pset1sol[1]

pset1sol[1] - 18.02 Spring 2009 Problem Set 1 Solutions 1....

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Unformatted text preview: 18.02 Spring 2009 Problem Set 1 Solutions 1. As in the hint, let ~ A =-- OP and ~ B =-- OR . Let M be the midpoint of the segment PQ . Then we have:-- OM = ~ A + 1 2 ~ B. (1) The diagonal vector- PR from P to R is- PR = ~ B- ~ A. (2) Let N be the point 1 / 3 of the way from P to R . Then:-- ON = ~ A + 1 3- PR = 2 3 ~ A + 1 3 ~ B = 2 3-- OM. (3) Because-- ON is a scalar multiple of-- OM , the vectors-- ON and-- OM are parallel. This shows that the point N lies on the line segment from O to M , which is what we wanted to show. 2. (a) The geometric definition of the dot product is ~ A ~ B = | ~ A || ~ B | cos . Because | cos | 1 for all , we have: | ~ A ~ B | = | ~ A || ~ B || cos | | ~ A || ~ B | . (4) Equality holds when | cos | = 1. This means that either = 0 or = . The case = 0 means that ~ A and ~ B point in the same direction, while the case = means ~ A and ~ B point in opposite directions. (b) The inequality in components becomes simpler if we square both sides: | ~ A ~ B | 2 | ~ A | 2 | ~ B | 2 (5) In 2D, where ~ A = a 1 + a 2 , ~ B = b 1 + b 2 : ( a 1 b 1 + a 2 b 2 ) 2 ( a 2 1 + a 2 2 )( b 2 1 + b 2 2 ) (6) In 3D, where ~ A = a 1 + a 2 + a 3 k , ~ B = b 1...
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pset1sol[1] - 18.02 Spring 2009 Problem Set 1 Solutions 1....

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