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IBMathsDPApplications & Interpretation (AI)HLRevision Notes3. Geometry & TrigonometryVector Equations of LinesShortest Distance Between a Point and a Line

Shortest Distance Between a Point and a Line (DP IB Applications & Interpretation (AI)): Revision Note

Amber

Written by: Amber

Reviewed by: Dan Finlay

Updated on

DP Applications & Interpretation (AI): HLDP Applications & Interpretation (AI): HLDP Analysis & Approaches (AA): HL

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Shortest Distance Between a Point and a Line

What is the shortest distance from a point to a line?

  • The shortest distance from any point to a line is always the perpendicular distance

    • Let l be a line with equation bold italic r equals bold italic a plus straight lambda bold italic b  

    • Let P be a point that does not lie on the line

  • The shortest distance between the point and the line is sometimes referred to as the length of the perpendicular

Diagram showing the shortest distance, labelled "S", from point P to line L. Point F, on line L, is the foot of the perpendicular.
Example of the shortest distance between a point and a line

How do I find the shortest distance from a point to a line?

  • For example, consider

    • the line l space colon space bold italic r equals open parentheses table row 3 row cell negative 1 end cell row cell negative 2 end cell end table close parentheses plus lambda open parentheses table row 1 row cell negative 2 end cell row 0 end table close parentheses

    • the point P open parentheses 10 comma space 5 comma space minus 10 close parentheses

  • STEP 1
    Sketch a diagram showing the point F on the line l that is closest to the point P

    • The vector stack F P with rightwards arrow on top will be perpendicular to the line l

    • The point F is sometimes called the foot of the perpendicular

  • STEP 2
    Use the equation of the line to find the position vector of the point F  in terms of lambda

    • stack O F with rightwards arrow on top equals open parentheses table row cell 3 plus lambda end cell row cell negative 1 minus 2 lambda end cell row cell negative 2 end cell end table close parentheses

  • STEP 3
    Use this to find the displacement vector stack F P with rightwards arrow on top in terms of lambda

    • stack F P with rightwards arrow on top equals open parentheses table row 10 row 5 row cell negative 10 end cell end table close parentheses minus open parentheses table row cell 3 plus lambda end cell row cell negative 1 minus 2 lambda end cell row cell negative 2 end cell end table close parentheses equals open parentheses table row cell 7 minus lambda end cell row cell 6 plus 2 lambda end cell row cell negative 8 end cell end table close parentheses

  • STEP 4
    Set the scalar product of the direction vector of the line l and the displacement vector stack F P with rightwards arrow on top equal to zero

    • open parentheses table row cell 7 minus lambda end cell row cell 6 plus 2 lambda end cell row cell negative 8 end cell end table close parentheses times open parentheses table row 1 row cell negative 2 end cell row 0 end table close parentheses equals 0 rightwards double arrow negative 5 minus 5 lambda equals 0

  • STEP 5
    Solve the equation to find the value of lambda

    • lambda equals negative 1

  • STEP 6
    Substitute lambda into stack F P with rightwards arrow on top and find the magnitude open vertical bar stack F P with rightwards arrow on top close vertical bar 

    • open vertical bar open parentheses table row cell 7 minus open parentheses negative 1 close parentheses end cell row cell 6 plus 2 open parentheses negative 1 close parentheses end cell row cell negative 8 end cell end table close parentheses close vertical bar equals 12

How can I use the vector product to find the shortest distance from a point to a line?

  • The vector product can be used to find the shortest distance from any point to a line on a 2-dimensional plane

  • The formula for the shortest distance is fraction numerator open vertical bar stack A P with rightwards arrow on top cross times b close vertical bar blank over denominator open vertical bar b close vertical bar end fraction

    • bold italic r equals bold italic a plus lambda bold italic b is an equation of the line

    • P is the point not on the line

    • A is any point on the line

Examiner Tips and Tricks

This formula is not given in the formula booklet.

Worked Example

Point A  has coordinates (1, 2, 0) and the line l has equation bold italic r equals open parentheses table row 2 row 0 row 6 end table close parentheses plus lambda open parentheses table row 0 row 1 row 2 end table close parentheses. 

Point B lies on the l such that open square brackets A B close square brackets  is perpendicular to l.

Find the shortest distance from A to the line l.

3-10-5-ib-aa-hl-short-distance-lines-we-1

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    Author
    Reviewer
    Amber

    Author: Amber

    Expertise: Maths Content Creator

    Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.