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

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Solution to Set 1 1. (a) ±a = 1 2 ± b therefore ± ± b (b) Let = c ± b ,then ± i + ± j = c ± j + c ± k . There is no c that can create an ± i component. Therefore, is not parallel to ± b . 2. · ± b =2 3 4= 5c o s θ = 5 3 29 = 0 . 536 θ . 137 radian = 122 . 4 3. Determine whether ± PQ and ± PR are perpendicular ± =(2 , 0 , 1) ( 1 , 3 , 0) = (3 , 3 , 1); ± =( 1 , 1 , 6) ( 1 , 3 , 0) = (0 , 2 , 6) ± · ± =6 6=0 4. × ± b = ± i +2 ± j ± k ± c · ( × ± b )=1+2+1=4 5. The diagonals are ±u + ±v and ² ( ). If their lengths are equal, then | + | 2 = | | 2 = ( ² ) · ( ² ). Therefore, 2 · = 2 · · =0 the Fgure is a rectangle. 6. Let ± P and ± A be the position vectors of points P and A . The directed line segments ± PO = ± P and
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This note was uploaded on 02/27/2010 for the course MATH MATH101 taught by Professor Chan during the Fall '09 term at HKUST.

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