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IBMathsDPAnalysis & Approaches (AA)HLRevision Notes3. Geometry & TrigonometryVector Equations of LinesShortest Distance Between Two Lines

Shortest Distance Between Two Lines (DP IB Analysis & Approaches (AA)): Revision Note

Amber

Written by: Amber

Reviewed by: Dan Finlay

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DP Analysis & Approaches (AA): HLDP Analysis & Approaches (AA): HL

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Shortest Distance Between Two Lines

What is the shortest distance between two skew lines?

  • The shortest distance between two lines is always the perpendicular to both lines

    • Let l subscript 1 be a line with equation bold italic r equals bold italic a subscript 1 plus lambda bold italic b subscript 1  

    • Let l subscript 2 be a line with equation bold italic r equals bold italic a subscript 2 plus mu bold italic b subscript 2  

  • The vector product of the two direction vectors is perpendicular to both lines

Diagram showing two skew lines, with explanations on calculating shortest distance, involving vector product and points on each line.
Example of the shortest distance between two skew lines

How do I find the shortest distance between two skew lines?

  • For example, consider the lines with equations:

    • l subscript 1 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

    • l subscript 2 space colon space bold italic r equals open parentheses table row 9 row 9 row cell negative 9 end cell end table close parentheses plus mu open parentheses table row cell negative 1 end cell row 4 row 1 end table close parentheses

Using the vector product

  • STEP 1
    Find the vector product of the direction vectors bold italic b subscript 1 and bold italic b subscript 2

    • bold italic d equals open parentheses table row 1 row cell negative 2 end cell row 0 end table close parentheses cross times open parentheses table row cell negative 1 end cell row 4 row 1 end table close parentheses equals open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 2 end table close parentheses

  • STEP 2
    Find an expression for the position vector of a point A on begin mathsize 16px style l subscript 1 end style and a point B on l subscript 2  in terms of lambda and mu

    • stack O A with rightwards arrow on top space 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

    • stack O B with rightwards arrow on top space equals open parentheses table row cell 9 minus mu end cell row cell 9 plus 4 mu end cell row cell negative 9 plus mu end cell end table close parentheses

  • STEP 3
    Find the displacement vector between the general points

    • stack A B with rightwards arrow on top equals open parentheses table row cell 9 minus mu end cell row cell 9 plus 4 mu end cell row cell negative 9 plus mu 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 6 minus mu minus lambda end cell row cell 10 plus 4 mu plus 2 lambda end cell row cell negative 7 plus mu end cell end table close parentheses

  • STEP 4
    Set the displacement vector parallel to the vector product and set up a system of three linear equations in terms of k comma blank lambda and mu

    • open parentheses table row cell 6 minus mu minus lambda end cell row cell 10 plus 4 mu plus 2 lambda end cell row cell negative 7 plus mu end cell end table close parentheses equals k open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 2 end table close parentheses rightwards double arrow table row cell negative mu minus lambda plus 2 k equals negative 6 end cell row cell 4 mu plus 2 lambda plus k equals negative 10 end cell row cell mu minus 2 k equals 7 end cell end table

  • STEP 5
    Solve the system to find the value of k

    • k equals negative 4

  • STEP 6
    Multiply the vector product by kand find the magnitude open vertical bar k bold italic d close vertical bar 

    • open vertical bar negative 4 open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 2 end table close parentheses close vertical bar equals 12

Using the scalar product

  • STEP 1
    Sketch a diagram showing the point F subscript 1 on the line l subscript 1 that is closest to the point F subscript 2 on the line l subscript 2

    • The vector stack F subscript 1 F subscript 2 with rightwards arrow on top will be perpendicular to both lines

  • STEP 2
    Use the equations of the lines to find the position vector of the point F subscript 1  in terms of lambda and F subscript 2 in terms of mu

    • stack O F subscript 1 with rightwards arrow on top space 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

    • stack O F subscript 2 with rightwards arrow on top space equals open parentheses table row cell 9 minus mu end cell row cell 9 plus 4 mu end cell row cell negative 9 plus mu end cell end table close parentheses

  • STEP 3
    Use this to find the displacement vector stack F subscript 1 F subscript 2 with rightwards arrow on top in terms of lambda and mu

    • stack F subscript 1 F subscript 2 with rightwards arrow on top equals open parentheses table row cell 9 minus mu end cell row cell 9 plus 4 mu end cell row cell negative 9 plus mu 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 6 minus mu minus lambda end cell row cell 10 plus 4 mu plus 2 lambda end cell row cell negative 7 plus mu end cell end table close parentheses

  • STEP 4
    Form two equations by setting the scalar product of the direction vector of each line and the displacement vector stack F subscript 1 F subscript 2 with rightwards arrow on top equal to zero

    • open parentheses table row cell 6 minus mu minus lambda end cell row cell 10 plus 4 mu plus 2 lambda end cell row cell negative 7 plus mu 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 9 mu minus 5 lambda equals 14

    • open parentheses table row cell 6 minus mu minus lambda end cell row cell 10 plus 4 mu plus 2 lambda end cell row cell negative 7 plus mu end cell end table close parentheses times open parentheses table row cell negative 1 end cell row 4 row 1 end table close parentheses equals 0 rightwards double arrow 18 mu plus 9 lambda equals negative 27

  • STEP 5
    Solve the equations simultaneously to find the values of lambda and mu

    • lambda equals negative 1 and mu equals negative 1

  • STEP 6
    Substitute lambda and mu into stack F subscript 1 F subscript 2 with rightwards arrow on top and find the magnitude open vertical bar stack F subscript 1 F subscript 2 with rightwards arrow on top close vertical bar 

    • open vertical bar open parentheses table row cell 6 minus open parentheses negative 1 close parentheses minus open parentheses negative 1 close parentheses end cell row cell 10 plus 4 open parentheses negative 1 close parentheses plus 2 open parentheses negative 1 close parentheses end cell row cell negative 7 plus open parentheses negative 1 close parentheses end cell end table close parentheses close vertical bar equals 12

How do I find the shortest distance between two parallel lines?

How do I find the shortest distance between a point on a line and an interesting line?

  • You can use the steps to find the shortest distance between a point and a line

  • It might be quicker to use right-angled trigonometry if you also know the point of intersection and/or the angle between the lines

    • The shortest length between the point and the line is perpendicular to the line

Worked Example

Consider the skew lines l subscript 1 and l subscript 2 as defined by:

l subscript 1: bold italic r equals open parentheses table row 6 row cell negative 4 end cell row 3 end table close parentheses plus lambda open parentheses table row 2 row cell negative 3 end cell row cell blank 4 blank end cell end table close parentheses

l subscript 2: bold italic r equals open parentheses table row cell negative 5 end cell row 4 row cell negative 8 end cell end table close parentheses plus mu open parentheses table row cell negative 1 end cell row 2 row cell blank 1 blank end cell end table close parentheses

Find the minimum distance between the two lines.

3-10-5-ib-aa-hl-short-distance-lines-we-2

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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.