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# s1 - Solution to Set 1 1 1(a a = b 2 therefore a b(b Let a...

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Solution to Set 1 1. (a) a = 1 2 b therefore a b (b) Let a = cb , then i + j = cj + ck . There is no c that can create an i component. Therefore, a is not parallel to b . 2. a · b = 2 3 4 = 5 cos θ = 5 3 29 = 0 . 536 θ = 2 . 137 radian = 122 . 4 3. Determine whether PQ and PR are perpendicular PQ = (2 , 0 , 1) ( 1 , 3 , 0) = (3 , 3 , 1); PR = ( 1 , 1 , 6) ( 1 , 3 , 0) = (0 , 2 , 6) PQ · PR = 6 6 = 0 4. a × b = i + 2 j k c · ( a × b ) = 1 + 2 + 1 = 4 5. The diagonals are u + v and ± ( u v ). If their lengths are equal, then | u + v | 2 = | u v | 2 = ( u ± v ) · ( u ± v ). Therefore, 2 u · v = 2 u · v u · v = 0 u v the figure is a rectangle. 6. Let P and A be the position vectors of points
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