Presentation on theme: "b a The Vector or x Product"— Presentation transcript:
1 b a The Vector or x Product In C4 the scalar product was covered a . b = |a||b|cosq Pronounced a dot babSuppose the angle between two vectors a and b isThe scalar product is written as a . b and is defined asThe 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 = 1as the angle between these unit vectors is zero and cos0 = 1i . j = 0 j . k = 0 i . k = 0j . i = 0 k . j = 0 k . i = 0So 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,810610
4 6 6 10 10 We also have different answers for the same size vectors at different angles.610610The 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 multiplyHowever, six of these are perpendicular.e.g. is along the x-axis and is along the y-axis.24The scalar product of these components is zero.The other three multiplications e.g 2 andinvolve parallel components so the angle between them is zero.
6 SUMMARYFor the scalar product of 2 column vectors,e.g.andwe multiply the “tops”,“middles”and “bottoms”and add the results. So,
7 e.g.1 Find the scalar product of the vectors andSolution:
8 Perpendicular Vectors If a.b = 0Theneither a = 0 orb = 0 ora = 0 or b = 0 are trivial cases as they mean the vector doesn’t exist. So, we must haveThe 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. isthe product of 2 vectors divided bythe product of the 2 magnitudes of the vectors
10 e.g. Find the angle between andSolution: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 byFind the cosine of angle ABC.Solution:We always sketch and label a triangleB( any shape triangle will do )CBUT the a and b of the formula are not the a and b of the question.AWe need the vectors and
14 Finding Angles between Lines e.g.aWith 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:andwhere
16 e.g. Find the acute angle, a, between the lines andSolution:whereand
17 e.g. Find the acute angle, a, between the lines andSolution:whereand
18 e.g. Find the acute angle, a, between the lines andSolution:whereand(nearest degree)Autograph
19 The Vector ProductThe vector product is defined as a b = |a| |b|sin 𝐧𝐧 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 ba b is out of the paper as thedirection of 𝐧 is outa b is into the paper as thedirection of 𝐧 is in
20 a b = |a||b|sin 𝐧Using this definition:i i = j j = k k =as the angle between these unit vectors is zero and sin0 = 0ij = jk = ki =ik = ji = kj =kij–j–k–i
21 a b = (a1i + a2j + a3k) (b1i + b2j + b3k) a b = |a||b|sin 𝐧a b = (a1i + a2j + a3k) (b1i + b2j + b3k)= a1b2k – a1b3j – a2b1k + a2b3i + a3b1j – a3b2i= (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)kThis result is on pg 4 of the formula book written as a column vectorLink to applet
22 So the vector is perpendicular to ExSo the vector is perpendicular toThis fact will be essential in understanding the equation of a planeLink to demoLink to worksheet