Angle Between Two Lines in Space

Real-World Hook: Satellite Dishes and Antenna Alignment

When installing a satellite dish, technicians must calculate the angle between the dish direction and the satellite signal path to get perfect reception. Similarly, structural engineers calculate angles between support beams to ensure buildings can withstand stress. These are practical applications of finding angles between lines in 3D space!

Angle Between Two Lines: The Formula

Using Direction Ratios

If two lines have direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂), the angle θ between them is:

$$\boxed{\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \cdot \sqrt{a_2^2+b_2^2+c_2^2}}}$$

Interactive Demo: Visualize Angles Between Lines

Explore how angles between lines are measured in 3D space.

Using Direction Cosines

If direction cosines are (l₁, m₁, n₁) and (l₂, m₂, n₂):

$$\boxed{\cos\theta = |l_1l_2 + m_1m_2 + n_1n_2|}$$

Memory Trick 🧠

“Dot Product Over Product of Magnitudes”

This is exactly the dot product formula for vectors!
u·v = |u||v| cos θ
Rearranging: cos θ = (u·v)/(|u||v|)
The absolute value ensures 0° ≤ θ ≤ 90° (acute angle convention)

Why Absolute Value?

Lines don’t have a “direction” - we always measure the acute angle between them.

Without |·|: might get obtuse angle (90° < θ < 180°)
With |·|: always get acute angle (0° ≤ θ ≤ 90°)

Special Cases: Parallel and Perpendicular Lines

Parallel Lines (θ = 0°)

Two lines are parallel if cos θ = 1, which means:

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

Or using direction cosines: l₁ = l₂, m₁ = m₂, n₁ = n₂

Key Insight: Direction ratios are proportional for parallel lines.

Perpendicular Lines (θ = 90°)

Two lines are perpendicular if cos θ = 0, which means:

$$\boxed{a_1a_2 + b_1b_2 + c_1c_2 = 0}$$

Or using direction cosines: l₁l₂ + m₁m₂ + n₁n₂ = 0

Key Insight: Dot product = 0 for perpendicular lines (just like perpendicular vectors!)

Common Mistake ⚠️

Confusing parallel and perpendicular conditions:

Parallel: RATIOS are equal (a₁/a₂ = b₁/b₂ = c₁/c₂)
Perpendicular: DOT PRODUCT is zero (a₁a₂ + b₁b₂ + c₁c₂ = 0)

Example: Lines (2, 4, 6) and (1, 2, 3)
Check parallel: 2/1 = 4/2 = 6/3 = 2  ✓ PARALLEL
Check perpendicular: 2(1) + 4(2) + 6(3) = 2 + 8 + 18 = 28 ≠ 0  ✗

❌ Common error: For perpendicular, checking 2/1 + 4/2 + 6/3 = 0
✓  Correct: Multiply corresponding components, then sum

Angle Between Lines in Different Forms

Lines in Symmetric Form

$$\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1} \quad \text{and} \quad \frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$$

Direction ratios: (a₁, b₁, c₁) and (a₂, b₂, c₂) (read directly from denominators!)

Then apply:

$$\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \cdot \sqrt{a_2^2+b_2^2+c_2^2}}$$

Lines in Vector Form

$$\vec{r} = \vec{a_1} + \lambda\vec{b_1} \quad \text{and} \quad \vec{r} = \vec{a_2} + \mu\vec{b_2}$$

Direction vectors: b₁ and b₂

Then:

$$\boxed{\cos\theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| \cdot |\vec{b_2}|}}$$

Lines in Parametric Form

$$x = x_1 + \lambda a_1, y = y_1 + \lambda b_1, z = z_1 + \lambda c_1$$ $$x = x_2 + \mu a_2, y = y_2 + \mu b_2, z = z_2 + \mu c_2$$

Direction ratios: (a₁, b₁, c₁) and (a₂, b₂, c₂) (coefficients of λ and μ)

Then apply the standard formula.

Worked Examples

Example 1: Basic Angle Calculation (JEE Main Level)

Problem: Find the angle between lines with direction ratios (1, 2, 2) and (2, 3, 6).

Solution:

$$\cos\theta = \frac{|1(2) + 2(3) + 2(6)|}{\sqrt{1^2+2^2+2^2} \cdot \sqrt{2^2+3^2+6^2}}$$ $$= \frac{|2 + 6 + 12|}{\sqrt{1+4+4} \cdot \sqrt{4+9+36}}$$ $$= \frac{20}{\sqrt{9} \cdot \sqrt{49}} = \frac{20}{3 \times 7} = \frac{20}{21}$$

Answer: θ = cos⁻¹(20/21) ≈ 18.19°

Example 2: Checking Perpendicularity (JEE Main Level)

Problem: Show that lines (x-1)/2 = (y-3)/3 = (z+1)/4 and (x-2)/1 = (y-4)/-5 = (z+3)/5 are perpendicular.

Solution: Direction ratios: (2, 3, 4) and (1, -5, 5)

Check: a₁a₂ + b₁b₂ + c₁c₂ = 2(1) + 3(-5) + 4(5) = 2 - 15 + 20 = 7 ≠ 0

Conclusion: Lines are NOT perpendicular (common trap question!)

Example 3: Checking Parallelism (JEE Main Level)

Problem: For what value of k are the lines (x-3)/1 = (y-4)/2 = (z-5)/k and (x-1)/k = (y-2)/4 = (z-3)/2 parallel?

Solution: For parallel lines: a₁/a₂ = b₁/b₂ = c₁/c₂

$$\frac{1}{k} = \frac{2}{4} = \frac{k}{2}$$

From b₁/b₂: 2/4 = 1/2 ✓

From a₁/a₂: 1/k = 1/2 → k = 2

From c₁/c₂: k/2 = 1/2 → k = 1

Contradiction! No single value of k makes them parallel.

Wait - let’s reconsider. For the ratios to be equal:

$$\frac{1}{k} = \frac{2}{4} \Rightarrow k = 2$$ $$\frac{k}{2} = \frac{2}{4} \Rightarrow k = 1$$

These give different values, so no value of k makes them parallel.

Alternative approach: Check if k=1 works: DR: (1, 2, 1) and (1, 4, 2) → 1/1 ≠ 2/4 ✗

Check if k=2 works: DR: (1, 2, 2) and (2, 4, 2) → 1/2 = 2/4 but 2/2 ≠ 2/4 ✗

Answer: No such value of k exists.

Example 4: Vector Form (JEE Advanced Level)

Problem: Find angle between lines r = (2i + 3j - k) + λ(i + 2j + 2k) and r = (3i + j + 2k) + μ(2i + j - 2k).

Solution: Direction vectors: b₁ = i + 2j + 2k, b₂ = 2i + j - 2k

$$\vec{b_1} \cdot \vec{b_2} = 1(2) + 2(1) + 2(-2) = 2 + 2 - 4 = 0$$

Since dot product = 0, the lines are perpendicular.

Answer: θ = 90°

Example 5: Finding Unknown Direction Ratio (JEE Advanced Level)

Problem: Find k if the angle between lines with direction ratios (1, 2, 3) and (2, 3, k) is 60°.

Solution: Given: cos 60° = 1/2

$$\frac{|1(2) + 2(3) + 3(k)|}{\sqrt{1^2+2^2+3^2} \cdot \sqrt{2^2+3^2+k^2}} = \frac{1}{2}$$ $$\frac{|2 + 6 + 3k|}{\sqrt{14} \cdot \sqrt{13+k^2}} = \frac{1}{2}$$ $$\frac{|8 + 3k|}{\sqrt{14(13+k^2)}} = \frac{1}{2}$$ $$2|8 + 3k| = \sqrt{14(13+k^2)}$$

Squaring both sides:

$$4(8 + 3k)^2 = 14(13 + k^2)$$ $$4(64 + 48k + 9k^2) = 182 + 14k^2$$ $$256 + 192k + 36k^2 = 182 + 14k^2$$ $$22k^2 + 192k + 74 = 0$$ $$11k^2 + 96k + 37 = 0$$

Using quadratic formula:

$$k = \frac{-96 \pm \sqrt{96^2 - 4(11)(37)}}{22}$$ $$= \frac{-96 \pm \sqrt{9216 - 1628}}{22} = \frac{-96 \pm \sqrt{7588}}{22}$$ $$= \frac{-96 \pm 87.11}{22}$$

k ≈ -0.40 or k ≈ -8.33

Practice Problems

Level 1: JEE Main Basics

Find the angle between lines with direction ratios (1, 1, 2) and (√3-1, -√3-1, 4).
Show that lines (x-1)/3 = (y-2)/2 = (z-3)/1 and (x+1)/6 = (y-1)/4 = (z-2)/2 are parallel.
Check if lines with direction ratios (2, -3, 4) and (1, 2, 1) are perpendicular.
Find the angle between the X-axis and line (x-1)/1 = (y-2)/2 = (z-3)/3.

Level 2: JEE Main Standard

Find the angle between lines r = i + 2j + k + λ(i + j + k) and r = 2i - j - k + μ(2i + j + 2k).
Find k if lines (x-1)/1 = (y-2)/2 = (z-3)/k and (x-2)/k = (y-3)/1 = (z-4)/2 are perpendicular.
Prove that the line joining (1, 2, 3) and (2, 3, 5) is perpendicular to the line joining (3, 5, 9) and (4, 3, 5).
Find the acute angle between the lines (x-2)/3 = (y+1)/-2 = z/1 and (x-1)/1 = (2y+3)/3 = (z+5)/2.

Level 3: JEE Advanced

Find the angle between any two diagonals of a cube.
Find the equation of a line passing through (1, 2, 3) and making equal angles with the coordinate axes.
A line makes angles α, β, γ with the coordinate axes. Another line makes angles α + 90°, β + 90°, γ + 90° with the axes. Find the angle between these lines.
Find the angle between lines:
- (x-1)/1 = (y-1)/2 = (z-1)/3
- (x-2)/-3 = (y-3)/-2 = (z-4)/-1

Quick Reference Table

Condition	Direction Ratios	Direction Cosines	Angle
Parallel	a₁/a₂ = b₁/b₂ = c₁/c₂	l₁ = l₂, m₁ = m₂, n₁ = n₂	0°
Perpendicular	a₁a₂ + b₁b₂ + c₁c₂ = 0	l₁l₂ + m₁m₂ + n₁n₂ = 0	90°
General	cos θ = \|a₁a₂+b₁b₂+c₁c₂\|/√(…)	cos θ = \|l₁l₂+m₁m₂+n₁n₂\|	0° to 90°

Algorithm: Finding Angle Between Lines

Step 1: Identify the form of line equations

Symmetric: Read DR from denominators
Parametric: Read DR from coefficients of parameter
Vector: Direction vector is coefficient of parameter

Step 2: Extract direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂)

Step 3: Check special cases first

Parallel: a₁/a₂ = b₁/b₂ = c₁/c₂ → θ = 0°
Perpendicular: a₁a₂ + b₁b₂ + c₁c₂ = 0 → θ = 90°

Step 4: If neither, apply formula