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H2 Mathematics · JC 1 and 2

Vectors: Lines and Planes

This topic uses vectors to describe lines and planes in three dimensions. You write equations of lines and planes in vector and cartesian form, find where they meet, and calculate angles, distances and the foot of a perpendicular. The scalar product detects perpendicular directions and the vector product gives a normal to a plane. Many questions are multi-part, with each result feeding the next, so you need a clear method for each standard task.

What the syllabus expects

  • Write the equation of a line in the form r = a + λd and in cartesian form, and determine whether two lines are parallel, intersecting or skew
  • Write the equation of a plane in the forms r = a + λb + μc, r·n = d and ax + by + cz = d
  • Use the scalar product to find the angle between two lines, between a line and a plane, or between two planes
  • Use the vector product to find a vector perpendicular to two given vectors, such as a normal to a plane
  • Find the point of intersection of a line and a plane, and the line of intersection of two planes
  • Find the foot of the perpendicular from a point to a line or to a plane, and the perpendicular distance
  • Find the reflection of a point in a line or a plane using the foot of the perpendicular

Equations of lines

A line through the point with position vector a in the direction d has vector equation r = a + λd, where λ is a real parameter. On this page vectors are written in the form (x, y, z). For example, the line through A(1, 2, 0) and B(3, 1, 3) has direction AB = (2, -1, 3) and equation r = (1, 2, 0) + λ(2, -1, 3). Writing x = 1 + 2λ, y = 2 - λ and z = 3λ, then eliminating λ, gives the cartesian form (x - 1)/2 = (y - 2)/(-1) = z/3. In three dimensions, two lines are parallel if their directions are scalar multiples, intersecting if equating the two position vectors gives a consistent solution for both parameters, and skew if they are neither. To test for intersection, solve two of the three component equations for the two parameters and check them in the third.

Equations of planes

A plane can be described by a point a and two non-parallel directions b and c in it: r = a + λb + μc. A normal vector n is perpendicular to every direction in the plane and can be found as the vector product n = b × c. Every point on the plane satisfies r·n = a·n, which is the scalar product form r·n = d. Writing r = (x, y, z) turns this into the cartesian form ax + by + cz = d. For example, if b = (1, 0, 2) and c = (0, 1, -1), then b × c = (0 × (-1) - 2 × 1, 2 × 0 - 1 × (-1), 1 × 1 - 0 × 0) = (-2, 1, 1). If the plane passes through (1, 1, 1), then d = -2 + 1 + 1 = 0, so the plane is -2x + y + z = 0. Check that n is perpendicular to both b and c: n·b = -2 + 0 + 2 = 0 and n·c = 0 + 1 - 1 = 0.

Angles

The scalar product a·b = |a||b| cos θ gives angles. The acute angle between two lines with directions d₁ and d₂ satisfies cos θ = |d₁·d₂|/(|d₁||d₂|). Between two planes, use their normals in the same way. Between a line and a plane, the angle θ satisfies sin θ = |d·n|/(|d||n|), because the angle between the line and the normal is 90° - θ. For example, the line with direction (2, -1, 2) and the plane x + 2y + 2z = 5, with normal (1, 2, 2), give d·n = 2 - 2 + 4 = 4, |d| = 3 and |n| = 3. So sin θ = 4/9 and θ = 26.4°. A zero scalar product means perpendicular directions, so a line is parallel to a plane, or lies in it, exactly when d·n = 0.

Intersections, feet of perpendiculars and distances

To find where the line r = a + λd meets the plane r·n = p, substitute: (a + λd)·n = p, solve for λ, and put λ back into the equation of the line. The same idea finds the foot N of the perpendicular from a point P to a plane: the line through P with direction n meets the plane at N, and the distance is |PN|. The distance from P to the plane r·n = p can also be found as |OP·n - p|/|n|. For a line r = a + λd, the foot N of the perpendicular from P satisfies PN·d = 0, which gives one equation in λ. The line of intersection of two non-parallel planes has direction n₁ × n₂ and passes through any point satisfying both equations. The reflection P' of P in a plane or a line comes from the midpoint: ON = (OP + OP')/2, so OP' = 2ON - OP.

Worked example

The plane Π has equation x + 2y + 2z = 12 and P is the point (1, 1, 0). Find (a) the foot of the perpendicular from P to Π, (b) the perpendicular distance from P to Π, (c) the reflection of P in Π, and (d) the acute angle between Π and the x-axis.

  1. A normal to Π is n = (1, 2, 2), with |n| = √(1 + 4 + 4) = 3.
  2. The line through P perpendicular to Π is r = (1, 1, 0) + λ(1, 2, 2), so a general point on it is (1 + λ, 1 + 2λ, 2λ).
  3. Substitute into Π: (1 + λ) + 2(1 + 2λ) + 2(2λ) = 12, so 3 + 9λ = 12 and λ = 1. The foot is N(2, 3, 2).
  4. Distance PN = |λ| × |n| = 1 × 3 = 3 units. Check with the formula: |(1, 1, 0)·(1, 2, 2) - 12|/3 = |3 - 12|/3 = 3.
  5. N is the midpoint of P and its reflection P', so OP' = 2ON - OP = (4, 6, 4) - (1, 1, 0) = (3, 5, 4).
  6. The x-axis has direction (1, 0, 0). The angle θ between it and Π satisfies sin θ = |(1, 0, 0)·(1, 2, 2)|/(1 × 3) = 1/3, so θ = 19.5°.

Answer: (a) N(2, 3, 2). (b) 3 units. (c) (3, 5, 4). (d) 19.5°.

Common mistakes

  • Using cos instead of sin for the angle between a line and a plane. The normal makes the angle 90° - θ with the line, so sin θ = |d·n|/(|d||n|).
  • Concluding that two non-parallel lines intersect without checking the third component equation. If the parameters found from two equations fail the third, the lines are skew.
  • Using the same letter for the parameters of two different lines. Use λ for one line and μ for the other, or the equations force a false link between them.
  • Making sign errors in the middle component of a vector product. Check the result by confirming that its scalar product with each original vector is zero.
  • Giving an obtuse angle when the question asks for the angle between two lines or two planes. Use the modulus of the scalar product to get the acute angle.
  • Measuring the distance from a point to a plane to some convenient point on the plane. The distance is the length of the perpendicular.

Exam tips

  • Draw a rough diagram for foot of perpendicular and reflection questions, labelling P, N, the line or plane and the direction vectors. It shows which scalar product should be zero.
  • You may use the graphing calculator to solve the simultaneous equations for a point on two planes, but write down the equations you solved.
  • Keep exact values such as √14 in intermediate steps, and give angles to 0.1° unless the question says otherwise.

Practice questions

1. Find the cartesian equation of the plane through the point (2, -1, 3) with normal (3, 1, -2).

Show answer

3x + y - 2z = -1. Using r·n = a·n, the constant is (2)(3) + (-1)(1) + (3)(-2) = 6 - 1 - 6 = -1.

2. The lines l₁ and l₂ have equations r = (1, 0, 2) + λ(1, 1, 0) and r = (3, 4, 0) + μ(0, 1, -1). Show that they intersect, find the point of intersection and find the acute angle between them.

Show answer

They meet at (3, 2, 2) at an angle of 60°. Equating components: 1 + λ = 3 gives λ = 2, and λ = 4 + μ gives μ = -2. Check the z components: 2 = 0 - μ = 2, which is consistent. Substituting λ = 2 gives (3, 2, 2). For the angle, (1, 1, 0)·(0, 1, -1) = 1 and both directions have length √2, so cos θ = 1/2 and θ = 60°.

3. Find a vector equation of the line of intersection of the planes x + y + z = 6 and x - y + 2z = 5.

Show answer

r = (7, 0, -1) + t(3, -1, -2). The direction is (1, 1, 1) × (1, -1, 2) = (1 × 2 - 1 × (-1), 1 × 1 - 1 × 2, 1 × (-1) - 1 × 1) = (3, -1, -2). For a point, put y = 0: x + z = 6 and x + 2z = 5, so z = -1 and x = 7. Check: 7 + 0 - 1 = 6 and 7 - 0 - 2 = 5.

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