Practice Maths

← Vectors in Three Dimensions › Vector Equations of Lines in 3D › Solutions

Vector Equations of Lines in 3D — Full Worked Solutions

  1. Vector equation through given point with given direction. Fluency

    (a) r = (1, 2, 3) + λ(2, −1, 4)

    (b) r = (0, −1, 5) + λ(1, 3, −2)

  2. Parametric equations through A = (3,0,−2) and B = (1,4,1). Fluency

    Direction vector: d = B − A = (1−3, 4−0, 1−(−2)) = (−2, 4, 3)

    Using point A as the base:

    x = 3 − 2λ,   y = 4λ,   z = −2 + 3λ

    (Any scalar multiple of (−2,4,3) is also a valid direction.)

  3. Convert r = (2,−3,1)+λ(3,2,−1) to Cartesian form. Fluency

    Parametric equations: x = 2+3λ, y = −3+2λ, z = 1−λ

    Solve for λ:

    λ = (x−2)/3 = (y+3)/2 = (z−1)/(−1)

    Cartesian form: (x−2)/3 = (y+3)/2 = (z−1)/(−1)

  4. Test whether point P lies on the given line. Fluency

    (a) P = (5,−1,7), r = (1,3,−1)+λ(2,−2,4):

    Need (5,−1,7) = (1,3,−1)+λ(2,−2,4)

    x: 5=1+2λ ⇒ λ=2; y: −1=3−4=−1 ✓; z: 7=−1+8=7 ✓

    P is on the line (λ=2).

    (b) P = (3,2,0), r = (1,1,2)+λ(1,0,−1):

    x: 3=1+λ ⇒ λ=2; y: 2=1+0=1 ≠ 2. Inconsistent.

    P is NOT on the line.

  5. Intersection of L₁: r=(1,2,0)+λ(1,−1,2) and L₂: r=(3,0,4)+μ(−1,1,0). Understanding

    Set r₁ = r₂:

    x: 1+λ = 3−μ …(i)

    y: 2−λ = μ …(ii)

    z: 2λ = 4 …(iii)

    From (iii): λ=2. From (ii): μ=2−2=0.

    Check (i): 1+2=3 and 3−0=3 ✓

    Intersection point: r = (1,2,0)+2(1,−1,2) = (3, 0, 4). Point: (3, 0, 4)

  6. Classify each pair of lines. Understanding

    (a) L₁: direction (1,2,1); L₂: direction (2,4,2) = 2(1,2,1). Proportional ⇒ parallel.

    Is (1,3,4) on L₁? (1,3,4) = (0,1,2)+λ(1,2,1): λ=1 from x; y: 1+2=3 ✓; z: 2+1=3≠4. Not the same line.

    Parallel and distinct.

    (b) L₁: r=(1,0,0)+λ(1,1,0); L₂: r=(0,1,1)+μ(1,0,1).

    Directions not proportional. Set equal:

    x: 1+λ=μ …(i); y: λ=1 …(ii, from 0+λ=1+0); z: 0=1+μ ⇒ μ=−1 …(iii)

    From (ii) λ=1; from (iii) μ=−1; check (i): 1+1=2 ≠ −1. Inconsistent.

    Skew.

  7. Distance between skew lines L₁: r=(1,0,0)+λ(1,1,0) and L₂: r=(0,1,0)+μ(0,1,1). Understanding

    n = d₁ × d₂ = (1,1,0)×(0,1,1):

    i: (1)(1)−(0)(1)=1; j: −[(1)(1)−(0)(0)]=−1; k: (1)(1)−(1)(0)=1

    n = (1,−1,1); |n|=√3

    b−a = (0,1,0)−(1,0,0) = (−1,1,0)

    (−1,1,0)·(1,−1,1) = −1−1+0 = −2

    Distance = |−2|/√3 = 2/√3 = 2√3/3 ≈ 1.15 units

  8. Line through P=(2,−1,4) and Q=(5,3,−2). Does R=(8,7,−8) lie on it? Understanding

    (a) Direction: d = Q−P = (3,4,−6)

    Vector equation: r = (2,−1,4)+λ(3,4,−6)

    (b) Test R=(8,7,−8):

    x: 8=2+3λ ⇒ λ=2; y: 7=−1+8=7 ✓; z: −8=4−12=−8 ✓

    R lies on the line (at λ=2).

  9. Distance between skew lines L₁: r=(0,0,0)+λ(1,0,1) and L₂: r=(1,1,0)+μ(0,1,1). Problem Solving

    Verify skew: Set equal. x: λ=1 …(i); y: 0=1+μ ⇒ μ=−1 …(ii); z: λ=μ ⇒ 1=−1. Inconsistent. Skew ✓

    n = (1,0,1)×(0,1,1):

    i: (0)(1)−(1)(1)=−1; j: −[(1)(1)−(1)(0)]=−1; k: (1)(1)−(0)(0)=1

    n = (−1,−1,1); |n|=√3

    b−a = (1,1,0)−(0,0,0) = (1,1,0)

    (1,1,0)·(−1,−1,1) = −1−1+0 = −2

    Distance = 2/√3 = 2√3/3 ≈ 1.15 units

  10. Aircraft on skew paths; find minimum distance between paths. Problem Solving

    L₁: r=(10,0,5)+λ(2,1,−1); L₂: r=(0,5,7)+μ(1,−1,0)

    Verify skew: Set equal. z: 5−λ=7 ⇒ λ=−2; y: −2=5−μ ⇒ μ=7; x: 10+2(−2)=6; 0+7=7. 6≠7. Skew. ✓

    n = (2,1,−1)×(1,−1,0):

    i: (1)(0)−(−1)(−1)=0−1=−1

    j: −[(2)(0)−(−1)(1)]=−[0+1]=−1

    k: (2)(−1)−(1)(1)=−2−1=−3

    n = (−1,−1,−3); |n|=√(1+1+9)=√11

    b−a = (0,5,7)−(10,0,5) = (−10,5,2)

    (−10,5,2)·(−1,−1,−3) = 10−5−6 = −1

    Minimum distance = |−1|/√11 = 1/√11 = √11/11 ≈ 0.30 km