EXAMPLE 1 The vector a is directed due north and |a| = 24. The vector b is directed due west and |b| = 7. Find |a + b| |a| = 24 |b| = 7 |a + b| |a + b| 2 = |a| 2 + |b| 2 |a + b| 2 = |a + b| 2 = 625 |a + b| = 25

EXAMPLE 2

We have to prove that is some multiple of As is a multiple of, and are parallel

EXAMPLE 3 We need to find and. If they are parallel, as they both start at B, the three points, B, P and M must be on the same line. and have identical vector parts. This means they are parallel. As they both start at B; B, P and M lie on same line.

EXAMPLE 4

= 0 b 7/4b7/4b

Equating coefficients of a and b a:a: b:b:

2 4 6 y x a b A B CARTESIAN COMPONENTS OF VECTORS This diagram shows three vectors: i j aa If a = xi + yj then a can be written as a If a = xi + yj then the a modulus (magnitude)of a is:

EXAMPLE abaijbij Two vectors, a and b are defined by a = 3i + 5j and b = 7i – 3j. abab Find(a)3a – 2b(b)|3a – 2b| ab (c)A unit vector in the same direction as 3a – 2b (a) ab 3a - 2b = ijij 3(3i + 5j) – 2(7i – 3j) ij = -5i + 21j (b) ab| |3a - 2b| = (c) ab Unit vector of |3a - 2b| =

x y z THREE DIMENSIONAL VECTORS i j k aa If a = xi + yj + zk then a can be written as a If a = xi + yj + zk then the a modulus (magnitude)of a is: A(3, 5, -2) B(-1, 2, 6) Find: (a) (b) (c)A unit vector in the same direction as (a) (b) (c)

EXAMPLE aijka Given that a = 5ti + 2tj + tk and that |a| = 3√10, find the possible values of t.

EXAMPLE The points A and B have position vectorsand respectively. (a)Find (b)Find, in terms of t,

(c)Find the value of t that makesa minimum (d)Find the minimum value of If is a minimum then is a minimum

SCALAR PRODUCT SCALAR PRODUCT SCALAR PRODUCT is one method of method of multiplying two vectors together. It also allows us to find the angle between two vectors. θ a b Notice that both vectors are pointing AWAY from X. When using this formula, the vectors should ALWAYS be pointing away from the point of intersection. X aijk If, a = a 1 i + a 2 j + a 3 k and bijk b = b 1 i + b 2 j + b 3 k a.b Then, a.b = a 1 b 1 + a 2 b 2 + a 3 b 3

SPECIAL CASES OF SCALAR PRODUCT a b a b θ = 90° cosθ = 0 a.b a.b = 0 θ = 0° cosθ = 1 a.b ab a.b = |a||b| Dot product = 0 when vectors are perpendicular Dot product is product of magnitudes of vectors when vectors are parallel

EXAMPLE aijkbijkab If, a = 6i – 7j + 12k and b = –2i + j + k, find the angle between a and b. a.b a.b =(6 x -2) + (-7 x 1) + (12 x 1)= -7 a= |a| = b= |b| =

EXAMPLE ijjk The angle between the vectors i + 3j and j + λk is 60°. Show that a.b a.b =(1 x 0) + (3 x 1) + (0 x λ)= 3 a= |a| = b= |b| =

EXAMPLE Given that the points A and B have coordinates (7, 4, 4) and (2, -2, -1) respectively, use a vector method to find the value of cosAOB, where O is the origin. Prove that the area of triangle AOB is A(7, 4, 4) B(2, -2, -1) O θ 9 3 Area = ½ ab sinθ

VECTOR EQUATION OF A STRAIGHT LINE Suppose a straight line passes through the point A a with position vector a d It is also parallel to the vector d O A R a d r You can find the position vector of any point on this line, say r point R (position vector r), using this information. ra To find r, you travel along a from d O to A and then some multiple, t, of d till you get to R. dtddtd So: r r = a d + td The VECTOR EQUATION OF A LINE

EXAMPLE Find a vector equation of the straight line which passes through the point A with position vector and is parallel to Position vector of point a Lying on line, a Direction vector of d line, d rad Vector equation of a line: r = a + td r r =

EXAMPLE Calculate to 1 decimal place, the distance between P, where t = 1, and the point Q where t = 5 on the line with equation: rikijk r = (2i + 5k) + t(-3i + 4j – k) pikijk p = (2i + 5k) + 1(-3i + 4j – k) ijk = -i + 4j + 4k qikijk q = (2i + 5k) + 5(-3i + 4j – k) ij = -13i + 20j qp q – p ijk = -12i + 16j – 4k

EXAMPLE Find a vector equation of the straight line which passes through ijk the point A with position vector -7i + 6j + 2k and the point B with ijk position vector 3i + j + 2k. A(-7,6,2) B(3,1,2) O a b d Vector equation of rad a line, r = a + td d ba d = b - a d ijkijk d = 3i + j + 2k – (-7i + 6j + 2k) d ij d = 10i – 5j d ij d = 5(2i – j ) We only need the vector part as this is the direction vector of the line Vector equation of the line: rijkij r = (-7i + 6j + 2k) + t(2i – j)

INTERSECTION OF TWO LINES EXAMPLE Determine whether or not each pair of lines intersects. If they do, determine the coordinate of the point of intersection. (a) and Write each vector equation as one vector:

Use two of your rows to form two simultaneous equations to find the value of t and s that might make the lines intersect t = s – 2 s – 4t = 3(1) 2t = 2s – 9 2s – 2t = 9(2) 2x(1)2s – 8t = 6(3) (1) – (3)6t = 3(3) t = ½ s = 5 Now use the 3 rd row to check these solutions work ( ½ ) = 12 – 5 7 = 7 Lines intersect when t = ½ and s = 5 Point of intersection: (3, 1, 7)

and (b) 3 + 2t = 3 + 6s 6s – 2t = 0(1) t – 3 = 4 – 4s 4s + t = 7(2) 2x(2)8s + 2t = 14(3) (1) + (3)14s = 14 s = 1 t = 3 1 – 4(3) = ≠ 3 These two lines DO NOT intersect

EXAMPLE Find the acute angle between the lines with the vector equations: r 1 ikijk r 1 = (2i + k) + t(11i + 5j – 3k) r 2 ijijk and r 2 = (i + j) + s(-3i + 5j + 4k) To find the angle between two lines, find the angle between their DIRECTION VECTORS. d 1 =ijk d 1 = 11i + 5j – 3k d 2 =ijk d 2 = -3i + 5j + 4k |d 1 | = √155 |d 2 | = √50 d 1. d 2 = d 1. d 2 = (11 x -3) + (5 x 5) + (-3 x 4) = -20 This is an OBTUSE angle. Acute angle = 76.9⁰