A LevelFurther MathsEdexcelCore PureRevision NotesFurther VectorsVector PlanesEquations of planes

Equations of planes (Edexcel A Level Further Maths): Revision Note

Exam code: 9FM0

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Amber

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Equation of a Plane in Vector Form

How do I find the vector equation of a plane?

  • A plane is a flat surface which is two-dimensional

    • Imagine a flat piece of paper that continues on forever in both directions

  • A plane in often denoted using the capital Greek letter Π

  • The vector form of the equation of a plane can be found using two direction vectors on the plane

    • The direction vectors must be

      • parallel to the plane

      • not parallel to each other

      • therefore they will intersect at some point on the plane

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

    • bold italic r equals bold italic a plus s bold italic b plus t bold italic c

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

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

      • b and c are two non-parallel direction (displacement) vectors parallel to the plane

      • s and t are scalars

  • The formula can also be written as

    • bold r equals bold a plus lambda left parenthesis bold b minus bold a right parenthesis plus mu left parenthesis bold c minus bold a right parenthesis equals left parenthesis 1 minus lambda minus mu right parenthesis bold a plus lambda bold b plus mu bold c

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

      • a, b, c are the position vectors of known points on the plane

      • λ and μ are scalars

    • These formulae are given in the formula booklet but you must make sure you know what each part means

  • As a could be the position vector of any point on the plane and b and c could be any non-parallel direction vectors on the plane there are infinite vector equations for a single plane

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

  • Given the equation of a plane 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 plus blank mu 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 then the point r with position vector blank open parentheses fraction numerator x over denominator table row y row z end table end fraction close parentheses is on the plane if there exists a value of λ and μ such that

    • open parentheses fraction numerator x over denominator table row y row z 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 plus blank mu 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

    • This means that there exists a single value of λ and μ that satisfy the three parametric equations:

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

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

      • z equals blank a subscript 3 plus lambda b subscript 3 blank plus blank mu c subscript 2

  • Solve two of the equations first to find the values of λ and μ that satisfy the first two equation and then check that this value also satisfies the third equation

  • If the values of λ and μ do not satisfy all three equations, then the point r does not lie on the plane

Examiner Tips and Tricks

  • The formula for the vector equation of a plane is given in the formula booklet, make sure you know what each part means

  • Be careful to use different letters, e.g. lambda and mu as the scalar multiples of the two direction vectors

Worked Example

The points A, B and C have position vectors bold italic a equals 3 bold i plus 2 bold j minus bold k, bold italic b equals bold i minus 2 bold j plus 4 bold k, and bold italic c equals 4 bold i minus bold j plus 3 bold k respectively, relative to the origin O.

(a) Find the vector equation of the plane.

3-11-1-ib-aa-hl-vector-plane-vector-form-we-solution-a

(b) Determine whether the point D with coordinates (-2, -3, 5) lies on the plane.

3-11-1-ib-aa-hl-vector-plane-vector-form-we-solution-b

Equation of a Plane in Cartesian Form

How do I find the vector equation of a plane in cartesian form?

  • The cartesian equation of a plane is given in the form

    • a x plus b y plus c z equals d

    • This is given in the formula booklet

  • A normal vector to the plane can be used along with a known point on the plane to find the cartesian equation of the plane

    • The normal vector will be a vector that is perpendicular to the plane

  • The scalar product of the normal vector and any direction vector on the plane will be zero

    • The two vectors will be perpendicular to each other

    • The direction vector from a fixed-point A to any point on the plane, R can be written as r – a

    • Then n ∙ (r – a) = 0 and it follows that (n ∙ r) – (n ∙ a) = 0

  • This gives the equation of a plane using the normal vector:

    • n ∙ r = a ∙ n

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

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

      • n is a vector that is normal to the plane

    • This is given in the formula booklet

  • If the vector r is given in the form open parentheses table row x row y row z end table close parentheses and a and n are both known vectors given in the form open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell row cell a subscript 3 end cell end table close parentheses and open parentheses table row cell n subscript 1 end cell row cell n subscript 2 end cell row cell n subscript 3 end cell end table close parentheses then the Cartesian equation of the plane can be found using:

    • bold italic n times bold italic r equals n subscript 1 x plus n subscript 2 y plus n subscript 3 z

    • bold italic a times bold italic n equals a subscript 1 n subscript 1 plus a subscript 2 n subscript 2 plus a subscript 3 n subscript 3

    • Therefore n subscript 1 x plus n subscript 2 y plus n subscript 3 z equals a subscript 1 n subscript 1 plus a subscript 2 n subscript 2 plus a subscript 3 n subscript 3

    • This simplifies to the form a x plus b y plus c z equals d

      • A version of this is given in the formula booklet

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

  • Given the equation of the plane bold r equals bold a plus lambda bold b plus mu bold c

    • Form three equations

      • x equals a subscript 1 plus lambda b subscript 1 plus mu c subscript 1

      • y equals a subscript 2 plus lambda b subscript 2 plus mu c subscript 2

      • z equals a subscript 3 plus lambda b subscript 3 plus mu c subscript 3

  • Choose a pair of equations and use them to form an equation without μ

  • Choose another pair and form another equation without μ

  • Use your two expressions to form an equation without μ and λ

  • Rewrite the equation in the form a x plus b y plus c z plus d equals 0

Examiner Tips and Tricks

  • In an exam, using whichever form of the equation of the plane to write down a normal vector to the plane is always a good starting point

Worked Example

A plane straight capital pi has equation bold r equals open parentheses table row 3 row 2 row cell negative 1 end cell end table close parentheses plus lambda open parentheses table row cell negative 2 end cell row cell negative 4 end cell row 5 end table close parentheses plus mu open parentheses table row 1 row cell negative 3 end cell row 4 end table close parentheses. Find the equation of the plane in its Cartesian form.

al-fm-6-2-1-equation-of-plane-in-cartesian-form-1
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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.

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