Vectors – Ch 11. What do you know? The basics … A B 6 3 a or a Column vector –a–a Negative of a vector a A B A B.

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Presentation transcript:

Vectors – Ch 11

What do you know?

The basics … A B 6 3 a or a Column vector –a–a Negative of a vector a A B A B

The magnitude A B 6 3

A B C D Equal Vectors Same magnitude and direction Component form

x y z 5 –3 –2 C D Which all work in 3d as well What is the column vector for DC? What is the vector in component form? What is the magnitude of CD? Why is modulus important in magnitude?

The magnitude of the vector is given by the length of the line. The direction of the vector is given by the arrow on the line. Interchanging between component form and magnitude/ direction Change the vector from component form into magnitude and direction Change the vector from magnitude and direction into component form (6, 30 o )

A B Answer (5, degrees)

A B (6, 30 o ) Answer 5.20i + 3j

Interchanging between component form and magnitude/ direction Change the vector from component form into magnitude and direction Change the vector from magnitude and direction into component form Read Example 11.3 on page 278 Do Exercise 11a in page 281/282

Page 282 Ex 11A

a 2a2a Multiplying by scalars A vector can be multiplied by a scalar. When a vector is multiplied by a scalar the resulting vector lies either parallel to the original vector or on the same line.

a b a + b Adding Vectors For example, suppose and. In general, if and, then.

The parallelogram law for adding vectors

Find vector c such that 2c + a = b. Suppose that and. Start by rearranging the equation to make c the subject. 2c + a = b 2c = b – a c = (b – a) Your turn

A grid of congruent parallelograms

Unit base vectors The horizontal unit base vector,, is called i.The vertical unit base vector,, is called j. A vector with a magnitude of 1 is called a unit vector. More terminology ….

Unit Vectors a

Page 287 Ex 11B

Page 287 Ex 11B

Page 287 Ex 11B

Using vectors

Position vectors Positional vectors start at the origin. Example Points L, M and N have coordinates (4,3) (-2,-1) and (2,2) Write down in component form the position vector of L and the vector What can you tell about lines OL and MN?

Example Points L, M and N have coordinates (4,3) (-2,-1) and (2,2) Write down in component form the position vector of L and the vector What can you tell about lines OL and MN? Lines OL and MN are parallel and of equal length

O q p Point P has coordinates (4, 6). Point Q has coordinates (3, –2). The position vectors are given by: Write the vector PQ as a column vector. Let M be the mid-point of the line PQ. What is the position vector of the point M ?

Write the vector PQ as a column vector. O P p q Q P –p q – p O q Q So

What is the position vector of the point M ? O P p q Q M and so

Page 297 Ex 11C Q 1

Vector Equation of a line O P p q Q What is the equation of the line passing through P and Q?

Vector Equation of a line O P p q Q General form of the Vector Equation of a line is given by

Page 297 Ex 11C Q 2

Converting to Cartesian form If we want to convert between we need to recognise that r is a column vector made up from x and y coordinates.

Converting from Cartesian form If we want to convert back then we need to start with a positional vector Now convert the gradient into a vector Combine to give: NOTE: Any Vector equation is valid if it is equivalent (our answers were different) It is therefore best to often simplify the vector where possible

Page 297 Ex 11C Q 3 + 4

Intersection of 2 vector lines How do we find the solution to 2 intersection lines (in Cartesian form)? Substituting in gives:

Page 297 Ex 11C Q 5,6,7

The angle between 2 vectors O A a b B Consider the angle between a and b. How would we calculate this? Re-arranged cosine rule gives: a - b If we allocate a and b with Cartesian coordinates we can show that: See Page 300 for proof

The angle between 2 vectors O A a b B a - b The numerator is known as the scalar product and can be re- written as:or

The angle between 2 vectors Calculate the angle between the vectors p and q: O P p q Q The scalar product is : NOTE: p and q are perpendicular as the scalar product is 0 For 3d ‘scalar products’ the same method is extended i.e.

Page 302 Ex 11d