TMJC H2 Chp 6A 3D Vector Geometry (Lines) Learning Package 2024

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Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 1 of 23 H2 Mathematics (9758) Chapter 6A 3D Vector Geometry (Lines) Core Concept Notes Success Criteria: Surface Learning Deep Learning Transfer Learning  Interpret and find equations of lines in the form r = a + d (vector equation) or x a y b z c l m n − − −== (cartesian equation)  Convert the equations from one form to another  Determine the relationship (i.e. intersecting, parallel or skew) between two lines  Explain that two lines are coplanar if they are intersecting or parallel  Find the angle between two lines  Find the point of intersection of two lines if they intersect  Find the length of projection of a vector onto a given line  Find the foot of the perpendicular and perpendicular distance from a point to a line  Find the reflection of a point in a line  Interpret given information in contextual question

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 2 of 23 Pre Reading Look through the following 2 examples BEFORE the first Independent Learning module on Chapter 6A in SLS. 1. Using the grid, draw the following vectors, marking the points R1, R2, R3, R4 and R5. (a) 1 1 1OR OA =+ − (b) 2 12 1OR OA =+ − (c) 3 13 1OR OA =+ − (d) 4 10 1OR OA =+ − (e) 5 12 1OR OA =− − (i) Join the points R1, R2, R3, R4 and R5. What do they form? (A line segment) (ii) What is the geometrical representation of the following equation? 1 1OR OA  =+ − ,   . Position vector of a point on the line that passes through point A and is parallel to 1 1  − . X 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 O X A X X X X 2R 1R 3R 4R 5R

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 3 of 23 2. The position vector of any point from the line is given in the form r = ac bd    +       , where   . (i) State 2 possible values of a b   . Just provide the column vector of ANY point on the line. Possible values of a b   are 0 3 6, , etc8 5 2                  (ii) State 2 possible values of c d   . What does d c represent? Possible values of c d   are 11 or 11 −       −    . d c represents the gradient of the straight line. 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 O X

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 4 of 23 §1 Equation of Line 1.1 Vector Equation of a Line In 3-dimensional space, a straight line is uniquely located in space if it has a known direction and passes through a known fixed point. Consider a straight line l passing through a fixed point A with position vector a and which is parallel to a given vector d. Let R be any point on the line, and r be the position vector of R. Since line l is parallel to d, AR = d for some   . Now, OR OA AR=+ .  equation of line :l r = a + d ,   . Note (i) Each value of  gives the position vector of a different point on the line. (ii) If a point R lies on the line l with equation =+r a d , then the position vector of the point R is given by: OR =+ad for some   . (iii) An equation of this form ,=rd represents a line passing through the origin. (iv) Is the vector equation of a line unique? No. Why? Vector equation of the line l in parametric form which passes through the point with position vector a and parallel to the vector d is given by: :l =+r a d , where  Position vector of ANY point on the line l. Position vector of a fixed point on the line l. Direction vector of the line (a vector that is parallel to the line l ).  is a real parameter r a A O d R l

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 5 of 23 Example 1 Find a vector equation of the line passing through ( )1, 1, 2A − and parallel to the vector 3. +ik Solution: A vector equation of the line is :l 13 1 0 , 21         = − +           r . Discussion: What are the equations of the x-axis, y-axis and z-axis? x-axis : 01 : 0 0 , 00 xl         = +           r y-axis : 00 : 0 1 , 00 yl         = +           r z-axis : 00 : 0 0 , 01 zl         = +           r The line passing through point A and parallel to vector b has equation ,r a b = +  The x-axis passes through point ( )0,0,0 and parallel to vector 1 0 0    .

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 6 of 23 Example 2 Find a vector equation of the line through ( )5, 2,7A and ( )3,6,3B − . Determine whether the points ( )7,8,1C − and ( )1, 2, 4D lie on the line. Solution: AB OB OA=− 82 4 4 1 41 −       = = − −      −     a direction vector for the line is 2 1 1  − Therefore, a vector equation of the line AB is 52 2 1 , 71         = + −           r . OR 58 2 4 , 74  −       = +        −    r OR 32 6 1 , 31 ss −       = + −           r OR 38 6 4 , 34 tt −−       = +        −    r 7 5 2 2 7 5 8 2 1 1 8 2 1 7 1 1 1 7 2 12 16 16   −−                       = + −  − = −                                  −        − =       −    Since 6 =− satisfies the equation, C lies on the line AB . 1 5 2 2 1 5 2 2 1 1 2 2 4 7 1 1 4 7 2 4 2 1 0 0 1 3 3                            = + −  − = −                                  − = −        − =  =       − = −    Since no consistent value of  satisfies the equation, D does not lie on the line AB . To determine if a point C lies on the line ,r a d = +  . Check if there exists a real value of  that satisfies the equation OC a d =+ . If Yes, C lies on the line. Otherwise, C does not lie on the line.

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 7 of 23 1.2 Parametric and Cartesian Forms Let x y z  =  r , 1 2 3 a a a  =  a and 1 2 3 d d d  =  d . Then from r = a + d,   , we have: 11 22 33 x a d y a d z a d             =+                ,   ---  This is known as the vector equation (form) of the line. From  if we equate the i, j and k components, we have: 1 1 2 2 3 3, , ,x a d y a d z a d   = + = + = +  ---  From  if we make