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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 scalar.

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Presentation on theme: "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 scalar."— Presentation transcript:

1 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 scalar product is sometimes called the “dot” product. The result of the scalar product is a scalar quantity not a vector ! The Vector or x Product In C4 the scalar product was covered a. b = |a||b|cos  Pronounced a dot b

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 = 0j  k = 0i  k = 0 j  i = 0 k  j = 0k  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,

4 the same size vectors at different angles. We also have different answers for 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 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. and The other three multiplications e.g 2 and 3... 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”, and add the results. So, “middles”and “bottoms”

7 e.g.1 Find the scalar product of the vectors and Solution:

8 Perpendicular Vectors Then either a = 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. b = 0 or If a.b = 0

9 the product of 2 vectors divided by 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 the 2 magnitudes of the vectors

10 Solution: e.g. Find the angle betweenand ( 3s.f. ) Tip: If at this stage you get zero, STOP. The vectors are perpendicular.

11 When solving problems, we have to be careful to use the correct vectors. e.g. The triangle ABC is given by Solution: ( any shape triangle will do ) We always sketch and label a triangle A B C BUT the a and b of the formula are not the a and b of the question. We need the vectors and Find the cosine of angle ABC.

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14 Finding Angles between Lines e.g. With lines instead of vectors, we have 2 possible angles. We usually give the acute angle.  We use the 2 direction vectors only since these define the angle. ( If the obtuse angle is found, subtract from 180. )

15 where Solution: and e.g. Find the acute angle, , between the lines

16 and Solution: whereand e.g. Find the acute angle, , between the lines

17 and Solution: whereand e.g. Find the acute angle, , between the lines

18 and Solution: whereand (nearest degree) Autograph

19 The Vector Product

20 Using this definition: i  i = j  j = k  k = as the angle between these unit vectors is zero and sin0 = 0 i  j = j  k = k  i = i  k = j  i = k  j = 0 ki j –j–k–i

21 a  b = (a 1 i + a 2 j + a 3 k)  (b 1 i + b 2 j + b 3 k) = (a 2 b 3 – a 3 b 2 )i + (a 3 b 1 – a 1 b 3 )j + (a 1 b 2 – a 2 b 1 )k This result is on pg 4 of the formula book written as a column vector = a 1 b 2 k – a1b3j a1b3j – a 2 b 1 k + a 2 b 3 i + a 3 b 1 j – a3b2ia3b2i Link to applet

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


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