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IBMathsDPApplications & Interpretation (AI)HLRevision Notes3. Geometry & TrigonometryVector Equations of LinesVector Equations of Lines

Vector Equations of Lines (DP IB Applications & Interpretation (AI)): Revision Note

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

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Equation of a Line in Vector Form

How do I find the vector equation of a line?

  • The formula for finding the vector equation of a line is

    • bold italic r equals bold italic a plus lambda bold italic b

      • Where r is the position vector of any point on the line

      • a is the position vector of a known point on the line

      • b is a direction (displacement) vector

      • lambda is a scalar

    • This is given in the formula booklet

    • This equation can be used for vectors in both 2- and 3- dimensions

  • This formula is similar to a regular equation of a straight line in the form y equals m x plus c but with a vector to show both a point on the line and the direction (or gradient) of the line

    • In 2D the gradient can be found from the direction vector

    • In 3D a numerical value for the direction cannot be found, it is given as a vector

  • As a could be the position vector of any point on the line and b could be any scalar multiple of the direction vector there are infinite vector equations for a single line

  • Given any two points on a line with position vectors a and b the displacement vector can be written as b - a

    • So the formula r = a + λ(b - a) can be used to find the vector equation of the line

    • This is not given in the formula booklet

How do I determine whether a point lies on a line?

  • Given the equation of a line bold italic r blank equals blank open parentheses fraction numerator bold italic a subscript 1 over denominator table row cell bold italic a subscript 2 end cell row cell bold italic a subscript 3 end cell end table end fraction close parentheses plus lambda open parentheses fraction numerator bold italic b subscript 1 over denominator table row cell bold italic b subscript 2 end cell row cell bold italic b subscript 3 end cell end table end fraction close parentheses the point c with position vectorblank open parentheses fraction numerator bold italic c subscript 1 over denominator table row cell bold italic c subscript 2 end cell row cell bold italic c subscript 3 end cell end table end fraction close parentheses is on the line if there exists a value of lambdasuch that

  • open parentheses fraction numerator bold italic c subscript 1 over denominator table row cell bold italic c subscript 2 end cell row cell bold italic c subscript 3 end cell end table end fraction close parentheses blank equals blank open parentheses fraction numerator bold italic a subscript 1 over denominator table row cell bold italic a subscript 2 end cell row cell bold italic a subscript 3 end cell end table end fraction close parentheses plus lambda open parentheses fraction numerator bold italic b subscript 1 over denominator table row cell bold italic b subscript 2 end cell row cell bold italic b subscript 3 end cell end table end fraction close parentheses

  • This means that there exists a single value of lambdathat satisfies the three equations:

    • c subscript 1 equals blank a subscript 1 plus lambda b subscript 1

    • c subscript 2 equals blank a subscript 2 plus lambda b subscript 2

    • c subscript 3 equals blank a subscript 3 plus lambda b subscript 3

  • A GDC can be used to solve this system of linear equations for

    • The point only lies on the line if a single value of lambda exists for all three equations

  • Solve one of the equations first to find a value of lambdathat satisfies the first equation and then check that this value also satisfies the other two equations

  • If the value of lambda does not satisfy all three equations, then the point c does not lie on the line

Examiner Tips and Tricks

  • Remember that the vector equation of a line can take many different forms

    • This means that the answer you derive might look different from the answer in a mark scheme

  • You can choose whether to write your vector equations of lines using unit vectors or as column vectors

    • Use the form that you prefer, however column vectors is generally easier to work with

Worked Example

a) Find a vector equation of a straight line through the points with position vectors a = 4i – 5k and b = 3i - 3k

M83a0TRO_3-10-1-ib-aa-hl-vector-equation-of-a-line-we-a

b) Determine whether the point C with coordinate (2, 0, -1) lies on this line.

3-10-1-ib-aa-hl-vector-equation-of-a-line-we-b

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Equation of a Line in Parametric Form

How do I find the vector equation of a line in parametric form?

  • By considering the three separate components of a vector in the x, y and z directions it is possible to write the vector equation of a line as three separate equations

    • Letting begin mathsize 16px style bold italic r equals blank open parentheses fraction numerator x over denominator table row y row z end table end fraction close parentheses end style then bold italic r equals bold italic a plus lambda bold italic b becomes

    • begin mathsize 16px style open parentheses fraction numerator x over denominator table row y row z end table end fraction close parentheses equals blank open parentheses fraction numerator x subscript 0 over denominator table row cell y subscript 0 end cell row cell z subscript 0 end cell end table end fraction close parentheses plus lambda open parentheses fraction numerator l over denominator table row m row n end table end fraction close parentheses end style

      • Where begin mathsize 16px style open parentheses fraction numerator x subscript 0 over denominator table row cell y subscript 0 end cell row cell z subscript 0 end cell end table end fraction close parentheses end style is a position vector and begin mathsize 16px style open parentheses fraction numerator l over denominator table row m row n end table end fraction close parentheses end style is a direction vector

    • This vector equation can then be split into its three separate component forms:

      • x equals blank x subscript 0 plus blank lambda l blank

      • y equals blank y subscript 0 plus blank lambda m blank

      • z equals blank z subscript 0 plus blank lambda n

    • These are given in the formula booklet

Worked Example

Write the parametric form of the equation of the line which passes through the point (-2, 1, 0) with direction vector begin mathsize 16px style open parentheses fraction numerator 3 over denominator table row 1 row cell negative 4 end cell end table end fraction close parentheses end style.

3-10-1-ib-aa-hl-parametric-we

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Angle Between Two Lines

How do we find the angle between two lines?

  • The angle between two lines is equal to the angle between their direction vectors

    • It can be found using the scalar product of their direction vectors

  • Given two lines in the form bold italic r equals bold italic a subscript 1 plus lambda bold italic b subscript 1 and bold italic r equals bold italic a subscript 2 plus lambda bold italic b subscript 2 use the formula

    • begin mathsize 16px style theta equals cos to the power of negative 1 end exponent invisible function application open parentheses fraction numerator bold italic b subscript 1 blank bullet blank bold italic b subscript 2 over denominator open vertical bar bold italic b subscript 1 close vertical bar open vertical bar blank bold italic b subscript 2 close vertical bar end fraction close parentheses end style

  • If you are given the equations of the lines in a different form or two points on a line you will need to find their direction vectors first

  • To find the angle ABC the vectors BA and BC would be used, both starting from the point B

  • The intersection of two lines will always create two angles, an acute one and an obtuse one

    • A positive scalar product will result in the acute angle and a negative scalar product will result in the obtuse angle

      • Using the absolute value of the scalar product will always result in the acute angle

Examiner Tips and Tricks

  • In your exam read the question carefully to see if you need to find the acute or obtuse angle

    • When revising, get into the practice of double checking at the end of a question whether your angle is acute or obtuse and whether this fits the question

Worked Example

Find the acute angle, in radians between the two lines defined by the equations:

begin mathsize 16px style l subscript 1 colon space space bold italic a equals open parentheses table row 2 row 0 row cell blank 3 blank end cell end table close parentheses plus lambda open parentheses table row 1 row cell negative 4 end cell row cell blank minus 3 blank end cell end table close parentheses end style and  begin mathsize 16px style l subscript 2 colon space space bold italic b equals open parentheses table row 1 row cell negative 4 end cell row 3 end table close parentheses plus mu open parentheses table row cell blank minus 3 blank end cell row 2 row 5 end table close parentheses end style

R_UZJlZ8_3-10-3-ib-aa-hl-angle-between-we-solution-2a

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    Amber

    Author: Amber

    Expertise: Maths

    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.