Angles Between two Lines in 3D Space

A line is a straight, one-dimensional figure that extends infinitely in both directions and has no thickness or endpoints.

In 3D space, straight lines are generally represented in two forms: Cartesian form and vector form. Therefore, the angle between two lines in 3D can also be determined using these two representations.

Cartesian Form of Line

L_1: \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}

L_2: \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}

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

• 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 + b_² j + c² k

Angle Between Lines in Cartesian Form

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

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

Example 1: Find the angles between the two lines in 3D space whose only direction ratios are given as 2, 1, 2 and 2, 3, 1.

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 2: (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)

Example 3: If (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

Vector Form of Line

L₁: {\vec {r}} = {\vec {a}}₁ + t \vec{\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, lying in 3D space in Vector F.rm.
• {\vec {b}}_{1$ & ${\vec} & {\vec {b}}₂ are the two vectors parallel to L₁ & L2, respectively
• And t and u are the parameters.

Angle Between Lines in Vector Form

Then the angle ∅ between the vectors {\vec {b}}₁ and {\vec {b}}₂ is equal to the angle between L₁ and L_{2, which} is given by

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

Example 1: If (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 2: If {\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

Example 3: {\vec {r}} = (i + j + k) + t × {(-√3 - 1) If {\vec {r}}_1 = i + (\sqrt{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