s2 - Solution to Set 2 1. Two of the sides of the triangle...

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Solution to Set 2 1. Two of the sides of the triangle are represented by the line segments ±a =(3 , 3 , 3) (2 , 2 , 2) = (1 , 2 , 1) and ± b =(5 , 1 , 2) (2 , 1 , 2) = (3 , 0 , 0). The area of the triangle is given by 1 2 | × ± b | . Since × ± b =( ± i +2 ± j + ± k ) × 3 ± i =3 ± j 6 ± k , the area is 1 2 36 + 9 = 3 2 5. 2. vector form: ± r = 2 ± i + ± j + t (3 ± i 2 ± j + ± k ) −∞ <t< parametric form: x = 2+3 t y =1 2 t z = t 3. ± r ± P 0 x +1) ± i +( y 1) ± j z 3) ± k ± N · ( ± r ± P 0 )= 2( x + 1) + 15( y 1) 1 2 ( z 3) = 0 equation: 4 x +30 y z 31 = 0 4. (5 , 2 , 5) (2 , 1 , 4) = (3 , 3 , 1) (2 , 1 , 3) (2 , 1 , 4) = (0 , 2 , 1) Let ± N ± i +3 ± j + ± k ) × (2 ± j ± k 5 ± i ± j +6 ± k equation: [( x 5) ± i y 2) ± j z 5) ± k ] · ( 5 ± i ± j ± k 5 x y z 11 = 0 5. We are going to show that the vector a ± i + b ± j + c ± k is perpendicular to any line lying on the plane (or any directed line segments on the plane).
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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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