Porisma
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Vectors: lines and perpendicular feet

Two checks to classify the lines

Compare nonzero direction vectors first. Parallel directions need a coincidence check; nonparallel directions need all three coordinate equations.

  1. Compare directions

    Start with two nonzero direction vectors. Are they scalar multiples?

    m1=km2(k≠0)\mathbf m_1=k\mathbf m_2\quad (k\ne0)

    Yes · parallel directionsCheck for coincidence

    Test whether a point on one line also lies on the other. A shared point means coincident lines; otherwise they are distinct parallel lines. If the lines are given as distinct, coincidence is excluded.

    No · nonparallel directionsContinue to the coordinate test

    Use independent parameters λ and μ for the two lines. Their directions alone cannot distinguish intersecting from skew lines.

  2. Check every coordinate

    For nonparallel directions, does the same pair (λ, μ) satisfy all three component equations?

    a1+λm1=a2+μm2\mathbf a_1+\lambda\mathbf m_1=\mathbf a_2+\mu\mathbf m_2

    Yes · consistent systemIntersecting lines

    The solution (λ0,μ0)(\lambda_0,\mu_0) gives the common point:

    p=a1+λ0m1\mathbf p=\mathbf a_1+\lambda_0\mathbf m_1

    p=a2+μ0m2\mathbf p=\mathbf a_2+\mu_0\mathbf m_2

    No · inconsistent systemSkew lines

    No point lies on both lines.

    Solve the complete system with G.C. PlySmlt2, or solve two independent equations and check the unused third equation. An arbitrary pair may be dependent.

Try a worked example

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

Case A · Parallel directions

Compare l1:r=(1,2,3)+λ(1,2,1)l_1:\mathbf r=(1,2,3)+\lambda(1,2,1) and l2:r=(1,3,4)+μ(2,4,2)l_2:\mathbf r=(1,3,4)+\mu(2,4,2). Are they distinct parallel lines or the same line?

Show reasoning

  1. 1. Compare nonzero directions

    (2,4,2)=2(1,2,1)(2,4,2)=2(1,2,1), so the directions are parallel. This alone does not rule out coincidence.

  2. 2. Test the base-point displacement

    The displacement (0,1,1) cannot be a scalar multiple of (1,2,1): its first coordinate would force the scalar to zero, contradicting its other coordinates. The lines are distinct parallel lines.

    (1,3,4)−(1,2,3)=(0,1,1)(1,3,4)-(1,2,3)=(0,1,1)
Continue with the lesson question

Case B · Nonparallel directions

Compare l1:r=(1,2,3)+λ(1,2,1)l_1:\mathbf r=(1,2,3)+\lambda(1,2,1) with l2:r=(0,1,5)+μ(1,3,4)l_2:\mathbf r=(0,1,5)+\mu(1,3,4). Predict intersecting or skew. Why must the parameters remain independent?

Show reasoning

  1. 1. Rule out parallel directions

    The two direction vectors are not scalar multiples. Use separate parameters λ and μ, since the same point need not have the same parameter on both lines.

  2. 2. Solve two independent coordinates

    Equate x and z. Their equations give μ=−1 and λ=−2.

    λ−μ=−1,λ−4μ=2\lambda-\mu=-1,\qquad\lambda-4\mu=2

  3. 3. Check the unused coordinate

    The y equation is also satisfied. Both lines give the point (−1,−2,1), so they intersect.

    2(−2)−3(−1)=−12(-2)-3(-1)=-1
Continue with the lesson question

Case C · Check the unused coordinate

Compare l1:r=(1,2,3)+λ(1,2,1)l_1:\mathbf r=(1,2,3)+\lambda(1,2,1) with l2:r=(0,1,2)+μ(1,3,4)l_2:\mathbf r=(0,1,2)+\mu(1,3,4). Its direction is unchanged from Case B. Can the relationship change?

Show reasoning

  1. 1. The directions are still nonparallel

    Directions alone cannot distinguish intersection from skew.

  2. 2. Use x and z to get a candidate

    These two independent equations give μ=0 and λ=−1.

    λ−μ=−1,λ−4μ=−1\lambda-\mu=-1,\qquad\lambda-4\mu=-1

  3. 3. Reject the candidate with y

    The y equation fails. No common point exists and parallel directions were already excluded, so the lines are skew.

    2(−1)−3(0)=−2≠−12(-1)-3(0)=-2\ne-1
Continue with the lesson question

Case D · Different equations, same line?

Extra illustration: compare l1:r=(1,2,3)+λ(1,2,1)l_1:\mathbf r=(1,2,3)+\lambda(1,2,1) with l2:r=(2,4,4)+μ(2,4,2)l_2:\mathbf r=(2,4,4)+\mu(2,4,2). The base point and direction both look different. Are the lines distinct?

Show reasoning

  1. 1. Check the base point

    The second base point lies on the first line at λ=1.

    (2,4,4)=(1,2,3)+(1,2,1)(2,4,4)=(1,2,3)+(1,2,1)

  2. 2. Compare complete parametrizations

    As μ runs through all real values, λ=1+2μ also runs through all real values. The lines are coincident, not distinct parallel lines.

    (2,4,4)+μ(2,4,2)=(1,2,3)+(1+2μ)(1,2,1)(2,4,4)+\mu(2,4,2)=(1,2,3)+(1+2\mu)(1,2,1)
Source two-line direction sketches and decision tree. Test parallel directions first. If nonparallel, solve three coordinate equations in two independent parameters using the calculator or two independent equations plus a third-coordinate check. A consistent solution gives an intersection; inconsistency gives skew lines. Coincident lines are separately qualified.
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