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Vectors: lines and perpendicular feet

Choose a point-to-line distance method

These are alternative routes to the same nonnegative perpendicular distance. Choose a known point A on the infinite line and a nonzero direction m. Reuse a foot or projection length when the question has already supplied it.

  • A verified foot is already available

    If F has already been found, reuse it. Confirm that F lies on the line and PF is perpendicular to its direction.

    f=OF→,p=OP→\mathbf f=\overrightarrow{OF},\quad \mathbf p=\overrightarrow{OP}

    Known footTake the displacement's magnitude

    h=∣f−p∣.h=|\mathbf f-\mathbf p|.

    A line parameter is not this length. When P lies on the line, F=P and h=0.

    Use the companion foot method if F has not been established. A point merely lying on the line is not necessarily the perpendicular foot.

  • A direction and line point are available

    Find AP and normalise the line direction. This route gives the height directly without first finding F.

    m^=m∣m∣,m≠0\widehat{\mathbf m}=\dfrac{\mathbf m}{|\mathbf m|},\quad\mathbf m\ne\mathbf0

    Cross productUse area divided by base
    h=∣AP→×m^∣=∣AP→×m∣∣m∣.\begin{aligned} h&=|\overrightarrow{AP}\times\widehat{\mathbf m}|\\ &=\frac{|\overrightarrow{AP}\times\mathbf m|}{|\mathbf m|}. \end{aligned}

    An unnormalised cross product gives an area, not the perpendicular distance.

    Reversing or rescaling a nonzero direction leaves the distance unchanged after normalisation.

  • A projection length is available

    Resolve AP along the line, then use Pythagoras for the perpendicular component. A signed component and a length are different quantities.

    s=AP→⋅m^s=\overrightarrow{AP}\cdot\widehat{\mathbf m}

    d=∣s∣d=|s|

    Projection and PythagorasSubtract squared lengths

    h=∣AP→∣2−d2.h=\sqrt{|\overrightarrow{AP}|^2-d^2}.

    The signed component s can be negative; d and h are nonnegative. Keep exact values until the final calculation.

    All three routes give distance to the infinite line. A foot outside a specified segment or ray requires a separate restriction check.

Try a worked example

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

Reuse a foot already found

The existing foot calculation gives F=(0,0,2)F=(0,0,2) for P=(1,1,−1)P=(1,1,-1). How can you deduce the shortest distance without another projection?

Show reasoning

  1. 1. Take the perpendicular displacement's magnitude

    The foot has already been verified. The requested distance is a nonnegative length, not the line parameter −1.

    PF→=(−1,−1,3),h=1+1+9=11.\begin{aligned}\overrightarrow{PF}&=(-1,-1,3),\\h&=\sqrt{1+1+9}\\&=\sqrt{11}.\end{aligned}
Continue with the lesson question

Use the cross product with the right scale

Use A=(1,2,3)A=(1,2,3), B=(2,4,4)B=(2,4,4) and P=(1,1,−1)P=(1,1,-1). Predict why ∣AP→×AB→∣|\overrightarrow{AP}\times\overrightarrow{AB}| alone is not the distance to the line AB.

Show reasoning

  1. 1. Find the displacements

    AB is a nonzero direction, but its length is √6 rather than 1.

    AB→=(1,2,1),AP→=(0,−1,−4),∣AB→∣=6.\begin{aligned}\overrightarrow{AB}&=(1,2,1),\\\overrightarrow{AP}&=(0,-1,-4),\\|\overrightarrow{AB}|&=\sqrt6.\end{aligned}

  2. 2. Expand the cross product

    Compute all three signed cross-product components before taking the magnitude. The middle component is negative.

    AP→×AB→=(7,−4,1),∣(7,−4,1)∣=66.\begin{gathered}\overrightarrow{AP}\times\overrightarrow{AB}\\=(7,-4,1),\\|(7,-4,1)|=\sqrt{66}.\end{gathered}

  3. 3. Divide by the direction's length

    The cross product’s magnitude is the parallelogram area. Dividing by the base length gives its perpendicular height.

    h=666=11.h=\frac{\sqrt{66}}{\sqrt6}=\sqrt{11}.
Continue with the lesson question

A negative component gives a positive length

For the same A=(1,2,3)A=(1,2,3), B=(2,4,4)B=(2,4,4) and P=(1,1,−1)P=(1,1,-1), compare signed projection with projection length. Must the perpendicular foot lie between A and B?

Show reasoning

  1. 1. Keep direction and length distinct

    The component is negative, so AF points opposite to AB. Its magnitude gives the projection length.

    s=(0,−1,−4)⋅(1,2,1)6=−6,d=∣s∣=6.\begin{aligned}s&=\frac{(0,-1,-4)\cdot(1,2,1)}{\sqrt6}\\&=-\sqrt6,\\d&=|s|=\sqrt6.\end{aligned}

  2. 2. Use the right triangle

    AP²=17 and d²=6. The remaining perpendicular length agrees with the cross-product result.

    h=17−6=11.h=\sqrt{17-6}=\sqrt{11}.

  3. 3. Check which object the question names

    F=(0,0,2) lies before A along AB, giving the order F,A,B. The source sketch is schematic. Distance to the infinite line permits this foot; distance to a segment would also need its endpoint constraints checked.

Continue with the lesson question

Compare the three geometric routes

Expand diagram

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Three source diagrams show a perpendicular height h: cross product of AP with a unit line direction, magnitude PF when the foot is known, or projection length d followed by Pythagoras. All unit directions are explicitly normalised.
Three source diagrams show a perpendicular height h: cross product of AP with a unit line direction, magnitude PF when the foot is known, or projection length d followed by Pythagoras. All unit directions are explicitly normalised.
Three source diagrams show a perpendicular height h: cross product of AP with a unit line direction, magnitude PF when the foot is known, or projection length d followed by Pythagoras. All unit directions are explicitly normalised.
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