H2 Chp 6B 3D Vector Geometry (Lines and Planes) Learning Package 2024

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Chapter 6B 3D Vector Geometry TMJC 2024 Page 1 of 28 H2 Mathematics (9758) Chapter 6B 3D Vector Geometry (Lines & Planes) Core Concept Notes Success Criteria: Surface Learning Deep Learning Transfer Learning  Interpret and find equations of planes in the form ax by cz d   or   0  r a n or    r a b c .  Convert equations of planes from one form to another.  Determine whether a line lies in a plane, is parallel to a plane, or intersects a plane.  Find the angle between a line and a plane, and between two planes.  Find the point of intersection of a line and a plane when it exists.  Find foot of the perpendicular from a point to a plane  Find perpendicular distance between a point and a plane, between a line and a plane, and between two planes.  Find the line of intersection and the angle between two non-parallel planes.  Find the reflection of a point in a plane  Interpret given information in contextual question.  Solve 3D vector geometry questions involving unknowns.

Chapter 6B 3D Vector Geometry TMJC 2024 Page 2 of 28 §1 Equation of Planes A plane is a flat, two-dimensional surface that extends infinitely in all directions. Some examples of planes are: In the rest of this section, we will diagrammatise planes in the shape of a parallelogram. 1.1 Vector Equation of a Plane Consider the plane , which contains a given fixed point A, with position vector a. Suppose that the plane  is parallel to the two non-parallel vectors m1 and m2. Let R be a general point on , and r be the position vector of R. Now, OR OA AR  r       Given two non-parallel vectors 1m and 2m on  , any vector on  may be expressed as a linear combination of those two vectors. Hence, AR   may be expressed as a linear combination of 1m and 2m . i.e 1 2AR   m m   . So,  1 2   r a m m , ,  are real parameters. Vector equation of a plane  which contains the point A with position vector a and parallel to the vectors m1 and m2 is given by: 1 2: , where       r a m m  R O a Position vector of a general point on the plane Position vector of a fixed point on the plane and are two non-parallel vectors that are parallel to the plane real parameters

Chapter 6B 3D Vector Geometry TMJC 2024 Page 3 of 28 Note: (i) The position vector of any point P on  can be expressed as 1 2OP    a m m   for some ,    . (ii) If a plane  contains three non-collinear points A, B and C, then a vector equation of the plane can be given by : , ,OA AB AC      r        The equation is not unique. E.g. Another possible equation is , ,OB AB BC      r        . Example 1 Find a vector equation of the plane that contains the points  2,1, 4A ,  4, 2, 4B and  1,1,9 .C Solution: Find (any) two non-parallel vectors that are parallel to the plane: 4 2 2 2 1 1 4 4 0 AB OB OA                                       and 1 2 1 1 1 0 9 4 5 AC OC OA                                       A vector equation of the plane is: r 2 2 1 1 1 0 4 0 5 λ                               , ,    . Note: 2 2 1 1 1 0 4 0 5 x y λ z                                          2 2 1 4 5 x y z            The equation of the plane in parametric form is 2 2 1 4 5 x y z            . Sketch a simple parallelogram to represent a plane and include points A, B and C on it to help with visualisation B C

Chapter 6B 3D Vector Geometry TMJC 2024 Page 4 of 28 1.2 Equation of a Plane in Scalar Product Form Consider a plane,  , which passes through a given fixed point A with position vector a, and is perpendicular to a given vector n. Let R be a general point on , and r be the position vector of R. Since the vector n is perpendicular to the plane,  , n is perpendicular to any vector that lies on  . Since points A and R lie on , the vector AR   lies on . Therefore, AR   is perpendicular to n.  0AR n     0  r a n   r n a n D  r n , where D a n is a scalar constant. This is known as the equation of the plane in scalar product form. Equation of the plane  in scalar product form which contains the point A with position vector a and perpendicular to the vector n is given by: :   r n a n Example 2 Find an equation of plane p in the form Dr n given that plane p passes through the point  1,1, 1 and is perpendicular to the vector 3 i j k . Solution: Equation of the plane p: 1 1 1 1 1 1 5 3 1 3                                r i.e. 1 1 5 3         r . Position vector of a general point on the plane . Normal vector (a vector perpendicular to the plane  ). Position vector of a fixed point on the plane . O Sketch a simple parallelogram to represent a plane and include the given point and the normal vector to help with visualisation p

Chapter 6B 3D Vector Geometry TMJC 2024 Page 5 of 28 1.2.1 Normal Vector to a Plane (A) (B) (C) (i) A normal to a plane is perpendicular to any line in the plane (A) or any line parallel to the plane (B). If a vector is perpendicular to any two non-parallel vectors, say b and c in the plane (C), then this vector must be a normal vector, n, to the plane. (ii) Normal to a plane is not unique. Thus, vector equation of a plane is not unique. If n is a normal to a plane, then 2n, n3 2 are also normal to the plane. (iii) The planes 1 2 and   are parallel  their normal vectors 1 2 and n n are parallel. n l n l n

Chapter 6B 3D Vector Geometry TMJC 2024 Page 6 of 28 (iv) From the equation of a plane in scalar product form, if we divide both sides of the equation of the plane by the magnitude of the normal vector n, we will get the following:   r n a n n nr an n     ˆ ˆ ˆ 'D   r n a n r n    where (a) ˆ nn n is a unit vector normal to the plane, (b) 'D na n  is a constant, Note: Constant 'D  na n is the perpendicular distance from the origin, O, to the plane. Distance from origin O to the plane  = OF = length of projection of ontoOA   n = ˆOA n    = na n = 'D (v) 0 r n the plane passes through the origin. (vi) If the point P lies on  , then the position vector of P will satisfy OP n a n     . (vii) In scalar product form, equations of (a) x-y plane : 1 0 0 0