Practice Maths

Solutions — Planes and Intersections

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  1. Q1 — Cartesian equation from point and normal

    (a) Point (2, 1, −3), normal n = (1, 2, −1)

    d = n · a = (1)(2) + (2)(1) + (−1)(−3) = 2 + 2 + 3 = 7

    Plane equation: x + 2y − z = 7

    Verify: substitute (2,1,−3): 2 + 2 − (−3) = 7 ✓

    (b) Point (0, 0, 4), normal n = (3, −2, 1)

    d = (3)(0) + (−2)(0) + (1)(4) = 4

    Plane equation: 3x − 2y + z = 4

    Verify: substitute (0,0,4): 0 + 0 + 4 = 4 ✓

  2. Q2 — Plane through three given points

    Points: A = (1, 0, 0), B = (0, 2, 0), C = (0, 0, 3)

    Two vectors in the plane:

    AB = B − A = (−1, 2, 0)

    AC = C − A = (−1, 0, 3)

    Normal n = AB × AC:

    i component: (2)(3) − (0)(0) = 6

    j component: −[(−1)(3) − (0)(−1)] = −(−3) = 3

    k component: (−1)(0) − (2)(−1) = 0 + 2 = 2

    n = (6, 3, 2)

    d = n · A = (6)(1) + (3)(0) + (2)(0) = 6

    Plane equation: 6x + 3y + 2z = 6 (equivalently: x/1 + y/2 + z/3 = 1)

    Verify B: 6(0) + 3(2) + 2(0) = 6 ✓;   C: 6(0) + 3(0) + 2(3) = 6 ✓

  3. Q3 — Distance from a point to a plane

    (a) P = (1, 2, 3), plane 2x − y + 2z = 5

    Formula: dist = |ax₀ + by₀ + cz₀ − d| / √(a² + b² + c²)

    = |2(1) − (2) + 2(3) − 5| / √(4 + 1 + 4)

    = |2 − 2 + 6 − 5| / √9

    = |1| / 3 = 1/3 units

    (b) P = (0, 0, 0), plane 3x + 4y − 12z = 26

    = |3(0) + 4(0) − 12(0) − 26| / √(9 + 16 + 144)

    = |−26| / √169

    = 26 / 13 = 2 units

  4. Q4 — Line-plane intersection point

    Line: r = (1, 0, 2) + λ(2, 1, −1)  ⇒  x = 1+2λ, y = λ, z = 2−λ

    Plane: x + 2y − z = 3

    Substitute into the plane equation:

    (1 + 2λ) + 2(λ) − (2 − λ) = 3

    1 + 2λ + 2λ − 2 + λ = 3

    5λ − 1 = 3

    5λ = 4  ⇒  λ = 4/5

    Substitute back:

    x = 1 + 2(4/5) = 1 + 8/5 = 13/5

    y = 4/5

    z = 2 − 4/5 = 6/5

    Intersection point: (13/5, 4/5, 6/5)

    Verify: 13/5 + 8/5 − 6/5 = 15/5 = 3 ✓

  5. Q5 — Line of intersection of two planes

    Planes: Π1: 2x + y − z = 4   and   Π2: x − y + 2z = 1

    Step 1: Direction vector

    n1 = (2, 1, −1),   n2 = (1, −1, 2)

    d = n1 × n2:

    i: (1)(2) − (−1)(−1) = 2 − 1 = 1

    j: −[(2)(2) − (−1)(1)] = −[4 + 1] = −5

    k: (2)(−1) − (1)(1) = −2 − 1 = −3

    d = (1, −5, −3)

    Step 2: Find a point on both planes

    Set z = 0:   2x + y = 4   and   x − y = 1

    Adding these equations: 3x = 5  ⇒  x = 5/3

    From x − y = 1: y = 5/3 − 1 = 2/3

    Point on line: (5/3, 2/3, 0)

    Line of intersection: r = (5/3, 2/3, 0) + λ(1, −5, −3)

  6. Q6 — Angle between two planes

    Planes: x + y + z = 6   and   2x − y + z = 3

    Normal vectors: n1 = (1, 1, 1)   and   n2 = (2, −1, 1)

    n1 · n2 = (1)(2) + (1)(−1) + (1)(1) = 2 − 1 + 1 = 2

    |n1| = √(1 + 1 + 1) = √3

    |n2| = √(4 + 1 + 1) = √6

    cosθ = |n1 · n2| / (|n1| |n2|) = 2 / (√3 × √6) = 2 / √18 = 2 / (3√2) = √2/3

    θ = cos−1(√2/3) ≈ 61.9°

  7. Q7 — Plane containing a point and a line

    Point P = (2, 1, 3); Line: r = (0, 1, 0) + λ(1, 0, 2)

    Step 1: Direction of the line: d = (1, 0, 2)

    Step 2: Vector from line point to P: v = (2, 1, 3) − (0, 1, 0) = (2, 0, 3)

    Both d and v lie in the required plane.

    Step 3: Normal n = d × v = (1, 0, 2) × (2, 0, 3):

    i: (0)(3) − (2)(0) = 0

    j: −[(1)(3) − (2)(2)] = −[3 − 4] = 1

    k: (1)(0) − (0)(2) = 0

    n = (0, 1, 0)

    Step 4: d = n · (0, 1, 0) = 1

    Plane equation: y = 1

    Verify: P=(2,1,3): y=1 ✓; Line point (0,1,0): y=1 ✓; direction (1,0,2) · (0,1,0)=0 (parallel to plane) ✓

  8. Q8 — Finding an unknown parameter on a plane

    Point (1, k, 2) lies on plane 3x − 2y + 4z = 7

    Substitute the point coordinates into the plane equation:

    3(1) − 2(k) + 4(2) = 7

    3 − 2k + 8 = 7

    11 − 2k = 7

    2k = 4

    k = 2

    Verify: 3(1) − 2(2) + 4(2) = 3 − 4 + 8 = 7 ✓

  9. Q9 — Plane through three points, then distance from origin

    Points: A = (2, 1, 0), B = (3, −1, 2), C = (1, 2, −1)

    Step 1: Edge vectors from A:

    AB = (3−2, −1−1, 2−0) = (1, −2, 2)

    AC = (1−2, 2−1, −1−0) = (−1, 1, −1)

    Step 2: Normal n = AB × AC:

    i: (−2)(−1) − (2)(1) = 2 − 2 = 0

    j: −[(1)(−1) − (2)(−1)] = −[−1 + 2] = −1

    k: (1)(1) − (−2)(−1) = 1 − 2 = −1

    n = (0, −1, −1)

    Step 3: d = n · A = (0)(2) + (−1)(1) + (−1)(0) = −1

    Plane: 0x − y − z = −1, i.e. y + z = 1

    Verify all three points:

    A(2,1,0): 1 + 0 = 1 ✓;   B(3,−1,2): −1 + 2 = 1 ✓;   C(1,2,−1): 2 + (−1) = 1 ✓

    Step 4: Distance from origin O = (0, 0, 0) to y + z = 1 (i.e. 0x + y + z = 1):

    dist = |0 + 0 − 1| / √(0 + 1 + 1) = 1/√2 = √2/2 ≈ 0.71 units

  10. Q10 — Perpendicular planes: find k, then line of intersection

    Planes: Π1: x + 2y − z = 3   and   Π2: 2x − y + kz = 5

    Part 1: Find k for perpendicular planes

    Perpendicular planes have perpendicular normals: n1 · n2 = 0

    n1 = (1, 2, −1),   n2 = (2, −1, k)

    (1)(2) + (2)(−1) + (−1)(k) = 0

    2 − 2 − k = 0  ⇒  k = 0

    Part 2: Line of intersection

    With k = 0, planes are: Π1: x + 2y − z = 3   and   Π2: 2x − y = 5

    Direction: d = n1 × n2 = (1, 2, −1) × (2, −1, 0):

    i: (2)(0) − (−1)(−1) = 0 − 1 = −1

    j: −[(1)(0) − (−1)(2)] = −[0 + 2] = −2

    k: (1)(−1) − (2)(2) = −1 − 4 = −5

    d = (−1, −2, −5) (or equivalently (1, 2, 5))

    Find a point: Set z = 0. Then x + 2y = 3 and 2x − y = 5.

    From Π2: y = 2x − 5. Substitute into Π1: x + 2(2x − 5) = 3 ⇒ 5x = 13 ⇒ x = 13/5, y = 1/5

    Line of intersection: r = (13/5, 1/5, 0) + λ(1, 2, 5)