SPTA Mathematics Higher Notes

Transcription

1 H Vectors SPT Mathematics Higher Notes Definitions: MGNITUDE - the SIZE of an object SCLR - quantities or objects that have magnitude only, i.e. Length, Weight, Number of Sharks in the sea, Height of a wall, etc. VECTOR - quantities or objects that have magnitude ND direction, i.e. plane flys 00km on a bearing of 15 Wind speed is 15mph Southeasterly. Vectors: vector can be named in ways meaning the same thing: o old, Italisised and underlined letter, u represents Vector u o capital letters with an arrow above,, represents the Directed Line Segment. vector can also be shown graphically as shown: u vector can be described using components, in: o D u = ( x y ), e.g. u = ( ) meaning RIGHT and DOWN. x 5 o D u = ( y), e.g. u = ( ) meaning 5 LEFT, IN and UP. z The MGNITUDE is written as u or and can be calculated using the components: u = x + y + z or = x + y + z, for D miss out the z! Vectors with the same components are equal vectors, regardless of their starting point, i.e. = ( 5 ), CD = ( 5 ) hence = CD Remember direction matters so

2 1. Write down the components of the Vectors below: D F G C E H = ( 6 ) DC = ( 6 ) EF = ( 1 ) GH = ( 10 ). Find the magnitude of the Directed Line Segments above. = x + y DC = x + y EF = x + y GH = x + y = 6 + DC = + ( 6) EF = ( 1) + GH = 10 + ( ) = DC = EF = GH = = 5 DC = 5 EF = 10 GH = 10 = 5 DC = 5 GH = 6 dding/subtracting Vectors & Multiplying by a Scalar: When you add or subtract Vectors you produce the RESULTNT Vector. When subtracting Vectors you actually add the negative Vector, i.e. u v = u + ( v) Vectors are added/subtracted using the Nose to Tail method as shown: u u v u v v v u + v u Their components can be added/subtracted as follows: a d a + d a d If u = ( b) and v = ( e) then u + v = ( b + e) and u v = ( b e) c f c + f c f x kx Multiplying a Vector by a Scalar looks like: If u = ( y) then ku = ( ky) z kz You cannot multiply a Vector by another Vector.

3 . For the example above find: a) u + v (b) u v (c) u (d) u + 5v u = ( 6 ) and v = ( ) so: a) u + v = ( ) (b) u v = ( + ( ) ( ) ) = ( 8 1 ) = ( 5 ) c) u = ( 6 ) (d) u + 5v = (6 ) + 5 ( ) = ( 18 9 ) = (18 9 ) + ( 10 0 ) = ( 8 11 ). For the Vectors below draw the resultant Vector: a) a + b (b) a b (c) b a b a) a + b a b b) a a b b c) b 5. State the components of the resultant Vectors above: a) a + b = ( 1 8 ) (b) a b = ( ) (c) b = ( 6 ) 6. In the square opposite name the equivalent vector to: a) (b) C (c) D = DC C = D D = D = C 7. In the same square name the resultant vector when: a) + C (b) C D C + C = C C = D

4 Parallel, Unit and Position Vectors: vector are is PRLLEL to another vector if it is a multiple of the first one, i.e. v = ku Direction is not important! UNIT VECTOR is a vector with magnitude ONE. x POSITION VECTOR is a Vector joing a point, (x, y, z), to the Origin, i.e. O = a = ( y) z The Vector joining any points, and, can be found using the position Vectors: = b a VERY important 1 8. Prove that the vectors, a = ( ) and b = ( ) are parallel. 8 1 a = ( ) = ( ) = b hence a is parallel to b 8 9. P and Q are the points (,, 5) and ( 8, 6, ) respectively. Find the magnitude of vector PQ. PQ = q p PQ = x + y + z 8 PQ = ( 6 ) ( ) PQ = ( 11) ( 1) 5 11 PQ = ( 8 1 ) PQ = PQ = Find the components of the unit vector parallel to the vector a = ( 0 ) 5 a = x + y + z a = ( 5) a = a = a = 1 So the unit vector is 1 ( 0 ) 1 5

5 Special Unit Vectors: ny Vector can be expressed in terms of the Unit Vectors i = ( 0), j = ( 1) and l = ( 0) a) Express the vector, a = ( ) in terms of i, j, k : i j 8k 8 5 b) Write the components of the vector a = 5i j + k : a = ( ) 7 c) Write the components of the vector b = 7i + 11k : b = ( 0 ) 11 d) Calculate the magnitude of the vector c = 5i + j k c = x + y + z c = ( 1) c = c = 0 dapting Known Formulae: The following formulae can be used in Dimensions by adding a z part to them: o Distance Formula: (x x ) + (y y ) + (z z ) o Midpoint Formula: ( x +x, y +y, z +z ) The Position Vector of the midpoint can be found using the formula: M = 1 (a + b)

6 D Coordinates: s seen earlier a point in Dimensions has coordinates in the form: (x, y,,z) The x-axis goes from Left to Right, the y-axis goes In & Out and the z-axis goes Up & Down. 1. cuboid is shown in the diagram opposite with Point U(,, ). a) State the coordinates of all the other points: O(0, 0, 0) P(, 0, 0) Q(,, 0) R(0,, 0) S(0, 0, ) T(, 0, ) V(0,, ) b) State the components of the FCE diagonal VT and the SPCE diagonal PV and calculate the magnitudes of these vectors. VT = t v PV = v p 0 0 VT = ( 0) ( ) PV = ( ) ( 0) 0 VT = ( ) PV = ( ) 0 VT = x + y + z VT = + ( ) + 0 VT = VT = 0 VT = 5 PV = x + y + z PV = ( ) + + PV = PV = 9 PV = 9 c) If OP = u, OR = v and OS = w express the vector QS in terms of u, v and w QS = QP + PT + TS QS = OR + OS OP QS = v + w u

7 d) Prove that LPRV is a right angle. PR = r p RV = v r PV = v p PR = ( ) ( 0) RV = ( ) ( ) PV = ( ) ( 0) PR = ( ) RV = ( 0) PV = ( ) 0 PR = x + y + z PR = ( ) PR = PR = 0 RV = x + y + z RV = RV = RV = 9 RV = PV = x + y + z PV = ( ) + + PV = PV = 9 Now use the CONVERSE of PYTHGORIS from National 5: PR = 0 RV = PV = 9 PR = ( 0) = 0 RV = = 9 so PR + RV = = 9 PV = ( 9) = 9 Since PR + RV = PV the angle LPRV is a right angle. We will see shortly a quicker way of proving that Vectors are Perpendicular! e) Find the coordinates of the midpoint of ST Midpoint Formula Position Vector Midpoint 0 M = ( x +x y, +y z, +z ) M = 1 [( 0) + ( 0)] M = ( ,, ) OR M = (, 0, 6 ) M = 1 ( 0) = ( 0) 6 M = (, 0, ) M = (, 0, ) Vector Components Coordinates

8 Collinearity: s seen earlier in the STRIGHT LINE topic, points are said to be COLLINER, i.e. lie on a straight line if the gradient between pairs of points are equal (PRLLEL) and if one of the points is common to both Gradients. The same is true for D Vectors, remember for Parallel Vectors: v = ku We cannot find Gradients in D, but v = ku is the equivalent. 1. a) Prove that the points (0,, 5), (,, ) and C(10, 8, 0) are collinear. = b a = ( ) ( 0 5 C = c b 10 ) C = ( 8 ) ( 0 6 = ( ) C = ( 6 ) = ( ) C = ( ) 1 1 so = C ) so = C, Therefore is parallel to C and since is a common point,, & C are collinear. Statement MUST be written b) Find the ratio C = C =, C so C is

9 Dividing Lines: point can split a line joing other points in a given ratio as follows: m T n So T splits the line in the ratio m : n If we know the coordinates of points we can find the third point in ways: o lgebraicly using the fact that = b a o Or Using the SECTION FORMUL: t = n m+n a + m m+n b The point T can either split the line INTERNLLY or EXTERNLLY as shown below: m T n m n T 1. The point T divides the line in the ratio :. For the coordinates (, 5, 1) & (, 10, 6) T find the coordinates of the point T. Section Formula lgebraicly t = n m+n a + m m+n b t = ( 5 ) + ( 10) t = ( 5 ) + ( 10) t = ( ) + ( 6 ) t = ( ) = ( ) So T ( 16,, 16) Coordinates Components T = T (t a) = (b t) t a = b t 5t = b + a 5t = ( 10) + ( 5 ) t = ( 0) + ( 10) t = ( 0) t = ( ) 16 So T ( 16,, 16)

10 15. The point T divides the line externally in the ratio 5 :. For the coordinates (, 1, ) & (6,, 1) find the coordinates of the point T. 5 T = T Scalar Product: (b a) = (t b) b a = t b t = 5b a 6 t = 5 ( ) ( 1) t = ( 0) ( ) 5 8 t = ( 18) 8 t = ( 6) So T (8, 6, 1) 1 5 = The SCLR PRODUCT is as close to multiplying vectors together as we can get! The Scalar Product is sometimes known as the DOT Product as it is written in the form: a. b The Scalar Product can be calculated in ways: o The component form: a. b = a 1 b 1 + a b + a b o The Modulus form: a. b = a b cos θ Where θ is the angle betweem the vectors a and b To use the Modulus form above both vectors must either be moving away from or towards the angle: oth moving oth moving a is moving towards away from angle towards the angle and b is moving away from the angle. We can reaarange the rd diagram to allow us to find the Dot Product: 180 θ

11 5 16. Find a. b given that a = ( ) and b = ( ) 1 a. b = a 1 b 1 + a b + a b a. b = 5 + ( ) + ( 1) a. b = 15 + ( 1) + ( ) = Find PQ. PR given that P(,, 9), Q(, 1, ) & R(6,, 1) PQ = q p PR = r p 6 PQ = ( 1) ( ) PR = ( ) ( ) PQ = ( ) PR = ( 0 ) 5 8 PQ. PR = a 1 b 1 + a b + a b PQ. PR = ( 5) ( 8) PQ. PR = = 18. For each diagram below calculate a. b : a) a. b = a b cos θ a. b = 7 cos 60 a. b = a. b = 10.5 b) 5π 6 = 150 b = x + y + z θ = b = + ( 1) + θ = 0 b = a. b = a b cos θ a. b = 1 cos 0 a. b = 1 a. b = 1 = b = 1

12 ngle etween Vectors: We can reaarange the Scalar Product to allow us to be able to find the ngle inbetween the vectors: cos θ = a. b a b = a 1 b 1 + a b + a b a b 19. Calculate the angle, θ, between the vectors, m = i + j k and n = i + j + k m = i + j k n = i + j + k m = ( ) n = ( 1) m = x + y + z n = x + y + z m = + + ( ) n = m = n = m = 9 n = 6 m. n = a 1 b 1 + a b + a b cos θ = m. n = ( ) cos θ = m. n = ( 6) = 10 m.n m n θ = cos θ = Find LDEF when D(, 5, 1), E(,, ) & F(1, 7, ) s both vectors MUST travel in the same direction start from the outsides of the angle s name and work in! DE = e d FE = e f 1 DE = ( ) ( 5) FE = ( ) ( 7) 1 5 DE = ( ) = a FE = ( 10) = b a = x + y + z b = x + y + z a = ( 5) + ( ) + b = ( ) a = b = a = 8 b = 10

13 a. b = a 1 b 1 + a b + a b cos θ = a. b = ( 5) ( ) + ( ) 10 + cos θ = a. b = 0 + ( 0) + 6 = 6 Perpendicular Vectors: a.b a b θ = cos θ = 8.97 If vectors a and b are perpendicular then a. b = 0 1. Show that vectors a and b are perpendicular when a = i 5j + k and b = i + j + k a = i 5j + k b = i + j + k a = ( 5) b = ( ) 1 a. b = a 1 b 1 + a b + a b a. b = + ( 5) + 1 a. b = 1 + ( 15) + = 0 Since a. b = 0 a is perpendicular to b.. Find the value of k if the direct line segments DE = ( ) and FG = ( 5) are perpendicular. k Since DE is perpendicular to FG DE. FG = 0 a 1 b 1 + a b + a b = 0 ( ) + ( ) ( 5) + k = k = 0 + k = 0 k = k = 1.5

14 Scalar Product Properties: For Vectors vectors a, b and c the following properties exist: o a. b = b. a o a. (b + c) = a. b + a. c o a. a = a. Given that vectors p and q are perpendicular and p =, q =, r = calculate the exact value of p. (q + r) ngle between vectors p and r is: = 0 p. (q + r) = p. q + p. r p r 60 q = 0 + p r cos θ Since vectors p and q are perpendicular. p. q = 0 = cos 0 = = = 9 =. For the diagram opposite and a =, b = calculate the exact value of a. (a + b + c) a a. (a + b + c) = a. a + a. b + a. c = + cos 0 + cos = = = = 8