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Vectors: planes, normals and distances · Describe a plane

One plane, three equivalent forms

Convert the same plane between three forms. Compare the methods, then try a worked case.

  1. Parametric → scalar product

    Check that the spanning directions are independent. Their cross product must be nonzero.

    r=a+su+tv\mathbf r=\mathbf a+s\mathbf u+t\mathbf v

    Find the normalPerpendicular to both directions

    n=u×v≠0\mathbf n=\mathbf u\times\mathbf v\ne\mathbf0. If the cross product is zero, these directions do not span a plane.

    Fix its positionKeep the matching constant

    d=a⋅nd=\mathbf a\cdot\mathbf n, so r⋅n=d\mathbf r\cdot\mathbf n=d. Scaling n\mathbf n requires scaling dd by the same nonzero factor.

    The parameters s and t are independent real numbers. A different point or independent spanning pair can describe the same plane.

  2. Scalar product ↔ Cartesian

    Expand a dot product, or read the Cartesian coefficients as a normal.

    r⋅(nx,ny,nz)=d\mathbf r\cdot(n_x,n_y,n_z)=d

    ExpandCartesian equation

    nxx+nyy+nzz=dn_xx+n_yy+n_zz=d. The normal’s components are the coefficients of x, y and z.

    RecogniseScalar-product equation

    With r=(x,y,z)\mathbf r=(x,y,z) and n=(nx,ny,nz)\mathbf n=(n_x,n_y,n_z), collect the terms as r⋅n=d\mathbf r\cdot\mathbf n=d.

    If all three coefficients are zero, the equation describes all space when d=0, or no points when d≠0; it is not a plane.

  3. Cartesian → parametric

    Solve for a coordinate with a nonzero coefficient and let the other two be free parameters.

    nxx+nyy+nzz=d,n≠0n_xx+n_yy+n_zz=d,\quad\mathbf n\ne\mathbf0

    Choose free coordinatesAvoid division by zero

    If the coefficient of z is zero, solve for x or y instead. A nonzero normal guarantees at least one valid choice.

    Verify the whole familySame plane, new parameters

    Check that the base point satisfies the equation and the two independent directions are perpendicular to the normal. Substitute arbitrary parameters to verify the full family.

    Do not expect to recover the original base point and directions. Equivalent parametrizations need not look identical.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Parametric → scalar product

Start with r=(1,2,3)+s(1,2,1)+t(1,3,4)\mathbf r=(1,2,3)+s(1,2,1)+t(1,3,4). Which vector is perpendicular to both directions? Predict a scalar-product equation before opening the working.

Show reasoning

  1. 1. Find a nonzero normal

    Cross the two independent spanning directions. Reversing the order reverses the normal’s sign, which is equally valid.

    n=(1,2,1)×(1,3,4)=(5,−3,1)\mathbf n=(1,2,1)\times(1,3,4)=(5,-3,1)

  2. 2. Use a point on the plane

    The normal alone does not fix the plane’s position. Use the base point a=(1,2,3)\mathbf a=(1,2,3) to find the constant.

    d=a⋅n=5−6+3=2d=\mathbf a\cdot\mathbf n=5-6+3=2

  3. 3. Write scalar-product form

    Every displacement within the plane is perpendicular to the normal.

    r⋅(5,−3,1)=2\mathbf r\cdot(5,-3,1)=2
Continue with the lesson question

Scalar product → Cartesian

The same plane has r⋅(5,−3,1)=2\mathbf r\cdot(5,-3,1)=2. What equation do the coordinates (x,y,z)(x,y,z) satisfy?

Show reasoning

  1. 1. Expand the dot product

    Write r=(x,y,z)\mathbf r=(x,y,z). The coefficients are the components of the normal, not a point on the plane.

    5x−3y+z=25x-3y+z=2

  2. 2. Check the original point

    Substitute the original base point; keep the constant consistent if the normal is rescaled.

    5(1)−3(2)+3=25(1)-3(2)+3=2
Continue with the lesson question

Cartesian → parametric

Return from 5x−3y+z=25x-3y+z=2 to a parametric plane. Will it need exactly the same point and directions as the starting equation?

Show reasoning

  1. 1. Choose two free coordinates

    The coefficient of z is nonzero. Set x=s and y=t, then solve for z. For another plane, solve for a coordinate with a nonzero coefficient.

    x=s,y=t,z=2−5s+3tx=s,\quad y=t,\quad z=2-5s+3t

  2. 2. Collect the two directions

    The point and directions differ from the original representation, but describe the same plane. Both directions are perpendicular to the original normal.

    r=(0,0,2)+s(1,0,−5)+t(0,1,3)\mathbf r=(0,0,2)+s(1,0,-5)+t(0,1,3)

  3. 3. Verify the complete family

    Substitute arbitrary s,t, not just one point. The free parameters cancel.

    5s−3t+(2−5s+3t)=25s-3t+(2-5s+3t)=2
Continue with the lesson question

What if the directions are dependent?

Illustration: r=(1,2,3)+s(1,2,1)+t(2,4,2)\mathbf r=(1,2,3)+s(1,2,1)+t(2,4,2). Does two-parameter notation guarantee a plane?

Show reasoning

  1. 1. Check independence first

    The second direction is twice the first, so their cross product is zero. This family traces a line, not a plane; do not use the zero vector as a plane normal.

    r=(1,2,3)+(s+2t)(1,2,1)\mathbf r=(1,2,3)+(s+2t)(1,2,1)
Source conversion flow: parametric equation leads to scalar-product form using the cross product of independent directions and d=a dot n. Scalar-product and Cartesian equations convert in both directions by expanding or recognizing the dot product.
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