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

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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 θ . Because | cos θ | ≤ 1 for all θ , we have: | A · B | = | A || B || cos θ | ≤ | A || B | . (4) Equality holds when | cos θ | = 1. This means that either θ = 0 or θ = π . The case θ = 0 means that A and B point in the same direction, while the case θ = π 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 ˆ ı + a 2 ˆ , B = b 1 ˆ ı + b 2 ˆ : ( 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 ˆ ı + a 2 ˆ + a 3 ˆ k , B = b 1 ˆ ı + b 2 ˆ + b 3 ˆ k : ( a 1 b 1 + a 2 b 2 + a 3 b 3 ) 2 ( a 2 1 + a 2 2 + a 2 3 )( b 2 1 + b 2 2 + b 2 3 ) (7) (c) Following the hint, define: f ( t ) = ( A + tB ) · ( A + tB ) . (8)
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