2026 Chp 1C (Student) - JPJC

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(c) Vectors − Planes (Students’ version) / Pg 1 Chapter 1(c) : Planes 1. Equation of a Plane A plane is commonly defined using (i) two vectors parallel to the plane and a point on the plane or (ii) three non-collinear points or (iii) the direction of the normal to the plane and one point on the plane. In this section, we will learn the three different forms of equation of a plane, namely 1.1 Vector equation (in parametric form): = + +r a b c , where ,   ; 1.2 Vector equation (in scalar product form): d•=rn , where d   ; and 1.3 Cartesian equation: ax by cz d+ + = , where , , , a b c d   . Recall: An equation of a line is of the form r = a + b ,    where r represents position vector of a point on the line a represents position vector of a given (known) point on the line b represents a vector parallel to the line 1.1 Vector Equation of a Plane in Parametric Form Refer to the diagrams below. A is a fixed point on a plane and R1, R2, R3, ……etc are different points on the same plane. b and c are 2 vectors parallel to the plane. Then, by vector addition, 1 1 1AR =+ bc  2 2 2AR =+ bc  3 3 3AR =+ bc  Generalising, for any point R lying on the plane, AR =+ bc  , where ,   . In fact, any vectors on the plane or parallel to the plane can also be expressed as  +bc . c b A R2 R3 c b A c b A R1

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(c) Vectors − Planes (Students’ version) / Pg 2 We can obtain the vector equation of a plane in parametric form for the following cases: ➢ Given a point on the plane and two vectors parallel to the plane The equation of the plane through any given point A, with position vector a, and parallel to vectors b and c can be found as follows. OR OA AR=+  where R is any variable point on the plane.  = + +r a b c , ,   Vector equation of the plane in parametric form: = + +r a b c , ,   Example 1 Write down the vector equation of the plane in parametric form (a) through point C(4, –1, 2) and parallel to the vectors i + 3j + k and –i + 4j – k, (b) through the points A(3, –2, 0), B(2, 0, 3) and C(1, −1, 1) (c) that contains the lines ( ) ( ) and 2ts= − + + = − + − −r j k i k r j k i j k Solution (a) 4 1 1 1 3 4 2 1 1  −           = − + +           −      r , ,   (b) OA AC CB= + +r  3 2 1 2 1 1 0 1 2  −           = − + +                r , ,   (c) 0 1 2 1 0 1 1 1 1             = + + −          −−      r , ,   O c b A R O R C O R O R

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(c) Vectors − Planes (Students’ version) / Pg 3 1.2 Vector Equation of Plane in Scalar Product Form We can obtain the equation of a plane in scalar product form for the following cases: ➢ Given the vector normal to the plane and one point on the plane The equation of a plane can also be given / found in a form involving the scalar product of vectors. If n is a vector normal to the plane, then we can find a simple expression for the equation of the plane. Let R be any point and A is a given point on the plane. Then AR ⊥ n  . 0 AR •=n  ( ) 0 OR OA− • = n  ( ) 0− • =r a n 0 ( ) d • − • = • = • = r n a n r n a n Vector equation of the plane in scalar product form: d•=rn . Example 2 Find the equation of the plane passing through the point (0, 1, 1) and perpendicular to the vector 2− + +i j k . Solution • = •r n a n 1 0 1 1 1 1 3 2 1 2 −−           • = • =                r Hence, the equation of the plane is 1 13 2 − •= r THINK: What does it imply if the RHS of the scalar equation of plane is zero? The origin lies on the plane. Replace a by 0, to obtain 0•=rn as the equation of plane O R A n

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(c) Vectors − Planes (Students’ version) / Pg 4 Example 3 Find the equation of the following planes in scalar product form (a) the plane passing through the points A(2, –1, 4), B(3, 2, –6) and C(4, 1, 5); (b) the plane containing the lines ( ) ( ) and 2 2= + + + − = − + + + − +r i j k i j r i j k i k , where  and  are parameters. Solution (a) Find n using vector product of AB  & AC  or other choices: 1 3 10 AB  = −  , 2 2 1 AC  =   1 2 23 3 2 21 10 1 4 AB AC             =  = −          −−       Let 23 21 4  =−− n 23 21 4 or take  −  =   n Equation of plane in scalar product form: 23 2 23 21 1 21 51 4 4 4            • − = − • − =          −−      r i.e. 23 21 51 4  • − =− r (b) Identify vectors // to plane: 1 1 1 0  =− v 2 2 0 1 − =  v Find n using vector product: 1 2 1 1 0 1 0 1 2 −−           −  = −           −      Let 1 1 2  =  n Equation of plane in scalar product form: 1 1 1 1 1 1 4 2 1 2            • = • =                r i.e. 1 14 2  •= r Think: How would you check your answer in (a) & (b)? For (a), we can substitute the position vector of points B or C into the equation to verify. For (b), we can substitute the position vector of point Q into the equation to verify. (–1, 1, 2) Q (1, 1, 1) R P n

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(c) Vectors − Planes (Students’ version) / Pg 5 1.3 Equation of Plane in Cartesian Form (an equation in x, y , z only) This is obtained from expanding the scalar product form of the plane’ s equation by writing the position vector of R as x y z  =  r . If the equation of a plane is given as 1 2 2, 3  • − = r  1 22 3 x y z        • − =           2 3 2x y z− + = Thus the Cartesian equation of the plane is 2 3 2x y z− + = . Example 3 (continued) Write down the equation of the planes in Cartesian form for (a) and (b) in the above example. Solution (a) Equation of plane in Cartesian form: 23 21 4 51x y z− − = (b) Equation of plane in Cartesian form: 24x y z+ + =

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(c) Vectors − Planes (Students’ version) / Pg 6 Example 4 Find the equation of the plane passing through a point (1, 0, 1) and containing the line ( )2= + + +r j k i k ,    (i) in scalar product form, and (ii) in Cartesian form. Solution Think: What is the normal vector of a plane that is parallel to the plane given by 1 23 0  • − = r ? Special Planes You may come across terms such as x-y plane, x-z plane and y-z plane. The x-y plane is a plane which contains both the x and y– axis. Think about the following for x-y plane: - What is the axis that is perpendicular to the x-y plane? - What is the normal of the x-y plane? - What is a known point lying on the x-y plane? - What is the equation of the x-y plane in scalar product form and Cartesian form? Apply the same quest ioning to obtain the equations of the x-z plane and y-z plane in scalar product form and Cartesian form. (i) 1 2 0 1  =  v       