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Vectors: planes, normals and distances · Compare two planes

Decide how two planes meet

Compare the normals before solving equations. Parallel normals still need a constant check; nonparallel normals give an intersection line.

  1. Compare the nonzero normals

    Both equations must describe genuine planes. Is one normal a nonzero multiple of the other?

    r⋅n1=d1,r⋅n2=d2,n1,n2≠0.\mathbf r\cdot\mathbf n_1=d_1,\quad\mathbf r\cdot\mathbf n_2=d_2,\quad\mathbf n_1,\mathbf n_2\ne\mathbf0.

    Parallel normalsCheck the same scaling on the constants

    Choose k≠0k\ne0 with n1=kn2\mathbf n_1=k\mathbf n_2. Continue to the constant check before deciding whether the planes coincide.

    Nonparallel normalsFind the intersection line

    A direction is m=n1×n2≠0\mathbf m=\mathbf n_1\times\mathbf n_2\ne\mathbf0. Find a point p\mathbf p that satisfies both plane equations, then write r=p+λm\mathbf r=\mathbf p+\lambda\mathbf m. A point in only one plane is insufficient.

    Alternatively solve the two plane equations together with a valid free coordinate. Not every coordinate can be freely chosen in every pair of equations.

  2. For parallel normals, compare the constants

    Use the same nonzero scale factor as in the normal comparison. Does it also match the constants?

    n1=kn2:d1=kd2 ?\mathbf n_1=k\mathbf n_2:\quad d_1=kd_2\ ?

    Matching constantsThe same plane

    Scaling the entire equation gives the same point set. There are infinitely many common points, forming a plane rather than a unique intersection line.

    Different constantsDistinct parallel planes

    Once the normals match, different constants make the equations inconsistent. There is no common point.

    If distinctness is explicitly given, coincidence is excluded. A zero cross product of the normals cannot supply an intersection-line direction.

Try a worked example

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

Different equations, same plane?

Illustration: compare x+y=2x+y=2 and −2x−2y=−4-2x-2y=-4. Predict whether the planes are disjoint or coincident.

Show reasoning

  1. 1. Scale the whole equation

    Divide the second equation by −2, including its constant.

    x+y=2x+y=2

  2. 2. Compare point sets

    Both equations impose the same condition on x and y; z is unrestricted. These are the same plane, even though the normals point in opposite directions.

Same normal direction, different position?

Illustration: compare x+y=2x+y=2 and −2x−2y=−6-2x-2y=-6. Does parallelism of the normals prove coincidence?

Show reasoning

  1. 1. Match the normals and constants

    Divide the entire second equation by −2. The equations then have identical left sides and different constants.

    x+y=2,x+y=3x+y=2,\qquad x+y=3

  2. 2. Check for a common point

    No point can satisfy both conditions. The planes are distinct and parallel. Normalising the normal alone would have missed the different constants.

Find and verify a common line

Illustration using the unit’s pair x+y=2x+y=2 and y+z=3y+z=3: predict a point and direction for their common line.

Show reasoning

  1. 1. Use a valid free coordinate

    Choose y=t, then solve both equations for x and z.

    r=(2,0,3)+t(−1,1,−1),t∈R\mathbf r=(2,0,3)+t(-1,1,-1),\quad t\in\mathbb R

  2. 2. Verify every point in both planes

    The base point satisfies both equations. Substitution cancels the free parameter in both, verifying the complete line.

    (2−t)+t=2,t+(3−t)=3(2-t)+t=2,\qquad t+(3-t)=3

  3. 3. Check the direction

    The direction is perpendicular to both normals. Reversing a line direction gives an equally valid parametrization.

    (1,1,0)×(0,1,1)=(1,−1,1)=−(−1,1,−1)(1,1,0)\times(0,1,1)=(1,-1,1)=-(-1,1,-1)