Topic 3

Equations of Planes

Every plane is a(x−x0)+b(y−y0)+c(z−z0)=0a(x - x_0) + b(y - y_0) + c(z - z_0) = 0. It needs a n=⟨a,b,c⟩\mathbf n = \langle a, b, c\rangle and a point (x0,y0,z0)(x_0, y_0, z_0). In ax+by+cz=dax + by + cz = d, the coefficients are the normal, so you can read it straight off. If you only have two vectors lying in the plane, their is a normal.

4 kinds of question. Open the one you are stuck on.

1

Find the equation of a plane from a point and a normal vector

e.g. “Find the equation of the plane that passes through the point … and has normal vector …”
  1. 1
    Find a normal vector n=⟨a,b,c⟩\mathbf n = \langle a, b, c\rangle

    Parallel to a plane: copy its coefficients. Perpendicular to a line: the line's direction. from AA and BB: B−AB - A.

  2. 2
    Find a point (x0,y0,z0)(x_0, y_0, z_0) on the plane

    Equidistant from AA and BB: their midpoint.

  3. 3
    Write a(x−x0)+b(y−y0)+c(z−z0)=0a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
  4. 4
    Simplify to ax+by+cz=dax + by + cz = d; check the point

Remember: Normal from the coefficients, point for the constant.

See it in a problem4 worked examples

Find the plane through (2,−1,4)(2, -1, 4) with normal vector ⟨3,1,−2⟩\langle 3, 1, -2\rangle.

  1. 1
    Find a normal vector n=⟨a,b,c⟩\mathbf n = \langle a, b, c\rangle

    n=⟨3,1,−2⟩\mathbf n = \langle 3, 1, -2\rangle.

  2. 2
    Find a point (x0,y0,z0)(x_0, y_0, z_0) on the plane

    (2,−1,4)(2, -1, 4).

  3. 3
    Write a(x−x0)+b(y−y0)+c(z−z0)=0a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

    3(x−2)+(y+1)−2(z−4)=03(x - 2) + (y + 1) - 2(z - 4) = 0.

  4. 4
    Simplify to ax+by+cz=dax + by + cz = d; check the point

    3x+y−2z=−33x + y - 2z = -3. Check: 6−1−8=−36 - 1 - 8 = -3.

Answer:

3x+y−2z=−33x + y - 2z = -3

Common mistakes3
  • Signs in a(x−x0)a(x - x_0): the point (2,−1,4)(2, -1, 4) gives 3(x−2)+(y+1)−2(z−4)3(x - 2) + (y + 1) - 2(z - 4).
  • Using a direction in the plane as its normal. The normal is perpendicular to the plane.
  • Calling the normal "the plane". The answer is an equation in xx, yy, zz.
2

Find the equation of a plane through three points

e.g. “Find the plane containing the given three points”
  1. 1
    Find two non-parallel vectors that lie in the plane

    Three points P,Q,RP, Q, R: PQ→\overrightarrow{PQ} and PR→\overrightarrow{PR}. A point and a line: the line's direction, and the vector from a point on the line to the given point. Two lines that meet: their directions. Parallel lines: one direction, and a vector from one line to the other. lie in no plane.

  2. 2
    Cross them: n=u×v\mathbf n = \mathbf u\times\mathbf v

    n=0\mathbf n = \mathbf 0 means the three points are on one line: infinitely many planes contain them.

  3. 3
    Plug in a point: ax+by+cz=n⋅Pax + by + cz = \mathbf n\cdot P
  4. 4
    Check the other points

Remember: Two vectors in the plane, cross, plug in.

See it in a problem3 worked examples

Find the plane containing P=(1,2,0)P = (1, 2, 0), Q=(3,1,1)Q = (3, 1, 1) and R=(0,1,2)R = (0, 1, 2).

  1. 1
    Find two non-parallel vectors that lie in the plane

    PQ→=⟨2,−1,1⟩\overrightarrow{PQ} = \langle 2, -1, 1\rangle, PR→=⟨−1,−1,2⟩\overrightarrow{PR} = \langle -1, -1, 2\rangle.

  2. 2
    Cross them: n=u×v\mathbf n = \mathbf u\times\mathbf v

    PQ→×PR→=⟨−2+1, −1−4, −2−1⟩=⟨−1,−5,−3⟩\overrightarrow{PQ}\times\overrightarrow{PR} = \langle -2 + 1,\ -1 - 4,\ -2 - 1\rangle = \langle -1, -5, -3\rangle. Use n=⟨1,5,3⟩\mathbf n = \langle 1, 5, 3\rangle.

  3. 3
    Plug in a point: ax+by+cz=n⋅Pax + by + cz = \mathbf n\cdot P

    x+5y+3z=1+10+0=11x + 5y + 3z = 1 + 10 + 0 = 11.

  4. 4
    Check the other points

    QQ: 3+5+3=113 + 5 + 3 = 11. RR: 0+5+6=110 + 5 + 6 = 11.

Answer:

x+5y+3z=11x + 5y + 3z = 11

Common mistakes3
  • Crossing the points themselves, like ⟨1,2,0⟩×⟨3,1,1⟩\langle 1, 2, 0\rangle\times\langle 3, 1, 1\rangle, instead of vectors between points.
  • Plane through a point and a line: the line's point alone is not a second vector. Use A−PA - P, from the line to the given point.
  • Skipping the check: one slipped sign in the cross product gives a plane that misses a point.
3

Find the distance from a point to a plane

e.g. “Find the distance from the given point to the given plane”
  1. 1
    Write the plane as ax+by+cz=dax + by + cz = d

    Only given points? Find the plane first (three-points recipe).

  2. 2
    Plug the point in: ax0+by0+cz0−dax_0 + by_0 + cz_0 - d

    Two parallel planes with the same a,b,ca, b, c: just subtract their right-hand sides. A line parallel to the plane: use any point on the line.

  3. 3
    Divide by ∣n∣|\mathbf n| and drop the sign

Remember: Plug in, divide by the normal's length.

See it in a problem3 worked examples

Find the distance from (3,−1,2)(3, -1, 2) to the plane 2x−y+2z=12x - y + 2z = 1.

  1. 1
    Write the plane as ax+by+cz=dax + by + cz = d

    Already in the form: n=⟨2,−1,2⟩\mathbf n = \langle 2, -1, 2\rangle, d=1d = 1.

  2. 2
    Plug the point in: ax0+by0+cz0−dax_0 + by_0 + cz_0 - d

    6+1+4−1=106 + 1 + 4 - 1 = 10.

  3. 3
    Divide by ∣n∣|\mathbf n| and drop the sign

    ∣n∣=3|\mathbf n| = 3, so D=103D = \dfrac{10}{3}.

Answer:

103\dfrac{10}{3}

Common mistakes3
  • Parallel planes with scaled normals: rewrite −2x+4y−4z=10-2x + 4y - 4z = 10 as x−2y+2z=−5x - 2y + 2z = -5 before subtracting.
  • Dividing by a2+b2+c2a^2 + b^2 + c^2 instead of its square root.
  • "All planes at distance 22": there are two, one on each side.
4

Find where a line intersects a plane

e.g. “Find the point where the line … meets the plane …”
  1. 1
    Write the line as x(t)x(t), y(t)y(t), z(t)z(t)

    For the point on a plane closest to DD, use the line D+t nD + t\,\mathbf n, perpendicular to the plane.

  2. 2
    Substitute into the plane's equation
  3. 3
    Solve for tt

    No solution: the line is parallel to the plane. True for every tt: the line lies in it.

  4. 4
    Plug tt back into the line

Remember: Substitute, solve for t, plug back.

See it in a problem3 worked examples

Where does the line ⟨1,2,−1⟩+t⟨2,−1,3⟩\langle 1, 2, -1\rangle + t\langle 2, -1, 3\rangle meet the plane x+y+z=8x + y + z = 8?

  1. 1
    Write the line as x(t)x(t), y(t)y(t), z(t)z(t)

    x=1+2tx = 1 + 2t, y=2−ty = 2 - t, z=−1+3tz = -1 + 3t.

  2. 2
    Substitute into the plane's equation

    (1+2t)+(2−t)+(−1+3t)=8(1 + 2t) + (2 - t) + (-1 + 3t) = 8.

  3. 3
    Solve for tt

    2+4t=82 + 4t = 8, so t=32t = \tfrac{3}{2}.

  4. 4
    Plug tt back into the line

    (1+3, 2−32, −1+92)=(4,12,72)\left(1 + 3,\ 2 - \tfrac{3}{2},\ -1 + \tfrac{9}{2}\right) = \left(4, \tfrac{1}{2}, \tfrac{7}{2}\right).

Answer:

(4,12,72)\left(4, \tfrac{1}{2}, \tfrac{7}{2}\right)

Common mistakes3
  • Stopping at tt: the question asks for a point.
  • A shadow from a light at OO: the line runs from OO through the point, so its direction is the point itself.
  • Thinking "2=52 = 5" must be a slip. If your algebra is right, it means the line never meets the plane.

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