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The Vector or x Product In C4 the scalar product was covered Pronounced a dot b a . b = |a||b|cos Suppose the angle between two vectors a and b is b a The scalar product is written as a . b and is defined as a . b a b cos 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. a . b a b cos 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.i=0 j.k=0 k.j=0 i.k=0 k.i=0 So an answer is obtained when the components of the vector are in the same direction. SUMMARY For the scalar product of 2 column vectors, e.g. a1 a a 2 a 3 we multiply the “tops”, and b1 b b2 b3 “middles” and “bottoms” and add the results. So, a . b a1 b1 a 2 b2 a 3 b3 e.g.1 Find the scalar product of the vectors 2 a 1 3 and 4 b 2 2 a . b a1 b1 a 2 b2 a 3 b3 Solution: 2 4 a .b 1 . 2 3 2 ( 2 )( 4 ) ( 1 )( 2 ) ( 3 )( 2 ) 8 2 6 12 Perpendicular Vectors If a.b = 0 a b cos 0 Then either a = 0 or b = 0 or cosθ= 0 a = 0 or b = 0 are trivial cases as they mean the vector doesn’t exist. So, we must have co sθ = 0 θ 90 The vectors are perpendicular. Finding Angles between Vectors The scalar product can be rearranged to find the angle between the vectors. a . b a b cos a .b cos ab 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 e.g. Find the angle between b 2i j k Solution: a i j k and a .b cos ab a . b ( 1 )( 2 ) ( 1 )( 1 ) ( 1 )( 1 ) 2 Tip: If at 2 this2 stage 2 you get zero, STOP. 2 2 2 a The vectors 1 1 are 1perpendicular. 3 , b 2 1 1 cos 2 3 6 61 9 ( 3s.f. ) 6 When solving problems, we have to be careful to use the correct vectors. e.g. The triangle ABC is given by A ( 2 , 1, 3 ) , B (1 , 1 , 0 ) , C ( 2 , 1, 3 ) Find the cosine of angle ABC. Solution: We always sketch and label a triangle B ( any shape triangle will do ) A a .b cos ab C BUT the a and b of the formula are not the a and b of the question. We need the vectors and AB CB A ( 2 , 1, 3 ) , B (1 , 1 , 0 ) , = ` − ` C ( 2 , 1, 3 ) 1 2 1 AB b a 1 1 2 0 3 3 1 2 1 CB b c 1 1 2 0 3 3 1 AB 2 3 1 CB 2 3 1 1 1 4 9 4 AB . CB 2 . 2 3 3 AB 12 22 32 14 , CB AB . CB cos cos AB CB cos 12 2 2 3 2 4 14 2 7 14 2 4 cos 14 7 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 180 . ) We use the 2 direction vectors only since these define the angle. e.g. Find the acute angle, a, between the lines 1 0 r 3 s 1 2 2 Solution: a .b cos ab and where 2 1 r 0 t 1 1 2 e.g. Find the acute angle, a, between the lines 1 0 r 3 s 1 2 2 Solution: a .b cos ab and where 2 1 r 0 t 1 1 2 0 a 1 and 2 e.g. Find the acute angle, a, between the lines 1 0 r 3 s 1 2 2 Solution: a .b cos ab and where 2 1 r 0 t 1 1 2 0 1 a 1 and b 1 2 2 e.g. Find the acute angle, a, between the lines 1 0 r 3 s 1 2 2 Solution: and 2 1 r 0 t 1 1 2 0 1 a .b cos where a 1 and b 1 ab 2 2 a . b ( 0 )( 1 ) ( 1 )( 1 ) ( 2 )( 2 ) 5 a 1 2 2 cos 2 5 5 6 5, b 1 2 156 1 2 (nearest degree) 2 2 a 24 Autograph 6 The Vector Product The vector product is defined as a b = |a| |b|sin is a unit vector perpendicular to both a and b. To determine the direction of 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 is out a b is into the paper as the direction of is in a b = |a||b|sin Using this definition: ii=jj=kk= 0 as the angle between these unit vectors is zero and sin0 = 0 ij = k jk = i ki = j ik = –j ji = –k kj = –i a b = |a||b|sin a b = (a1i + a2j + a3k) (b1i + b2j + b3k) = a1b2k – a1b3j – a2b1k + a2b3i + a3b1j – a3b2i = (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)k a 2b 3 - a 3b 2 a 3 b 1 - a 1b 3 ab - a b 2 1 1 2 This result is on pg 4 of the formula book written as a column vector Link to applet Ex a1 a2 a 3 2 1 2 b 1 a 2b 3 - a 3b 2 b 2 a 3 b 1 - a 1b 3 b a b - a b 2 1 3 1 2 0 1 3 ( 2) 1 1 ( 2) 0 2 3 3 2 1 1 0 So the vector 5 6 2 5 6 5i 6 j 2k 2 is perpendicular to 2 1 2 and 0 1 3 This fact will be essential in understanding the equation of a plane Link to demo Link to worksheet