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IBMathsDPAnalysis & Approaches (AA)HLRevision Notes3. Geometry & TrigonometryVector Equations of LinesEquation of a Line in Vector Form

Equation of a Line in Vector Form (DP IB Analysis & Approaches (AA)): Revision Note

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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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    Amber

    Author: Amber

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    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.