IBMaths: AA HLDPRevision Notes3. Geometry & Trigonometry3.11 Vector Planes3.11.1 Vector Equations of Planes

Vector Equations of Planes (DP IB Maths: AA HL)

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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
    • If both direction vectors lie on the plane then they will intersect at a point
  • The formula for finding the vector equation of a plane is
    • bold italic r equals bold italic a plus lambda bold italic b plus mu 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
      • λ and μ are scalars
    • The formula is 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 3
  • 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

Exam Tip

  • 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 the 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 a row b row c end table close parentheses then the Cartesian equation of the plane can be found using:
    • bold italic n times bold italic r equals a x plus b y plus c z
    • bold italic a times bold italic n equals a subscript 1 a plus a subscript 2 b plus a subscript 3 c
    • Therefore a x plus b y plus c z equals a subscript 1 a plus a subscript 2 b plus a subscript 3 c
    • This simplifies to the form a x plus b y plus c z equals d

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

  • The Cartesian equation of a plane can be found if you know
    • the normal vector and
    • a point on the plane
  • The vector equation of a plane can be used to find the normal vector by finding the vector product of the two direction vectors
    • A vector product is always perpendicular to the two vectors from which it was calculated
  • The vector a given in the vector equation of a plane is a known point on the plane
    • Once you have found the normal vector then the point a can be used in the formula n ∙ r = a ∙ n to find the equation in Cartesian form
  • To find a x plus b y plus c z equals d given bold italic r equals bold italic a plus lambda bold italic b plus mu bold italic c :
    • Let bold italic n equals open parentheses table row a row b row c end table close parentheses equals bold italic b cross times bold italic c then d equals bold italic n times bold italic a

Exam Tip

  • 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 contains the point Aleft parenthesis 2 comma space 6 comma negative 3 right parenthesis and has a normal vector open parentheses table row 3 row cell negative 1 end cell row 4 end table close parentheses.

a)
Find the equation of the plane in its Cartesian form.

JMMFvkWx_3-11-1-ib-aa-hl-vector-plane-cartesian-we-solution-a

b)
Determine whether point B with coordinates left parenthesis negative 1 comma space 0 comma negative 2 right parenthesis lies on the same plane.

ghYsvxS~_3-11-1-ib-aa-hl-vector-plane-cartesian-we-solution-b

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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.