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Vectors: lines and perpendicular feet

Two ways to find a foot on a line

Choose either method for the same foot F. A is a known point on the line, p is P's position vector, and m is a nonzero line direction. The foot must lie on the line and have a perpendicular displacement from P.

  • Put the foot on the line

    Use this route when a line parameter is convenient. Write a candidate foot on the line before imposing perpendicularity.

    r=a+λm\mathbf r=\mathbf a+\lambda\mathbf m

    a=OA→,m≠0\mathbf a=\overrightarrow{OA},\quad \mathbf m\ne\mathbf0

    Parameter routeSolve one scalar equation

    Put f=a+λm\mathbf f=\mathbf a+\lambda\mathbf m. Then

    (f−p)⋅m=0.(\mathbf f-\mathbf p)\cdot\mathbf m=0.

    Hence

    λ=(p−a)⋅mm⋅m.\lambda=\frac{(\mathbf p-\mathbf a)\cdot\mathbf m}{\mathbf m\cdot\mathbf m}.

    Substitute back into f=a+λm\mathbf f=\mathbf a+\lambda\mathbf m to find F.

    Check both line membership and perpendicularity. The parameter can be negative; its sign is not a distance.

  • Project from a known line point

    Use the vector projection of AP onto the line direction. Keep its sign when locating the foot.

    m^=m∣m∣\widehat{\mathbf m}=\dfrac{\mathbf m}{|\mathbf m|}

    AP→=p−a\overrightarrow{AP}=\mathbf p-\mathbf a

    Projection routeAdd the projection vector to OA

    First find

    AF→=(AP→⋅m^)m^.\overrightarrow{AF}=(\overrightarrow{AP}\cdot\widehat{\mathbf m})\widehat{\mathbf m}.

    Then f=a+AF→\mathbf f=\mathbf a+\overrightarrow{AF}. A negative component puts F opposite to the chosen direction from A. Taking an absolute value here would put F on the wrong side.

    Use a unit direction, or divide by m·m as in the parameter formula. If P is on the line, F=P and the distance is zero.

Try a worked example

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

The foot is behind the line point

For P=(1,1,−1)P=(1,1,-1) and r=(1,2,3)+λ(1,2,1)\mathbf r=(1,2,3)+\lambda(1,2,1), predict which side of A=(1,2,3)A=(1,2,3) contains the foot F. Compare the two methods on the same geometry.

Show reasoning

  1. 1. Put F on the line

    Use the same parameter in all three coordinates, then require the displacement from P to F to be perpendicular to the line direction.

    PF→=(λ,1+2λ,4+λ),0=λ+2(1+2λ)+(4+λ)=6λ+6,λ=−1,F=(0,0,2).\begin{aligned}\overrightarrow{PF}&=(\lambda,1+2\lambda,4+\lambda),\\0&=\lambda+2(1+2\lambda)\\&\quad+(4+\lambda)\\&=6\lambda+6,\\\lambda&=-1,\\F&=(0,0,2).\end{aligned}

  2. 2. Check by signed projection

    AP has dot product −6 with m, while m·m=6. Keep the negative sign: AF points opposite to m, so F is behind A. A schematic’s left-to-right order cannot override the coordinates.

    AP→=(0,−1,−4),AF→=−66(1,2,1),OF→=(1,2,3)−(1,2,1)=(0,0,2).\begin{aligned}\overrightarrow{AP}&=(0,-1,-4),\\\overrightarrow{AF}&=\frac{-6}{6}(1,2,1),\\\overrightarrow{OF}&=(1,2,3)-(1,2,1)\\&=(0,0,2).\end{aligned}

  3. 3. Verify both conditions

    F lies on the line at λ=−1 and PF has zero dot product with its nonzero direction. A negative parameter is allowed on an infinite line; it is not a negative distance.

    PF→⋅m=−1−2+3=0.\begin{aligned}\overrightarrow{PF}\cdot\mathbf m&=-1-2+3\\&=0.\end{aligned}
Continue with the lesson question

Illustration: P already lies on the line

Boundary-case illustration: keep r=(1,2,3)+λ(1,2,1)\mathbf r=(1,2,3)+\lambda(1,2,1) but choose P=(3,6,5)P=(3,6,5). Predict the foot and distance. Use this extra comparison to check the zero-distance boundary.

Show reasoning

  1. 1. Check membership

    At λ=2 the line already reaches P. Its foot is P itself.

    (1,2,3)+2(1,2,1)=(3,6,5).\begin{aligned}(1,2,3)&+2(1,2,1)\\&=(3,6,5).\end{aligned}

  2. 2. Accept the zero displacement

    The perpendicular-displacement equation still holds when PF=0. The point-to-line distance is zero; do not require a nonzero displacement to accept this boundary.

    F=P,PF→=0,h=0.\begin{aligned}F&=P,\\\overrightarrow{PF}&=\mathbf0,\\h&=0.\end{aligned}

Compare the geometric constructions

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Two source constructions: F lies on l and PF is perpendicular to its direction m; alternatively AF is the vector projection of AP onto the line direction. A general schematic places A to the left of F.
Two source constructions: F lies on l and PF is perpendicular to its direction m; alternatively AF is the vector projection of AP onto the line direction. A general schematic places A to the left of F.
Two source constructions: F lies on l and PF is perpendicular to its direction m; alternatively AF is the vector projection of AP onto the line direction. A general schematic places A to the left of F.
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