Porisma
Back to reading

Vectors: planes, normals and distances · Find perpendicular feet and distances

Find the foot, then measure the distance

Move along the plane normal to find the foot. Check both its position and direction before taking a length; the line parameter is not itself a distance.

  1. Meet the plane along its normal

    Use a genuine plane with nonzero normal n. Start at P and move along n. Does P already satisfy the plane equation?

    r⋅n=d.\mathbf r\cdot\mathbf n=d.

    r=p+λn.\mathbf r=\mathbf p+\lambda\mathbf n.

    Yes · the point is on the planeThe foot is the original point

    When p⋅n=d\mathbf p\cdot\mathbf n=d, take λ=0\lambda=0 and f=p\mathbf f=\mathbf p. The point-to-plane distance is zero. This boundary case needs no nonzero displacement.

    No · the point is off the planeIntersect the normal line with the plane

    Impose (p+λn)⋅n=d(\mathbf p+\lambda\mathbf n)\cdot\mathbf n=d. Since n≠0\mathbf n\ne\mathbf0, the denominator is positive:

    λ=d−p⋅nn⋅n.\lambda=\frac{d-\mathbf p\cdot\mathbf n}{\mathbf n\cdot\mathbf n}.

    Substitute this parameter into f=p+λn\mathbf f=\mathbf p+\lambda\mathbf n to find the foot F.

    A negative parameter is valid: it means moving opposite to the chosen normal. An in-plane direction cannot supply this perpendicular construction.

  2. Check the foot before reporting a distance

    Both the plane-membership and normal-displacement checks must hold. If either fails, revisit the calculation before taking a length.

    h=∣f−p∣.h=|\mathbf f-\mathbf p|.

    1 · MembershipSubstitute the foot into the plane

    Verify f⋅n=d\mathbf f\cdot\mathbf n=d. A point somewhere in the plane is not enough unless it also satisfies the direction check.

    2 · PerpendicularityCheck displacement along the normal

    Verify f−p=λn\mathbf f-\mathbf p=\lambda\mathbf n. For P off the plane, this nonzero displacement is perpendicular to every in-plane direction. For P on the plane, F=P already settles the zero-distance case.

    3 · LengthTake a nonnegative magnitude

    The two distance calculations agree:

    h=∣λ∣ ∣n∣=∣d−p⋅n∣∣n∣.h=|\lambda|\,|\mathbf n|=\frac{|d-\mathbf p\cdot\mathbf n|}{|\mathbf n|}.

    Use the magnitude of the displacement, not the signed parameter alone. Multiplying both n and d by the same nonzero constant leaves F and h unchanged.

    The entire normal line intersects the plane, so its line-to-plane distance is zero. The original point’s distance can still be positive: identify which object the question asks about.

Try a worked example

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

Find the foot, then extend to a distance

Use the existing foot question: A=(1,3,−1)A=(1,3,-1) and r⋅(0,−1,2)=5\mathbf r\cdot(0,-1,2)=5. Predict the normal-line parameter and foot N. After the original foot question, extend the calculation to A’s distance from the plane.

Show reasoning

  1. 1. Intersect the normal line

    The normal is nonzero. Substitute the line through A along this normal into the plane equation.

    r=(1,3,−1)+λ(0,−1,2),−5+5λ=5,λ=2.\begin{aligned}\mathbf r&=(1,3,-1)+\lambda(0,-1,2),\\-5+5\lambda&=5,\qquad\lambda=2.\end{aligned}

  2. 2. Find and verify the foot

    The candidate lies on the plane, and its displacement from A is along the normal. Both checks are needed.

    N=(1,1,3),(1,1,3)⋅(0,−1,2)=−1+6=5,AN→=(0,−2,4)=2(0,−1,2).\begin{aligned}N&=(1,1,3),\\(1,1,3)\cdot(0,-1,2)&=-1+6=5,\\\overrightarrow{AN}&=(0,-2,4)=2(0,-1,2).\end{aligned}

  3. 3. Extend to the point-to-plane distance

    The parameter 2 multiplies a normal of length √5. Taking the displacement’s magnitude gives a positive distance.

    h=∣AN→∣=25.h=|\overrightarrow{AN}|=2\sqrt5.
Continue with the lesson question

A negative parameter: which distance?

In the existing distance question, A=(3,1,2)A=(3,1,2) and r⋅(2,−1,1)=4\mathbf r\cdot(2,-1,1)=4. Predict whether the normal-line parameter must be positive. Then distinguish the point’s distance from the entire normal line’s distance to the plane.

Show reasoning

  1. 1. Keep the parameter's sign

    A has dot product 7 with the normal. Reaching the plane requires moving opposite to that normal; a negative parameter is expected.

    λ=4−76=−12.\lambda=\frac{4-7}{6}=-\frac12.

  2. 2. Check the common point

    B is both the perpendicular foot from A and an intersection of the normal line with the plane.

    B=(2,32,32),2(2)−32+32=4,AB→=(−1,12,−12)=−12(2,−1,1).\begin{aligned}B&=(2,\tfrac32,\tfrac32),\\2(2)-\tfrac32+\tfrac32&=4,\\\overrightarrow{AB}&=(-1,\tfrac12,-\tfrac12)=-\tfrac12(2,-1,1).\end{aligned}

  3. 3. Answer for the requested object

    The entire line shares B with the plane, giving line-to-plane distance zero. A itself is off the plane, giving a positive point-to-plane distance.

    hA=∣−12∣6=62,hline=0.h_A=\left|-\frac12\right|\sqrt6=\frac{\sqrt6}{2},\qquad h_{\text{line}}=0.
Continue with the lesson question

Already on the plane?

The existing membership question uses Q=(1,2,3)Q=(1,2,3) and 3x−3y+2z=33x-3y+2z=3. As a boundary-case extension, predict Q’s perpendicular foot and distance.

Show reasoning

  1. 1. Check membership first

    The original question establishes that Q is already in the plane.

    3(1)−3(2)+2(3)=3.3(1)-3(2)+2(3)=3.

  2. 2. Use the zero-distance boundary case

    The normal is nonzero, but no movement is needed. Q is its own foot; do not require a nonzero displacement to accept this case.

    λ=3−322=0,F=Q,h=0.\lambda=\frac{3-3}{22}=0,\qquad F=Q,\qquad h=0.
Continue with the lesson question