MathematicsGrade 11–13

Lines and planes in space

A point travels along a line through space, two vectors span a plane. Parametric, normal and coordinate forms all describe the same thing.

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What is it about?

A line in space is given by a point AA and a direction. In parametric form: g ⁣:x⃗=a⃗+t⋅u⃗g\colon \vec x = \vec a + t \cdot \vec u, where a⃗\vec a is the position vector of AA and u⃗≠o⃗\vec u \ne \vec o the direction vector. Every value of tt gives exactly one point XX. For a⃗=(1,2,3)T\vec a = (1, 2, 3)^T, u⃗=(2,−1,1)T\vec u = (2, -1, 1)^T and t=2t = 2 you get X(5,0,5)X(5, 0, 5).

  • Trace points are where the line pierces a coordinate plane, e.g. x3=0x_3 = 0 gives t=−3t = -3 and S12(−5,5,0)S_{12}(-5, 5, 0). If, say, u3=0u_3 = 0, the line is parallel to the x1x2x_1x_2 plane or lies in it, so there is no trace point in that plane (or the whole line lies in it).
  • Point test: PP lies on gg if all three rows of p⃗=a⃗+tu⃗\vec p = \vec a + t \vec u give the same tt.
  • A plane needs two non-parallel spanning vectors: E ⁣:x⃗=a⃗+ru⃗+sv⃗E\colon \vec x = \vec a + r \vec u + s \vec v.
  • The normal vector n⃗=u⃗×v⃗\vec n = \vec u \times \vec v is perpendicular to the plane. Normal form n⃗⋅(x⃗−a⃗)=0\vec n \cdot (\vec x - \vec a) = 0, coordinate form n1x1+n2x2+n3x3=dn_1x_1 + n_2x_2 + n_3x_3 = d with d=n⃗⋅a⃗d = \vec n \cdot \vec a.
  • Example: 2x1+3x2+6x3=122x_1 + 3x_2 + 6x_3 = 12 has the intercepts S1(6,0,0)S_1(6, 0, 0), S2(0,4,0)S_2(0, 4, 0), S3(0,0,2)S_3(0, 0, 2) – the trace triangle.

Try it

  1. Drag XX along the line or move the slider tt. Watch the arrow chain a⃗+tu⃗\vec a + t \vec u. Load example
  2. Make u⃗\vec u parallel to the x3x_3 axis. How many trace points are left? Parallel to the x₃ axis
  3. Press “Convert” and follow the four steps to the trace triangle. Convert
  4. Make v⃗\vec v a multiple of u⃗\vec u. u and v parallel

Tasks

Task 1: Point test

Does R(7,−1,6)R(7, -1, 6) lie on g ⁣:x⃗=(1,2,3)T+t⋅(2,−1,1)Tg\colon \vec x = (1, 2, 3)^T + t \cdot (2, -1, 1)^T? Load task

Show solution

1+2t=71 + 2t = 7, 2−t=−12 - t = -1, 3+t=63 + t = 6 all give t=3t = 3, so R∈gR \in g. Show in simulation

Task 2: Coordinate form

Find a coordinate form and the intercepts of E ⁣:x⃗=(1,0,0)T+r⋅(−1,1,0)T+s⋅(−1,0,1)TE\colon \vec x = (1, 0, 0)^T + r \cdot (-1, 1, 0)^T + s \cdot (-1, 0, 1)^T. Load task

Show solution

n⃗=u⃗×v⃗=(1,1,1)T\vec n = \vec u \times \vec v = (1, 1, 1)^T, d=1d = 1: x1+x2+x3=1x_1 + x_2 + x_3 = 1 with S1(1,0,0)S_1(1, 0, 0), S2(0,1,0)S_2(0, 1, 0), S3(0,0,1)S_3(0, 0, 1). Show in simulation

Task 3: Trace triangle

Find the area of the trace triangle of 2x1+3x2+6x3=122x_1 + 3x_2 + 6x_3 = 12.

Show solution

S1S2→×S1S3→=(−6,4,0)T×(−6,0,2)T=(8,12,24)T\overrightarrow{S_1S_2} \times \overrightarrow{S_1S_3} = (-6, 4, 0)^T \times (-6, 0, 2)^T = (8, 12, 24)^T with length 2828, so the area is 1414. Show in simulation

Ideas for the classroom

  • Use: upper secondary analytic geometry (grade 12/13), introducing parametric forms or practising intercepts.
  • Parameter as time: XX travels through space with velocity u⃗\vec u – this explains why tt may be negative.
  • Common stumbling blocks:
    • Checking only two rows in the point test – all three must give the same tt.
    • Losing a sign in the middle component of the cross product; the check n⃗⋅u⃗=0\vec n \cdot \vec u = 0 reveals it.
    • Confusing trace points of a line (with coordinate planes) and of a plane (with the axes).