ACJC 2025 Vectors Summary

VECTORS (in MF15) A. Basic Results Equal vectors: =  =a b a b and a, b are in the same direction Parallel vectors: // for some , 0   =  a b a b Collinear points: A, B and C are collinear  AB AC= for some ,0 Ratio Theorem: If P divides AB in the ratio : ( : : or APAP PB PB  == ), then OA OBOP   += + . (in MF27) If P is the midpoint of AB, then ( )1 2OP OA OB=+ . O A B P  μ B. 3-D Vectors Position vector of point P (relative to O) is OP . If OP is given by a a b c b c  = + + =  r i j k , then Modulus of OP , 2 2 2OP a b c= = + +r Unit vector // to OP , 2 2 2 1ˆ aOP b abc cOP  == ++  r C. Scalar Product (Dot Product) Definition: cos=a.b a b where  is the angle between a and b. If 1 2 3 a a a  = a , 1 2 3 b b b  = b , then 11 2 2 1 1 2 2 3 3 33 ab a b a b a b a b ab        = = + +       a.b . (i) 2 =a.a a (ii) =a.b b.a (iii) 0⊥  =a b a.b a b  a b  a b  D. Vector Product (Cross Product) Definition: ( ) ˆsin=a b a b n and sin=a b a b where ˆn is a unit vector perpendicular to a and b  is the angle measured from a to b. i.e. ab is a vector perpendicular to both a and b. If 1 2 3 a a a  = a , 1 2 3 b b b  = b , then 2 3 3 211 2 2 3 1 1 3 33 1 2 2 1 a b a bab a b a b a b ab a b a b −        =  = −     −     ab (in MF27) (i)  =− a b b a (ii) //ab   = a b 0  =a a 0 E. Some Applications 1. Angle between a and b: cos = a.b ab Eg: cos BA BCABC BA BC = . 2. Projection/Perpendicular: Length of projection of a on d = ˆa.d Length of component of a ⊥ d = ˆad Vector component of a in the direction of d = ( )ˆˆa.d d 3. Areas: Area of parallelogram ABCD = AB AD Area of ABD = 1 2 AB AD a d  A B C D a b ab SUMMARY: Vectors Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Vectors / Summary / Page 1 of 4

VECTORS (LINES & PLANES) Vector form: Parametric form: Cartesian form: Equations of Lines Equations of Planes Parametric form: Scalar Product form: b O P A r a ⊥ distance from O to plane Cartesian form: P a O A r c b A (a) Point lies on a Line Given: L: To show (2, 2, 4) lies on L, sub. and show that  is consistent (same value) for all 3 equations. (b) Point lies on a Plane Given: : To show (1, 1, 1) lies on , sub. and show that LHS = RHS. To show: A Point lies on a (a) Line (b) Plane B (c) between 2 Planes Given: (a) between 2 Lines Angle  between (a) 2 Lines (b) Line & Plane (c) 2 Planes Given: Given: (b) bet. Line & Plane or Let  be the angle bet. L & n of . C (a) Between 2 Lines Put (1) = (2) & solve 2 eqns for  & μ. Check that  & μ satisfy the 3rd eqn. Sub.  into (1) [or μ into (2)] to obtain p.v. of point. (b) Between Line & Plane Sub. (1) into (2) to find . Then put  into (1) to obtain p.v. of point. (c) Between 2 Planes M1: Equation of line of intersection is [a is p.v. of any point on & (e.g. set z = 0 and solve eqns of & )] M2: Convert eqns of & to Cartesian form. Solve using GC. Intersection between (a) 2 Lines (b) Line & Plane (c) 2 Planes D Foot of Perpendicular from Point to (a) Line (b) Plane (a) Foot of ⊥ from Point to Line Given: L: p.v. of point P is p (b) Foot of ⊥ from Point to Plane Given: : p.v. of point P is p F M1: (i) N lies on line L  for some real value of  (ii)  Solve for  M2:  M1: N is point of intersection of & . Sub. (1) into (2) to find . M2: where A is any pt. on  P n N A  P b A N L (a) ⊥ Distance from Point to Line Given: L: p.v. of point P is p Perpendicular Distance from Point to (a) Line (b) Plane (b) ⊥ Distance from Point to Plane Given: : p.v. of point P is p M1: Find p.v. of N, foot of ⊥ from point P to line L.  ⊥ distance = M2: ⊥ distance = (See diagram below) G M1: Find p.v. of N, foot of ⊥ from point P to plane .  ⊥ distance = M2: ⊥ distance = (See diagram below) (a) Reflection of point in Line Given: L: p.v. of point P is p Reflection of Point in (a) Line (b) Plane (b) Reflection of point in Plane Given: : p.v. of point P is p P P A N Step 1: Find where N is foot of perpendicular from P to L or  Step 2: M1: Since , M2:  Find P N P A H Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Vectors / Summary / Page 2 of 4 (a) Parallel and (b) Line on plane and (c) Line intersect plane at 1 point E Relationship between Line & Plane

EXAMPLES (LINES & PLANES) Eqn of plane passing through A(3,−2,0), B(2,0,3) and C(1,−1,1): Parametric form: Eqn of line passing through A(2, −1, 3) and B(5, 3, 3) : Vector form: i.e. Parametric form: Equations of Lines Cartesian form: Equations of Planes A Cartesian form: Scalar Product form: A vector normal to plane is Equation of plane is (a) Point lies on a Line Given: L: To show (2, 2, 4) lies on L, sub. and show that  is consistent (same value) for all 3 eqns. (b) Point lies on a Plane Given: : To show (1, 1, 1) lies on , sub. and show that LHS = RHS. To show: A Point lies on a (a) Line (b) Plane B (c) Angle between 2 Planes (a) Angle between 2 Lines Angle  between (a) 2 Lines (b) Line & Plane (c) 2 Planes (b) Angle between Line & Plane C (a) Between 2 Lines Intersection between (a) 2 Lines (b) Line & Plane (c) 2 Planes (b) Between Line & Plane D Put Solving , Sub. , into 3rd eqn: Since LHS = RHS, and intersect. p.v. of point of intersection is Sub. (1) into (2), Then put into (1) p.v. of point of intersection is . (c) Between 2 Planes M1: A vector parallel to line of intersection is p.v. of point that lies on both planes is [set z = 0 and solve Cartesian eqns of for x & y.]  Eqn of line of intersection is M2: Convert eqns of to Cartesian form. Solve using GC. M3: Sub. (1) into (2):  Eqn of line of intersection is Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Vectors / Summary / Page 3 of 4

(a) Foot of ⊥ from Point to Line Foot of Perpendicular from Point to (a) Line (b) Plane (b) Foot of ⊥ from Point to Plane M1: (i) N lies on line L  for some real value of  (ii) M2: where M1: N is point of intersection of & . Sub. (1) into (2): M2: A(1,0,0) lies on plane. where P(3,−6,1) A(7,1,−5) N L P(7,0,3) N A  F Given: Point Given: Point (a) ⊥ Distance from Point to Line Perpendicular Distance from Point to (a) Line (b) Plane (b) ⊥ Distance from Point to Plane G Given: Point M1: Let N be foot of ⊥ from point P to line L. From (F),  ⊥ distance = M2: ⊥ distance Given: Point M1: Let N be foot of ⊥ from point P to plane . From (F),  ⊥ distance M2: ⊥ distance where A(1,0,0) lies on  Reflection of Point in (a) Line (b) Plane (b) Reflection of a point in Plane Let N be foot of ⊥ from point P to line L. From (F), M1: Since , M2: H Given: Point Given: Point Let N be foot of ⊥ from point P to plane . From (F), M1: Since , M2: (a) Reflection of a point in Line P P(3,−6,1) A N P(7,0,3) N P A Anglo-Chinese Junior College 2024 H2 Mathematics 9758: Vectors / Summary / Page 4 of 4 Relationship between Line and Plane E Given (a) Parallel: and (b) Line on Plane: and (c) Line intersects Plane at 1 point: ,