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C. Geometry and Algebra of Vectors · Lines and Planes

Vector and Parametric Equations of a Plane

C4.5 C4.6 C4.3About 70 minutes
Learning goals
  • I can write vector and parametric equations of a plane from a point and two directions, or from three points.
  • I can convert a vector or parametric equation of a plane to a scalar equation using a cross product.
  • I can convert a scalar equation of a plane to vector and parametric form.
  • I can test whether a point lies on a plane given in any form.

Words to know

vector equation of a plane
r = r₀ + sa + tb, s, t ∈ ℝ, where a and b are non-parallel vectors parallel to the plane.
parametric equations of a plane
x = x₀ + sa₁ + tb₁, y = y₀ + sa₂ + tb₂, z = z₀ + sa₃ + tb₃.
coplanar
Lying in the same plane.
direction vectors (of a plane)
Two non-parallel vectors parallel to the plane. Each is perpendicular to the normal.

A plane needs two directions

A line is "a point plus any amount of one direction". A sheet of paper lets you move in two independent directions: along its width and along its height. So a plane is "a point plus any amount of two non-parallel directions".

The scalar equation from the last lesson is great for testing points. The vector and parametric equations of this lesson are great for producing points and for working with lines that lie in a plane. This lesson builds them and shows how to switch between all three forms.

Vector and parametric equations of a plane

Let P₀ be a point on the plane with position vector r₀, and let a and b be two non-parallel vectors parallel to the plane. Then every point of the plane has position vector

r = r₀ + s a + t b, s, t ∈ ℝ.

Writing r₀ = [x₀, y₀, z₀], a = [a₁, a₂, a₃] and b = [b₁, b₂, b₃], the parametric equations are

x = x₀ + sa₁ + tb₁, y = y₀ + sa₂ + tb₂, z = z₀ + sa₃ + tb₃.

There are two parameters because a plane is two-dimensional. If a and b were parallel, all you could reach would be a line.

Through three points A, B, C: use r₀ = OA, a = AB and b = AC.

P₀ (s = 0, t = 0)abs = 2, t = 2r = r₀ + s a + t b
Two non-parallel direction vectors a and b make a grid on the plane. Every point of the plane is P₀ plus some amount s of a and some amount t of b. The marked point is P₀ + 2a + 2b.

Switching forms

Vector or parametric → scalar. A normal is n = a × b. Then substitute P₀ to find D.

Scalar → vector or parametric. Two ways:

  1. Find three points on the plane (intercepts are often easiest), then use the three-point method.
  2. Find a point, and two non-parallel vectors perpendicular to n = [A, B, C]. Handy choices are [B, −A, 0], [C, 0, −A] and [0, C, −B]: each has dot product 0 with n. Use two that are non-zero and not parallel.

Testing a point against a vector equation: either convert to scalar form and substitute, or solve two of the parametric equations for s and t and check the third.

Three points A, B, Cnot on one lineVectorr = r₀ + s a + t bParametricx, y, z in s and tScalarAx + By + Cz + D = 0n = a × bthree points, or twovectors ⟂ n
How the forms of a plane connect. Down: the cross product of the direction vectors is a normal. Up: find three points on the plane (intercepts are easy), or two non-parallel vectors perpendicular to n.
Worked example

Example 1: All three forms from three points

Find vector, parametric and scalar equations of the plane through A(3, 2, 5), B(0, −2, 2) and C(1, 3, 1).

  1. AB = [−3, −4, −3] and AC = [−2, 1, −4]. They are not parallel, so the points are not collinear.
  2. Vector: r = [3, 2, 5] + s[−3, −4, −3] + t[−2, 1, −4].
  3. Parametric: x = 3 − 3s − 2t, y = 2 − 4s + t, z = 5 − 3s − 4t.
  4. Normal: AB × AC = [(−4)(−4) − (−3)(1), (−3)(−2) − (−3)(−4), (−3)(1) − (−4)(−2)] = [19, −6, −11].
  5. Scalar: 19x − 6y − 11z + D = 0. Substitute A: 57 − 12 − 55 + D = 0, so D = 10.
  6. Check B: 0 + 12 − 22 + 10 = 0 ✓. Check C: 19 − 18 − 11 + 10 = 0 ✓.

r = [3, 2, 5] + s[−3, −4, −3] + t[−2, 1, −4]; x = 3 − 3s − 2t, y = 2 − 4s + t, z = 5 − 3s − 4t; 19x − 6y − 11z + 10 = 0.

Worked example

Example 2: Vector to scalar

Write the plane r = [2, 1, 0] + s[1, −1, 3] + t[2, 0, −5] as a scalar equation.

  1. n = [1, −1, 3] × [2, 0, −5] = [(−1)(−5) − (3)(0), (3)(2) − (1)(−5), (1)(0) − (−1)(2)] = [5, 11, 2].
  2. 5x + 11y + 2z + D = 0. Substitute (2, 1, 0): 10 + 11 + D = 0, so D = −21.
  3. Check with s = 1, t = 0, the point (3, 0, 3): 15 + 0 + 6 − 21 = 0 ✓.

5x + 11y + 2z − 21 = 0

Worked example

Example 3: Scalar to vector

Find vector and parametric equations of the plane 2x − y + 3z = 6.

  1. Intercepts: (3, 0, 0), (0, −6, 0) and (0, 0, 2).
  2. Directions from (3, 0, 0): [−3, −6, 0] and [−3, 0, 2]. Divide the first by −3 to get [1, 2, 0]. Change the sign of the second to get [3, 0, −2].
  3. Check against n = [2, −1, 3]: [1, 2, 0] · n = 2 − 2 + 0 = 0 ✓ and [3, 0, −2] · n = 6 + 0 − 6 = 0 ✓.
  4. Vector: r = [3, 0, 0] + s[1, 2, 0] + t[3, 0, −2]. Parametric: x = 3 + s + 3t, y = 2s, z = −2t.

r = [3, 0, 0] + s[1, 2, 0] + t[3, 0, −2] (many correct answers).

Worked example

Example 4: Is the point on the plane?

Decide whether P(1, 4, 0) and Q(5, 0, −2) lie on r = [2, 1, 0] + s[1, −1, 3] + t[2, 0, −5].

  1. Use the scalar form from Example 2, 5x + 11y + 2z − 21 = 0.
  2. P: 5 + 44 + 0 − 21 = 28 ≠ 0, so P is not on the plane.
  3. Q: 25 + 0 − 4 − 21 = 0, so Q is on the plane.
  4. Find s and t for Q: from y, 1 − s = 0 gives s = 1. From x, 2 + 1 + 2t = 5 gives t = 1. Check z: 0 + 3 − 5 = −2 ✓.

P is not on the plane; Q is (s = 1, t = 1).

Just for fun

A line needs one parameter. A plane needs two. If you ever meet an equation with three free parameters in three-space, it is not a plane any more: it has taken over the whole room.

Hands-on activity

Lattice on a tilted sheet

You need: a sheet of paper, a ruler, a pencil, and a clipboard or a hardcover book.

  1. Near the lower left of the sheet, mark a point P₀. From it, draw two arrows that are not parallel: a about 4 cm long to the right, and b about 3 cm long, slanting up and to the right.
  2. Using a ruler, mark the points P₀ + sa + tb for s = 0, 1, 2, 3 and t = 0, 1, 2. (Copy the arrows tip to tail.) Join them to make a slanted grid.
  3. Put the sheet on a clipboard and tilt it at any angle. Every grid point stays on the sheet: the sheet is the plane, and s and t are its two parameters.
  4. Mark a random point X on the sheet between grid lines. Estimate its s and t (for example, s ≈ 1.5, t ≈ 0.7). Check by measuring.
  5. Now draw a third arrow c = a + b. Does it lie on the sheet? (Yes: sums of vectors in a plane stay in the plane.)

What to look for: every point of the sheet can be reached with some s and t, including fractions and negatives.

Your turn

1

Write vector and parametric equations of the plane through (1, 2, 3) with direction vectors [1, 0, 2] and [0, 1, −1].

2

Write a vector equation of the plane through A(1, 0, 0), B(0, 2, 0) and C(0, 0, 3).

3

Find a scalar equation of the plane in the previous question.

4

Convert r = [0, 0, 1] + s[1, 1, 0] + t[0, 1, 1] to a scalar equation.

5

Write a vector equation of the plane x + 2y − z = 4.

6

Do the point (0, 0, 0) and directions [1, 2, 3] and [2, 4, 6] give a vector equation of a plane?

7

Find the point on r = [1, 1, 1] + s[2, 0, 1] + t[0, 3, −1] with s = 2 and t = −1.

More practice: warm-up

1

How many direction vectors does a vector equation of a plane use?

2

Which point on r = [3, −1, 2] + s[1, 1, 1] + t[1, 0, −1] has s = t = 0?

3

What plane is r = s[1, 0, 0] + t[0, 1, 0]?

4

Find a normal from the directions [1, 0, 0] and [0, 1, 0].

5

Can [2, −1, 0] be a direction vector of the plane x + 2y + 5z = 1?

More practice: on your own

1

Find vector and scalar equations of the plane through P(2, −1, 1) parallel to [1, 3, −2] and [0, 1, 1].

2

Find vector and scalar equations of the plane that contains the line r = [1, 2, 0] + t[1, −1, 1] and is parallel to the line r = [0, 0, 5] + u[2, 1, 0].

3

Is (4, 3, 2) on r = [1, 1, 1] + s[1, 1, 0] + t[1, 0, 1]?

4

Write parametric equations of the plane 3x − z = 6.

5

Find a scalar equation of the plane through (0, 0, 0), (1, 2, 3) and (2, −1, 1).

Challenge

1

Show that r = [1, 0, 0] + s[1, 1, 0] + t[0, 1, 1] and r = [2, 2, 1] + u[1, 0, −1] + v[1, 2, 1] describe the same plane.

2

The lines r = [1, 2, 3] + t[1, 0, 1] and r = [1, 2, 3] + u[2, 1, 0] meet at (1, 2, 3). Find a scalar equation of the plane containing both.

3

Are A(1, 0, 1), B(2, 1, 3), C(0, 1, 0) and D(3, 3, 6) coplanar?

Check your understanding

Quiz

1.

Which is a vector equation of a plane?

2.

Which vector is a normal to r = [2, 1, 0] + s[1, −1, 3] + t[2, 0, −5]?

3.

What is a scalar equation of r = [0, 0, 1] + s[1, 0, 0] + t[0, 1, 0]?

4.

Which pair could be the two direction vectors of the plane 2x − y + 3z = 6?

5.

How many parameters do the parametric equations of a plane contain?

6.

Which point is on r = [1, 0, 2] + s[1, 1, 0] + t[0, 2, 1] when s = 1 and t = 2?

For parents and teachers

Big idea (C4.5, C4.6): a plane is "a point plus any combination of two non-parallel directions": r = r₀ + sa + tb. The cross product a × b turns this into the scalar form; intercepts or vectors perpendicular to the normal turn the scalar form back.

How to help: Ask your teen to write the same plane in all three forms and check that a point produced by the vector form (pick s and t) satisfies the scalar form. That single check catches almost every arithmetic slip.

Common difficulties:

  • Using two parallel direction vectors, or a direction vector and the normal. Both directions must be perpendicular to n and not multiples of each other.
  • Using the two points themselves (instead of AB and AC) as directions.
  • Cross product arithmetic. Encourage a written layout and a dot-product check: n · a = 0 and n · b = 0.
  • Expecting a unique answer: vector and parametric equations of a plane are never unique.