algebraically, it leverages the standard inner product (dot product) in $\mathbb{R}^3$, projecting one direction vector onto another while ignoring orientation through the absolute value operator.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-zero Direction Vectors	`Vector norms $\vec{u} \neq \vec{0}, \vec{v} \neq \vec{0}$`	Strictly Mandatory / Necessary	If either direction vector is the zero vector, the components $a_i, b_i, c_i$ vanish, making the denominator zero and the dot product ratio undefined.
Real Direction Ratios / Cosines	`Scalars $a_i, b_i, c_i \in \mathbb{R}$`	Strictly Mandatory	Imaginary or non-real direction components invalidate the vector magnitude computation and Cauchy-Schwarz inequality bounds.
Acute Angle Range Constraint	$\theta \in [0, \frac{\pi}{2}]$	Essential for definition	Omitting the absolute value in the numerator can yield an obtuse angle, violating the standard geometric convention where the angle between lines is taken as acute.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Orthogonality (Perpendicularity) Condition

$$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$

Rule: If the sum of products of direction ratios equals zero, lines are perpendicular ($\theta = 90^\circ$). Instantly check dot product without evaluating square roots.

Application: Use this to identify perpendicular lines among 4 given options.

Parallelism (Collinearity of Direction) Condition

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

Rule: Check proportional direction ratios directly. If proportional, lines are parallel ($\theta = 0^\circ$ or $180^\circ$).

Direction Cosines Dot Product

$\cos \theta = \left|l_1 l_2 + m_1 m_2 + n_1 n_2\right|$

Rule: If lines are given in terms of direction cosines $(l_i, m_i, n_i)$, simply multiply and sum since magnitude is already unity.

Standard Axis / Coordinate Line Angles

$$\cos \theta = \frac{|a|}{\sqrt{a^2+b^2+c^2}}$$

Rule: Angle of a line with direction vector $(a,b,c)$ with the coordinate axes ($x$-axis) is obtained by taking its $x$-component divided by its magnitude.

Application: Direct conceptual questions in UP TGT.

Cross Product Magnitude Alternative

$$\sin \theta = \frac{|\vec{u} \times \vec{v}|}{|\vec{u}| |\vec{v}|}$$

Rule: Use sine formulation if cross product components are simpler to compute than dot product.

Application: Useful when options are given in terms of $\sin \theta$.

Rapid Elimination via Numerator Sign

$$a_1 a_2 + b_1 b_2 + c_1 c_2$$

Rule: Calculate only the numerator dot product. If it is positive, angle is acute ($<90^\circ$); if negative, obtuse before taking absolute value.

Application: Saves 30 seconds in matching options with $\cos \theta > 0$ or $\cos \theta < 0$.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Forgetting the Absolute Value in Numerator

Common Pitfall: Candidates compute $\frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{|\vec{u}||\vec{v}|}$ without absolute value, resulting in a negative cosine and selecting an obtuse angle when acute is required.

Counterexample: Lines with direction ratios $(1, 2, 2)$ and $(-1, -2, -2)$

Why it fails: Leads to $\cos \theta < 0$ ($\theta > 90^\circ$), whereas the geometric angle between lines must lie in $[0, 90^\circ]$.

🛡️ Exam Defense: Always enclose the dot product in absolute value bars: $|\vec{u} \cdot \vec{v}|$.

⚠️ COMMON ERROR #02

Unnormalized Symmetric Equations

Common Pitfall: Using coefficients of $x, y, z$ in symmetric form when they are not equal to $1$ (e.g., $2x - 1 = 3y$).

Counterexample: $\frac{2x-1}{2} = \frac{y}{3}$

Why it fails: Direction ratios are not $2$ and $3$; coefficients of variables must be reduced to unity first yielding $\frac{x - 1/2}{1} = \frac{y}{3}$ so direction ratio is $1$, not $2$.

🛡️ Exam Defense: Always rewrite symmetric equations in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ where $x, y, z$ have unit coefficients.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Angle between two lines	Non-zero direction vectors $\vec{u}, \vec{v}$	$$\cos \theta = \frac{\|\vec{u} \cdot \vec{v}\|}{\|\vec{u}\|\|\vec{v}\|}$$	Acute angle between directional trajectories	Given symmetric or vector equations of lines
Perpendicularity condition	Dot product of direction vectors is zero	$$a_1a_2 + b_1b_2 + c_1c_2 = 0$$	Lines intersect or are skew at $90^\circ$	Keywords: 'perpendicular', 'orthogonal'
Parallelism condition	Proportional direction ratios	$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$	Lines are strictly parallel or coincident	Keywords: 'parallel', 'direction ratios proportional'
Angle between line and plane	Normal vector of plane & direction of line	$$\sin \theta = \frac{\|a_1 a_2 + b_1 b_2 + c_1 c_2\|}{\sqrt{\Sigma a_i^2}\sqrt{\Sigma a_2^2}}$$	Complement of angle between line and plane normal	Keywords: 'line and plane', 'sine formulation'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the angle between the lines given by $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{6}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{2}$.

(A) $\cos^{-1}\left(\frac{20}{21}\right)$(B) $\cos^{-1}\left(\frac{10}{21}\right)$(C) $\cos^{-1}\left(\frac{1}{21}\right)$(D) $\cos^{-1}\left(\frac{2}{3}\right)$

📚 Method A: Formal Textbook Derivation

Step 1: Identify direction ratios: $(a_1, b_1, c_1) = (2, 3, 6)$ and $(a_2, b_2, c_2) = (1, 2, 2)$. Step 2: Compute magnitudes: $\sqrt{2^2+3^2+6^2} = \sqrt{4+9+36} = 7$ and $\sqrt{1^2+2^2+2^2} = \sqrt{1+4+4} = 3$. Step 3: Compute dot product: $(2)(1) + (3)(2) + (6)(2) = 2 + 6 + 12 = 20$. Step 4: Apply formula: $\cos \theta = \frac{20}{7 \times 3} = \frac{20}{21}$.

⚡ Method B: 25-Second Exam Shortcut

Instantly compute dot product $2+6+12=20$, denominators $7 \times 3 = 21$. Match with option (A).

💡 Tip: Check if direction ratios are already in simplest integer form to avoid extra simplification.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the lines $\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$ are perpendicular to each other, find the value of $k$.

(A) $\frac{10}{7}$(B) $-\frac{10}{7}$(C) $\frac{7}{10}$(D) $-\frac{7}{10}$

📚 Method A: Formal Textbook Derivation

Step 1: Direction ratios are $(-3, 2k, 2)$ and $(3k, 1, -5)$. Step 2: Use perpendicular condition $a_1a_2 + b_1b_2 + c_1c_2 = 0$. Step 3: $(-3)(3k) + (2k)(1) + (2)(-5) = 0 \implies -9k + 2k - 10 = 0 \implies -7k = 10 \implies k = -\frac{10}{7}$. Wait, re-evaluating: $-9k + 2k - 10 = 0 \implies -7k = 10 \implies k = -10/7$. Let's check options: (B) is $-\frac{10}{7}$.

⚡ Method B: 25-Second Exam Shortcut

Set $-9k + 2k - 10 = 0 \implies -7k = 10 \implies k = -10/7$.

💡 Tip: Watch out for sign errors when multiplying negative direction ratios.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The angle between the coordinate axes and a line with direction cosines $(l, m, n)$ satisfies $l^2 + m^2 + n^2 = $

A 0

B 1

C $\frac{1}{2}$

D 2

Derivation & Analysis:

By definition, the sum of squares of direction cosines of any line in 3D space is always equal to $1$. Thus $l^2 + m^2 + n^2 = 1$.

MCQ 2: If two lines have direction ratios $(1, 1, 2)$ and $(\sqrt{3}-1, --\sqrt{3}-1, 4)$, the angle between them is:

A $0^\circ$

B $30^\circ$

C $45^\circ$

D $60^\circ$

Derivation & Analysis:

Dot product $= 1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3}-1-\sqrt{3}-1+8 = 6$. Magnitudes: $\sqrt{1+1+4} = \sqrt{6}$ and $\sqrt{(4-2\sqrt{3}) + (4+2\sqrt{3}) + 16} = \sqrt{24} = 2\sqrt{6}$. Thus $\cos \theta = \frac{6}{\sqrt{6}(2\sqrt{6})} = \frac{6}{12} = \frac{1}{2}$, so $\theta = 60^\circ$.

MCQ 3: The lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x-1}{-1} = \frac{y-2}{2} = \frac{z-3}{-1}$ are:

A Perpendicular

B Parallel

C Inclined at $45^\circ$

D Inclined at $60^\circ$

Derivation & Analysis:

Check dot product: $1(-1) + 2(2) + 3(-1) = -1 + 4 - 3 = 0$. Since dot product is zero, the lines are perpendicular.

MCQ 4: If the angle between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{2}$ is $\theta$, then $\cos \theta =$?

A $\frac{8}{9}$

B $\frac{1}{3}$

C $\frac{2}{3}$

D $\frac{4}{9}$

Derivation & Analysis:

Direction ratios are $(2,2,1)$ and $(1,2,2)$. Dot product $= 2(1) + 2(2) + 1(2) = 2 + 4 + 2 = 8$. Magnitudes: $\sqrt{4+4+1} = 3$ and $\sqrt{1+4+4} = 3$. Thus $\cos \theta = \frac{8}{3 \times 3} = \frac{8}{9}$.

MCQ 5: The projection of a line segment joining $(1, 2, 3)$ and $(4, 5, 6)$ on a line with direction ratios $(2, -2, 1)$ is:

A 2

B 3

C 4

D 6

Derivation & Analysis:

Vector joining points is $(3, 3, 3)$. Direction vector is $(2, -2, 1)$ with magnitude $3$. Projection is $\frac{|3(2) + 3(-2) + 3(1)|}{3} = \frac{3}{3} = 1$. Wait, let's recompute: $(6 - 6 + 3)/3 = 3/3 = 1$. Let's check options... option A.

MCQ 6: For what value of $lambda$ are the lines $\frac{x-1}{2} = \frac{y-3}{3} = \frac{z}{1}$ and $\frac{x-2}{lambda} = \frac{y+1}{2} = \frac{z-2}{3}$ perpendicular?

A $-\frac{9}{2}$

B $\frac{9}{2}$

C $-\frac{2}{9}$

D $\frac{2}{9}$

Derivation & Analysis:

Perpendicular condition: $2(\lambda) + 3(2) + 1(3) = 0 \implies 2\lambda + 6 + 3 = 0 \implies 2\lambda = -9 \implies \lambda = -\frac{9}{2}$.

MCQ 7: If a line makes angles $\alpha, \beta, \gamma$ with the coordinate axes, then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = $

A 1

B 2

C 3

D 0

Derivation & Analysis:

Since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$, we have $(1-\sin^2 \alpha) + (1-\sin^2 \beta) + (1-\sin^2 \gamma) = 1 \implies 3 - (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 1 \implies \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2$.

MCQ 8: Direction cosines of a line equally inclined to the coordinate axes are:

A $(\pm 1, \pm 1, \pm 1)$

B $(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}Dios)$

C $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$

D $(\pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}}, 0)$

Derivation & Analysis:

If equally inclined, $l = m = n$. Since $l^2+m^2+n^2 = 1$, $3l^2 = 1 \implies l = \pm \frac{1}{\sqrt{3}}$.

MCQ 9: The angle between the lines whose direction ratios are $(0, 1, -1)$ and $(-\sqrt{3}, 2, 2)$ is:

A $30^\circ$

B $45^\circ$

C $60^\circ$

D $90^\circ$

Derivation & Analysis:

Dot product $= 0(-\sqrt{3}) + 1(2) + (-1)(2) = 0 + 2 - 2 = 0$. Wait, dot product is 0, so angle is $90^\circ$.

Q10

MCQ 10: If the lines $\frac{x}{1} = \frac{y}{-2} = \frac{z}{3}$ and $\frac{x}{3} = \frac{y}{p} = \frac{z}{-2}$ are perpendicular, then $p = $

A $-\frac{3}{2}$

B $\frac{3}{2}$

C $-\frac{2}{3}$

D $\frac{9}{2}$

Derivation & Analysis:

Dot product $= 1(3) + (-2)(p) + 3(-2) = 0 \implies 3 - 2p - 6 = 0 \implies -2p = 3 \implies p = -\frac{3}{2}$. Wait, let's re-verify: $3 - 2p - 6 = -3 - 2p = 0 \implies 2p = -3 \implies p = -3/2$.

Common questions

No. The angle between two lines depends solely on their direction ratios or direction cosines, not on the specific points they pass through.

Failing to convert symmetric equations into standard form (where coefficients of $x, y, z$ are $+1$) before extracting direction ratios.

Memorize the direct dot product formula, immediately check for orthogonality ($a_1a_2+b_1b_2+c_1c_2=0$), and use magnitude shortcuts.

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