DP IB Maths: AA HL

Revision Notes

3.10.1 Vector Equations of Lines

Test Yourself

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 begin mathsize 16px style 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 end style the point c with position vectorbegin mathsize 16px style blank 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 end style is on the line if there exists a value of lambdasuch that
    • begin mathsize 16px style 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 end style
    • 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

Exam Tip

  • 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

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

Equation of a Line in Cartesian Form

  • The Cartesian equation of a line can be found from the vector equation of a line by
    • Finding the vector equation of the line in parametric form
    • Eliminating lambda from the parametric equations
      • lambda can be eliminated by making it the subject of each of the parametric equations
      • For example: blank x equals blank x subscript 0 plus blank lambda l blankgives blank lambda equals blank fraction numerator blank x minus blank x subscript 0 over denominator l end fraction blank
  • In 2D the cartesian equation of a line is a regular equation of a straight line simply given in the form
    •  begin mathsize 16px style y equals m x plus c end style
    • begin mathsize 16px style a x plus b y plus d equals 0 end style
    • begin mathsize 16px style fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction equals fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction end style by rearranging begin mathsize 16px style y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses end style
  • In 3D the cartesian equation of a line also includes z and is given in the form
    • begin mathsize 16px style fraction numerator x minus blank x subscript 0 over denominator l end fraction equals blank fraction numerator y minus blank y subscript 0 over denominator m end fraction equals blank fraction numerator z minus blank z subscript 0 over denominator n end fraction end style
    • where 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
    • This is given in the formula booklet
  • If one of your variables does not depend on lambda then this part can be written as a separate equation
    • For example: begin mathsize 16px style m equals 0 blank rightwards double arrow y equals blank y subscript 0 blank end stylegives begin mathsize 16px style fraction numerator x minus blank x subscript 0 over denominator l end fraction equals blank fraction numerator z minus blank z subscript 0 over denominator n end fraction blank comma blank y equals blank y subscript 0 end style

How do I find the vector equation of a line given the Cartesian form?

  • If you are given the Cartesian equation of a line in the form
    • begin mathsize 16px style fraction numerator x minus blank x subscript 0 over denominator l end fraction equals blank fraction numerator y minus blank y subscript 0 over denominator m end fraction equals blank fraction numerator z minus blank z subscript 0 over denominator n end fraction end style
  • A vector equation of the line can be found by
    • STEP 1: Set each part of the equation equal to lambdaindividually
    • STEP 2: Rearrange each of these three equations (or two if working in 2D) to make x, y, and z the subjects
      • This will give you the three parametric equations
      • 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 blank
    • STEP 3: Write this in the vector form 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
    • STEP 4: Set r  to equal begin mathsize 16px style open parentheses fraction numerator x over denominator table row y row z end table end fraction close parentheses end style
  • If one part of the cartesian equation is given separately and is not in terms of lambda then the corresponding component in the direction vector is equal to zero

Worked example

A line has the vector equation begin mathsize 16px style r blank equals blank open parentheses table row 1 row 0 row 2 end table close parentheses plus lambda open parentheses table row 4 row cell negative 2 end cell row 1 end table close parentheses end style. Find the Cartesian equation of the line.

3-10-1-ib-aa-hl-cartesianwe

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Amber

Author: Amber

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.