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Topic 9

Tangent Planes & Normal Lines

Every tangent plane needs a point and a n\mathbf n. Then the plane is n⋅⟨x−x0,y−y0,z−z0⟩=0\mathbf n \cdot \langle x - x_0, y - y_0, z - z_0\rangle = 0. For a F(x,y,z)=cF(x,y,z) = c, the normal is the ∇F\nabla F. For a graph z=f(x,y)z = f(x,y), it is ⟨fx,fy,−1⟩\langle f_x, f_y, -1\rangle.

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

1

Find the tangent plane and normal line to a graph z = f(x, y)

e.g. “Find the tangent plane to the surface z=f(x,y)z = f(x,y) at the point …”
  1. 1
    Find the point: z0=f(x0,y0)z_0 = f(x_0, y_0)

    If only (x0,y0)(x_0, y_0) is given, this height is the missing coordinate. If the full point is given, check it.

  2. 2
    Compute fxf_x and fyf_y at (x0,y0)(x_0, y_0)

    These two numbers are the slopes of the plane in the xx and yy directions.

  3. 3
    Write z−z0=fx(x−x0)+fy(y−y0)z - z_0 = f_x(x - x_0) + f_y(y - y_0)

    Expand to z=ax+by+cz = ax + by + c only if the question asks for that form.

  4. 4
    Normal line: ⟨x0,y0,z0⟩+t⟨fx,fy,−1⟩\langle x_0, y_0, z_0\rangle + t\langle f_x, f_y, -1\rangle

    ⟨fx,fy,−1⟩\langle f_x, f_y, -1\rangle is ∇F\nabla F for F=f(x,y)−zF = f(x,y) - z, so it is normal to the plane.

Remember: Height, two slopes, plane.

See it in a problem3 worked examples

Find the tangent plane and the normal line to the graph of f(x,y)=x2y−3y2+2xf(x,y) = x^2y - 3y^2 + 2x at (x,y)=(1,2)(x,y) = (1, 2).

  1. 1
    Find the point: z0=f(x0,y0)z_0 = f(x_0, y_0)

    z0=f(1,2)=2−12+2=−8z_0 = f(1, 2) = 2 - 12 + 2 = -8, so the point is (1,2,−8)(1, 2, -8).

  2. 2
    Compute fxf_x and fyf_y at (x0,y0)(x_0, y_0)

    fx=2xy+2⇒fx(1,2)=6f_x = 2xy + 2 \Rightarrow f_x(1,2) = 6.  fy=x2−6y⇒fy(1,2)=−11\ f_y = x^2 - 6y \Rightarrow f_y(1,2) = -11.

  3. 3
    Write z−z0=fx(x−x0)+fy(y−y0)z - z_0 = f_x(x - x_0) + f_y(y - y_0)

    z+8=6(x−1)−11(y−2)z + 8 = 6(x - 1) - 11(y - 2), that is z=6x−11y+8z = 6x - 11y + 8.

  4. 4
    Normal line: ⟨x0,y0,z0⟩+t⟨fx,fy,−1⟩\langle x_0, y_0, z_0\rangle + t\langle f_x, f_y, -1\rangle

    ⟨x,y,z⟩=⟨1,2,−8⟩+t⟨6,−11,−1⟩\langle x, y, z\rangle = \langle 1, 2, -8\rangle + t\langle 6, -11, -1\rangle.

Answer:

Plane z=6x−11y+8z = 6x - 11y + 8; normal line ⟨1,2,−8⟩+t⟨6,−11,−1⟩\langle 1, 2, -8\rangle + t\langle 6, -11, -1\rangle

Common mistakes3
  • Dropping the −1-1: the normal to z=f(x,y)z = f(x,y) is ⟨fx,fy,−1⟩\langle f_x, f_y, -1\rangle, not ⟨fx,fy⟩\langle f_x, f_y\rangle.
  • Leaving xx and yy in the slopes. fx(x0,y0)f_x(x_0, y_0) and fy(x0,y0)f_y(x_0, y_0) are numbers.
  • Using the wrong height: z0=f(x0,y0)z_0 = f(x_0, y_0), not 00.
2

Find the tangent plane to a level surface F(x, y, z) = c

e.g. “Find the tangent plane to the surface (an equation in xx, yy and zz) at the point …”
  1. 1
    Write the surface as F(x,y,z)=cF(x,y,z) = c

    Put every term with a variable on one side. Check that the point satisfies the equation.

  2. 2
    Compute ∇F=⟨Fx,Fy,Fz⟩\nabla F = \langle F_x, F_y, F_z\rangle
  3. 3
    Evaluate at the point: that is the normal n\mathbf n

    Divide out a common factor to keep the numbers small.

  4. 4
    Write n⋅⟨x−x0, y−y0, z−z0⟩=0\mathbf n \cdot \langle x - x_0,\ y - y_0,\ z - z_0\rangle = 0

    Normal line, if asked: ⟨x0,y0,z0⟩+t n\langle x_0, y_0, z_0\rangle + t\,\mathbf n. For the form z=ax+by+cz = ax + by + c, solve for zz.

Remember: The gradient is the normal.

See it in a problem3 worked examples

Find the tangent plane and the normal line to the ellipsoid x2+4y2+z2=18x^2 + 4y^2 + z^2 = 18 at (1,−2,1)(1, -2, 1).

  1. 1
    Write the surface as F(x,y,z)=cF(x,y,z) = c

    F=x2+4y2+z2=18F = x^2 + 4y^2 + z^2 = 18. Check: 1+16+1=181 + 16 + 1 = 18.

  2. 2
    Compute ∇F=⟨Fx,Fy,Fz⟩\nabla F = \langle F_x, F_y, F_z\rangle

    ∇F=⟨2x,8y,2z⟩\nabla F = \langle 2x, 8y, 2z\rangle.

  3. 3
    Evaluate at the point: that is the normal n\mathbf n

    ∇F(1,−2,1)=⟨2,−16,2⟩\nabla F(1, -2, 1) = \langle 2, -16, 2\rangle; use n=⟨1,−8,1⟩\mathbf n = \langle 1, -8, 1\rangle.

  4. 4
    Write n⋅⟨x−x0, y−y0, z−z0⟩=0\mathbf n \cdot \langle x - x_0,\ y - y_0,\ z - z_0\rangle = 0

    (x−1)−8(y+2)+(z−1)=0(x - 1) - 8(y + 2) + (z - 1) = 0, that is x−8y+z=18x - 8y + z = 18. Normal line: ⟨1,−2,1⟩+t⟨1,−8,1⟩\langle 1, -2, 1\rangle + t\langle 1, -8, 1\rangle.

Answer:

Plane x−8y+z=18x - 8y + z = 18; normal line ⟨1,−2,1⟩+t⟨1,−8,1⟩\langle 1, -2, 1\rangle + t\langle 1, -8, 1\rangle

Common mistakes3
  • Treating zz as a function of xx and yy here. In FF, all three variables are independent: ∂∂x(x2z)=2xz\dfrac{\partial}{\partial x}(x^2z) = 2xz.
  • Using ∇F\nabla F before plugging in the point. The normal is a vector of numbers.
  • Sign slips in the plane: the point (1,−2,1)(1, -2, 1) gives (x−1)(x - 1), (y+2)(y + 2), (z−1)(z - 1).
3

Find the points where the tangent plane is parallel to a given plane

e.g. “Find all points on the surface … where the tangent plane is parallel to the plane …”
  1. 1
    Write the surface's normal at a general point

    Level surface: ∇F(x,y,z)\nabla F(x,y,z). Graph z=f(x,y)z = f(x,y): ⟨fx,fy,−1⟩\langle f_x, f_y, -1\rangle.

  2. 2
    Read off the target normal from the plane

    ax+by+cz=dax + by + cz = d has normal ⟨a,b,c⟩\langle a, b, c\rangle. A horizontal plane has normal ⟨0,0,1⟩\langle 0, 0, 1\rangle.

  3. 3
    Set the surface normal equal to λ\lambda times the target

    One equation per component. For a graph the zz-components fix λ\lambda at once; horizontal means fx=fy=0f_x = f_y = 0.

  4. 4
    Solve together with the surface equation
  5. 5
    List every point

    A sphere or ellipsoid always gives two opposite points.

Remember: Normals parallel, then stay on the surface.

See it in a problem3 worked examples

Find all points on the ellipsoid 2x2+y2+3z2=212x^2 + y^2 + 3z^2 = 21 where the tangent plane is parallel to the plane 4x+y−6z=04x + y - 6z = 0.

  1. 1
    Write the surface's normal at a general point

    ∇F=⟨4x,2y,6z⟩\nabla F = \langle 4x, 2y, 6z\rangle.

  2. 2
    Read off the target normal from the plane

    Target normal: ⟨4,1,−6⟩\langle 4, 1, -6\rangle.

  3. 3
    Set the surface normal equal to λ\lambda times the target

    4x=4λ4x = 4\lambda, 2y=λ2y = \lambda, 6z=−6λ6z = -6\lambda, so x=λx = \lambda, y=λ2y = \tfrac{\lambda}{2}, z=−λz = -\lambda.

  4. 4
    Solve together with the surface equation

    2λ2+λ24+3λ2=21⇒214λ2=21⇒λ=±22\lambda^2 + \tfrac{\lambda^2}{4} + 3\lambda^2 = 21 \Rightarrow \tfrac{21}{4}\lambda^2 = 21 \Rightarrow \lambda = \pm 2.

  5. 5
    List every point

    λ=2\lambda = 2: (2,1,−2)(2, 1, -2).  λ=−2\ \lambda = -2: (−2,−1,2)(-2, -1, 2).

Answer:

(2,1,−2)(2, 1, -2) and (−2,−1,2)(-2, -1, 2)

Common mistakes3
  • Setting the normals equal. ⟨4,2,−6⟩\langle 4, 2, -6\rangle and ⟨2,1,−3⟩\langle 2, 1, -3\rangle are parallel but not equal: you need the λ\lambda.
  • Skipping step 4: the point must lie on the surface, and that is what pins down its coordinates.
  • Keeping only one root of λ2=4\lambda^2 = 4: both λ=2\lambda = 2 and λ=−2\lambda = -2 give points.
4

Find the tangent line to the curve of intersection of two surfaces

e.g. “Find a vector tangent to the curve of intersection of the surfaces … and … at …”
  1. 1
    Check the point lies on both surfaces
  2. 2
    Find both normals at the point: ∇F\nabla F and ∇G\nabla G

    Write each surface as F=cF = c first: a graph z=f(x,y)z = f(x,y) becomes f(x,y)−z=0f(x,y) - z = 0.

  3. 3
    Cross them: d=∇F×∇G\mathbf d = \nabla F \times \nabla G

    The curve lies in both tangent planes, so its direction is perpendicular to both normals.

  4. 4
    Write the line, or rescale d\mathbf d as asked

    Line: ⟨x0,y0,z0⟩+t d\langle x_0, y_0, z_0\rangle + t\,\mathbf d. A vector of length LL: ±L d∣d∣\pm L\,\dfrac{\mathbf d}{|\mathbf d|}.

Remember: Two normals, one cross product.

See it in a problem3 worked examples

Find the tangent line to the curve of intersection of the cylinder x2+y2=25x^2 + y^2 = 25 and the plane x+y+z=10x + y + z = 10 at (3,4,3)(3, 4, 3).

  1. 1
    Check the point lies on both surfaces

    9+16=259 + 16 = 25 and 3+4+3=103 + 4 + 3 = 10.

  2. 2
    Find both normals at the point: ∇F\nabla F and ∇G\nabla G

    ∇F=⟨2x,2y,0⟩=⟨6,8,0⟩\nabla F = \langle 2x, 2y, 0\rangle = \langle 6, 8, 0\rangle, use ⟨3,4,0⟩\langle 3, 4, 0\rangle.  ∇G=⟨1,1,1⟩\ \nabla G = \langle 1, 1, 1\rangle.

  3. 3
    Cross them: d=∇F×∇G\mathbf d = \nabla F \times \nabla G

    ⟨3,4,0⟩×⟨1,1,1⟩=⟨4−0, 0−3, 3−4⟩=⟨4,−3,−1⟩\langle 3, 4, 0\rangle \times \langle 1, 1, 1\rangle = \langle 4 - 0,\ 0 - 3,\ 3 - 4\rangle = \langle 4, -3, -1\rangle.

  4. 4
    Write the line, or rescale d\mathbf d as asked

    ⟨x,y,z⟩=⟨3,4,3⟩+t⟨4,−3,−1⟩\langle x, y, z\rangle = \langle 3, 4, 3\rangle + t\langle 4, -3, -1\rangle.

Answer:

⟨x,y,z⟩=⟨3,4,3⟩+t⟨4,−3,−1⟩\langle x, y, z\rangle = \langle 3, 4, 3\rangle + t\langle 4, -3, -1\rangle

Common mistakes3
  • Adding the two normals, or using just one. Only the cross product is perpendicular to both.
  • Cross product slips. ⟨a1,a2,a3⟩×⟨b1,b2,b3⟩=⟨a2b3−a3b2, a3b1−a1b3, a1b2−a2b1⟩\langle a_1, a_2, a_3\rangle \times \langle b_1, b_2, b_3\rangle = \langle a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3,\ a_1b_2 - a_2b_1\rangle; check that d⋅∇F=0\mathbf d \cdot \nabla F = 0.
  • "A vector of length LL" has two answers, ±\pm.

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