Vectors Lesson 2 recap Aims: • To know the Scalar Product formula and be able to use it to determine if two vector are perpendicular or parallel. • To be able to use the scalar product to find the angle between vectors and direction vectors when a line is given in vector form. Starter – terminology match cards in pairs

In general, if a = a1i + a2j + a3k and b = b1i + b2j + b3k then The scalar product Two vectors can be multiplied together to give the scalar product, also known as the dot product. The result of this multiplication is a s_____________ quantity, hence the name. The scalar product of two vectors a and b is defined as: where θ is the angle between a and b when they are placed tail to tail. Note that two vectors can also be multiplied together to give the vector product (or cross product). However, this is not covered in this course. Also, if two vectors are placed ‘nose to nose’ the same value for θ will be given. Note that θ is always taken to be between 0° and 180°. In general, if a = a1i + a2j + a3k and b = b1i + b2j + b3k then a.b = a1b1+ a2b2 + a3b3.

Perpendicular and parallel vectors We have seen that, since cos 90° = ______: If a.b = 0 then a and b are perpendicular, as long as neither are the zero vector. In particular, for the unit base vectors i, j and k: i.j = i.k = j.i = j.k = k.i = k.j = Also, if two vectors are parallel the angle between them is taken as 0°. Since cos 0° = ______ we can conclude from the scalar product that: If the scalar product of two non-zero vectors a and b is 0 then a and b are perpendicular. If two vectors a and b are parallel, a.b = |a||b|. In particular, for the unit base vectors i, j and k: i.i = j.j = k.k = 1×1 = 1

Scalar products of vectors in component form Find the scalar product of a.b = Prove that the vectors a = –3i + j + 2k and b = 2i + 8j – k are perpendicular. 10 a.b = Since a and b are both non-zero vectors, and their scalar product is _____, they must be p______________________.

On w/b Find the scalar product of a.b = Find the scalar product of a = 4i +3j + 2k and b = 4i - 2j – 5k and state whether or not they are perpendicular. 10 a.b =

Finding the angle between two vectors A useful application of the scalar product is in finding the angle between two vectors. To do this we can write the scalar product in the form: For example, Find the angle between the vectors a and b where and Note that two vectors can also be multiplied together to give the cross product. However, this is not covered in this course. Also, is two vectors are placed ‘nose to nose’ the same value for θ will be given.

On w/b Find the acute angle between the vectors a and b where and Note that two vectors can also be multiplied together to give the cross product. However, this is not covered in this course. Also, is two vectors are placed ‘nose to nose’ the same value for θ will be given.

Exam Questions 1. The position vectors of three points A, B and C relative to an origin O are given respectively by = 7i + 3j – 3k, = 4i + 2j – 4k = 5i + 4j – 5k. (i) Find the angle between AB and AC. [6] (ii) Find the area of triangle ABC. [2]

Exam Question (ii)

On w/bs The position vectors of three points A, B and C relative to an origin O are given respectively by = 3i + 2j – 3k, = 2i + j – 4k = i + 4j – k. Find the angle between AB and AC.

Finding the angle between two vectors lines To find the angle between two vector lines we only need consider the direction of the lines. Re-call: r = a + t(b – a) 2. Lines L1, L2 and L3 have vector equations L1: r = (5i – j – 2k) + s(–6i + 8j – 2k), L2: r = (3i – 8j) + t( i + 3j + 2k), L3: r = (2i + j + 3k) + u(3i + cj + k). (i) Calculate the acute angle between L1 and L2. [4] (ii) Given that L1 and L3 are parallel, find the value of c. [2] Let p = and q =

Finding the angle between two vectors lines So the angle between them will be (ii) Since the scalar product of a and b is negative we expect the angle between them to be obtuce. If the lines are parallel then they must have the same direction.

On w/b Lines L1 and L2 have vector equations L1: r = (i – j – 9k) + s(–6i + 8j – 4k), L2: r = (2i – 8j) + t( 3i + 2j + 1k), Calculate the acute angle between L1 and L2. Do exercise 9G page 127. Qu 2,4,6 and exercise 9H page 131. Qu’s 3, 4 & 5.

Exam-style question last 10 mins Relative to a fixed origin O, the points A and B have position vectors –2i – 2j – k and mi + 4j – 7k respectively, where m is a constant. C is a point such that OABC is a rectangle. Find the value of the constant m. Write down the coordinates of the point C. Show the ratio of the rectangle’s height to its width is 1 : 3. a) Since OABC is a rectangle must be perpendicular to . So = where =

Exam-style question b) OC is the side opposite AB and so the position vector of the point C is the same the vector .  The coordinates of C are ( ). c) The height of the rectangle is given by the magnitude of the vector and the width of the rectangle is given by the magnitude of the vector .