Angles Between two Lines in 3D Space | Solved Examples

A line in mathematics and geometry is a fundamental concept representing a straight, one-dimensional figure that extends infinitely in both directions. Lines are characterized by having no thickness and being perfectly straight. Here are some important aspects and definitions related to lines:

Key Characteristics of Lines

• Infinite Length: A line extends without end in both directions. It has no beginning or endpoints.
• No Thickness: A line is considered to have no width or depth; it is purely one-dimensional.
• Straightness: A line does not curve or bend. It is the shortest distance between any two points lying on it.

Straight Lines in 3D space are generally represented in two forms: Cartesian Form and Vector Form. Hence, the angles between any two straight lines in 3D space are also defined in terms of both the forms of the straight lines. Let's discuss the methods of finding the angle between two straight lines in both forms one by one.

Cartesian Form of Line

L₁: (x - x₁) / a₁ = (y - y₁) / b₁ = (z - z₁) / c₁

L₂: (x - x₂) / a₂ = (y - y₂) / b₂ = (z - z₂) / c₂

Here L₁ & L₂ represent the two straight lines passing through the points (x₁, y₁, z₁) and (x₂, y₂, z₂) respectively, in 3D space in Cartesian Form. 

• Direction ratios of line L₁ are a₁, b₁, c₁ then a vector parallel to L₁ is {\vec {m}}₁ = a₁ i + b₁  j + c₁ k
• Direction ratios of line L₂ are a₂, b₂, c₂ then a vector parallel to L₂ is {\vec {m}}₂ = a₂ i + b₂  j + c₂ k

Angle Between Lines in Cartesian Form

Then the angle ∅ between L₁ and L₂ is given by:

∅ = cos^-1{({\vec {m}}₁ . {\vec {m}}₂) / (|{\vec {m}}₁| × |{\vec {m}}₂|)}

Examples

Example 1: (x - 1) / 1 = (2y + 3) / 3 = (z + 5) / 2 and (x - 2) / 3 = (y + 1) / -2 = (z - 2) / 0 are the two lines in 3D space then the angle ∅ between them is given by:

Solution:

Given: {\vec {m}}₁ = 1 i + (3 / 2)  j + 2 k and {\vec {m}}₂ = 3 i - 2 j + 0 k

⇒ |{\vec {m}}₁| = √(1² + (3/2)² + 2²) = √(29 / 2), and

⇒ |{\vec {m}}₂| = √(3² + 2² + 0²) = √(13)

Thus, ∅  = cos^-1{(1×3 + (3/2)×(-2) + (2)×0 ) / ((√(29) / 2) × √(13))}

⇒ ∅  = cos^-1{0 / ((√(29) / 2) × √(13))}

⇒ ∅  = cos^-1(0)

⇒ ∅  = π / 2

Example 2: Find the angles between the two lines in 3D space whose only direction ratios are given 2, 1, 2 and 2, 3, 1. In the question, equations of the 2 lines are not given, only their DRs are given. So the angle ∅ between the 2 lines is given by:

Solution:

{\vec {m}}₁ = Vector parallel to the line having DRs 2, 1, 2 = (2 i + j + 2 k)

v|{\vec {m}}₁| = √(2² + 1² + 2²) = √9 = 3

{\vec {m}}₂ = Vector parallel to the line having DRs 2, 3, 1 = (2 i + 3 j + k)

⇒ |{\vec {m}}₂| = √(2² + 3² + 1²) = √(14)

Thus, ∅  = cos^-1{(2×2 + 1×3 + 2×1) / (3 × √(14))}

⇒ ∅  = cos^-1{(4 + 3 + 2) / (3 × √(14))}

⇒ ∅  = cos^-1{9 / (3 × √(14))}

⇒ ∅  = cos^-1(3 / √(14))

Example 3: (x - 1) / 2 = (y - 2) / 1 = (z - 3) / 2 and (x - 2) / 2 = (y - 1) / 2 = (z - 3) / 1 are the two lines in 3D space then the angle ∅ between them is given by:

Solution:

{\vec {m}}₁ = 2 i + j + 2 k

⇒ |{\vec {m}}₁| = √(2² + 1² + 2²) = √9 = 3

{\vec {m}}₂ = 2 i + 2 j + k

⇒ |{\vec {m}}₂| = √(2² + 2² + 1²) = √9 = 3

Thus, ∅  = cos^-1{(2×2 + 1×2 + 2×1 ) / (3 × 3)}

⇒ ∅  = cos^-1{(4 + 2 + 2) / 9}

⇒ ∅  = cos^-1(8 / 9)

Vector Form of Line

L₁: {\vec {r}} = {\vec {a}}₁ + t . {\vec {b}}₁

L₂: {\vec {r}} = {\vec {a}}₂ + u . {\vec {b}}₂

Where,

• L₁ & L₂ represent the two straight lines passing through the points whose position vectors are {\vec {a}}₁ and {\vec {a}}₂ respectively in 3D space in Vector Form,
• {\vec {b}}₁ & {\vec {b}}₂ are the two vectors parallel to L₁ and L₂ respectively
• and t & u are the parameters.

Angle Between Lines in Vector Form

Then the angle ∅ between the vectors {\vec {b}}₁ and {\vec {b}}₂ is equals to the angle between L₁ and L₂ is given by:

∅ = cos^-1{({\vec {b}}₁ . {\vec {b}}₂) / (|{\vec {b}}₁| × |{\vec {b}}₂|)}

Solved Examples on Angles Between two Lines in 3D Space

Example 1: {\vec {r}} = (i + j + k) + t × {(-√3 - 1) i + (√3 - 1) j + 4 k} and {\vec {r}} = (i + j + k) + u × (i +  j + 2 k)  are the two lines in 3D space then the angle ∅ between them is given by:

Solution:

{\vec {b}}₁ = (-√3 - 1) i + (√3 - 1) j + 4 k

⇒ |{\vec {b}}₁| = √{(-√3 - 1)² + (√3 - 1)² + 4²)} = √(24)

{\vec {b}}₂ = i +  j + 2 k

⇒ |{\vec {b}}₂| = √(1² + 1² + 2²) = √6

Thus, ∅  = cos^-1{(-√3 - 1)×1 + (√3 - 1)×1 + 4×2 ) / (√(24) × √6)}

⇒ ∅  = cos^-1{6 / (√(24) × √6)}

⇒ ∅  = cos^-1(½)

⇒ ∅  = π / 3

Example 2: (i + 2 j + 2 k) and (3 i + 2 j + 6 k) are the two vectors parallel to the two lines in 3D space then the angle ∅ between them is given by:

Solution:

{\vec {b}}₁ = i + 2 j + 2 k

⇒ |{\vec {b}}₁| = √(1² + 2² + 2²)} = √9 = 3

{\vec {b}}₂ = 3 i + 2 j + 6 k

⇒ |{\vec {b}}₂| = √(3² + 2² + 6²) = √(49) = 7

Thus, ∅  = cos^-1{(1×3 + 2×2 + 2×6) / (7 × 3)}

⇒ ∅  = cos^-1{(3 + 4 + 12) / 21}

⇒ ∅  = cos^-1(19 / 21)

Example 3: {\vec {r}} = (3 i + 5 j + 7 k) + s × {(i + 2 j - 2 k} and {\vec {r}} = (4 i + 3 j + k) + t × (2 i + 4 j - 4 k)  are the two lines in 3D space then the angle ∅ between them is given by:

Solution:

{\vec {b}}₁ = i + 2 j - 2 k

⇒ |{\vec {b}}₁| = √(1² + 2² + (-2)²)} = √9 = 3

{\vec {b}}₂ = 2 i + 4 j - 4 k

⇒ |{\vec {b}}₂| = √(2² + 4² + (-4)²) = √(36) = 6

Thus, ∅  = cos^-1{(1×2 + 2×4 + (-2)×(-4)) / (3 × 6)}

⇒ ∅  = cos^-1{(2 + 8 + 8) / 18}

⇒ ∅  = cos^-1(18 / 18)

⇒ ∅  = cos^-1(1) = 0

Read More,

• Equation of Plane
• Equation of Line
• Distance Formula