Download presentation

1
**b a The Vector or x Product**

In C4 the scalar product was covered a . b = |a||b|cosq Pronounced a dot b a b Suppose the angle between two vectors a and b is The scalar product is written as a . b and is defined as The dot must NEVER be missed out. The result of the scalar product is a scalar quantity not a vector ! The scalar product is sometimes called the “dot” product.

2
**Using this definition:**

i . i = j . j = k . k = 1 as the angle between these unit vectors is zero and cos0 = 1 i . j = 0 j . k = 0 i . k = 0 j . i = 0 k . j = 0 k . i = 0 So an answer is obtained when the components of the vector are in the same direction.

3
Notice that the scalar product uses the magnitudes, a and b, of the vectors as well as the angle between them, so, we get a different answer for: different size vectors at the same angle, 8 10 6 10

4
**6 6 10 10 We also have different answers for**

the same size vectors at different angles. 6 10 6 10 The scalar product was defined so that the answer is unique to any 2 vectors. It will enable us easily to find the angle between the vectors, even in 3 dimensions.

5
**2 3 -1 4 -3 -1 2 4 The Scalar Product of 2 Column Vectors Suppose and**

Then to form , there are quantities to multiply However, six of these are perpendicular. e.g. is along the x-axis and is along the y-axis. 2 4 The scalar product of these components is zero. The other three multiplications e.g 2 and involve parallel components so the angle between them is zero.

6
SUMMARY For the scalar product of 2 column vectors, e.g. and we multiply the “tops”, “middles” and “bottoms” and add the results. So,

7
**e.g.1 Find the scalar product of the vectors**

and Solution:

8
**Perpendicular Vectors**

If a.b = 0 Then either a = 0 or b = 0 or a = 0 or b = 0 are trivial cases as they mean the vector doesn’t exist. So, we must have The vectors are perpendicular.

9
**Finding Angles between Vectors**

The scalar product can be rearranged to find the angle between the vectors. Notice how careful we must be with the lines under the vectors. The r.h.s. is the product of 2 vectors divided by the product of the 2 magnitudes of the vectors

10
**e.g. Find the angle between**

and Solution: Tip: If at this stage you get zero, STOP. The vectors are perpendicular. ( 3s.f. )

11
**When solving problems, we have to be careful to use the correct vectors.**

e.g. The triangle ABC is given by Find the cosine of angle ABC. Solution: We always sketch and label a triangle B ( any shape triangle will do ) C BUT the a and b of the formula are not the a and b of the question. A We need the vectors and

12
𝐀𝐁 =𝐁`𝐬 −𝐀`𝐬 𝐀𝐁𝐁𝐀 𝐑𝐮𝐥𝐞

14
**Finding Angles between Lines**

e.g. a With lines instead of vectors, we have 2 possible angles. We usually give the acute angle. ( If the obtuse angle is found, subtract from ) We use the 2 direction vectors only since these define the angle.

15
**e.g. Find the acute angle, a, between the lines**

Solution: and where

16
**e.g. Find the acute angle, a, between the lines**

and Solution: where and

17
**e.g. Find the acute angle, a, between the lines**

and Solution: where and

18
**e.g. Find the acute angle, a, between the lines**

and Solution: where and (nearest degree) Autograph

19
The Vector Product The vector product is defined as a b = |a| |b|sin 𝐧 𝐧 is a unit vector perpendicular to both a and b. To determine the direction of 𝑛 use the right hand rule where finger 1 is a, finger 2 is b and the thumb is a b a b is out of the paper as the direction of 𝐧 is out a b is into the paper as the direction of 𝐧 is in

20
a b = |a||b|sin 𝐧 Using this definition: i i = j j = k k = as the angle between these unit vectors is zero and sin0 = 0 ij = jk = ki = ik = ji = kj = k i j –j –k –i

21
**a b = (a1i + a2j + a3k) (b1i + b2j + b3k)**

a b = |a||b|sin 𝐧 a b = (a1i + a2j + a3k) (b1i + b2j + b3k) = a1b2k – a1b3j – a2b1k + a2b3i + a3b1j – a3b2i = (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)k This result is on pg 4 of the formula book written as a column vector Link to applet

22
**So the vector is perpendicular to**

Ex So the vector is perpendicular to This fact will be essential in understanding the equation of a plane Link to demo Link to worksheet

Similar presentations

OK

55: The Vector Equation of a Plane © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules.

55: The Vector Equation of a Plane © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on special types of chromosomes male Ppt on first conditional game Download ppt on conservation of resources Ppt on charge coupled device manufacturer Ppt on power system transient stability Ppt on conference call etiquette you need to know Cathode ray tube display ppt on tv Ppt on latest gadgets in market Ppt on networking related topics about global warming Ppt on hard copy devices