HW6_sol - Chapter 5 SECTION 1 14 1(c cos = 10.65 221 46(d...

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Chapter 5 SECTION 1 1. (c) cos θ = 14 221 , θ 10 . 65 (d) cos θ = 4 6 21 , θ 62 . 19 3. (b) p =(4 , 4) T , x - p =( - 1 , 1) T p T ( x - p )= - 4+4=0 (d) p - 2 , - 4 , 2) T , x - p , - 1 , 2) T p T ( x - p - 8+4+4=0 4. If x and y are linearly independent and θ is the angle between the vectors, then | cos θ | < 1 and hence | x T y | = ± x ±± y ±| cos θ | < 6 8. (b) - 3( x - 4) + 6( y - 2) + 2( z +5)=0 11. (a) x T x = x 2 1 + x 2 2 0 (b) x T y = x 1 y 1 + x 2 y 2 = y 1 x 1 + y 2 x 2 = y T x 75

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76 Chapter 5 (c) x T ( y + z )= x 1 ( y 1 + z 1 )+ x 2 ( y 2 + z 2 ) =( x 1 y 1 + x 2 y 2 )+( x 1 z 2 + x 2 z 2 ) = x T y + x T z 12. The inequality can be proved using the Cauchy-Schwarz inequality as follows: ± u + v ± 2 u + v ) T ( u + v ) = u T u + v T u + u T v + v T v = ± u ± 2 +2 u T v + ± v ± 2 = ± u ± 2 ± u ±± v ± cos θ + ± v ± 2 ≤± u ± 2 ± u v ± + ± v ± 2 ± u ± + ± v ± ) 2 Taking square roots, we get ± u + v ±≤± u ± + ± v ± Equality will hold if and only if cos θ = 1. This will happen if one of the vectors is a multiple of the other. Geometrically one can think of ± u ± and ± v ± as representing the lengths of two sides of a triangle. The length of the third side of the triangle will be ± u + v ± . Clearly the length of the third side must be less than the sum of the lengths of the Frst two sides. In the case of equality the triangle degenerates to a line segment. 13. No. ±or example, if x 1 = e 1 , x 2 = e 2 , x 3 =2 e 1 , then x 1 x 2 , x 2 x 3 , but x 1 is not orthogonal to x 3 . 14. (a) By the Pythagorean Theorem α 2 + h 2 = ± a 1 ± 2 where α is the scalar projection of a 1 onto a 2 . It follows that α 2 = ( a T 1 a 2 ) 2 ± a 2 ± 2 and h 2 = ± a 1 ± 2 - ( a T 1 a 2 ) 2 ± a 2 ± 2 Hence h 2 ± a 2 ± 2 = ± a 1 ± 2 ± a 2 ± 2 - ( a T 1 a 2 ) 2 (b) If a 1 a 11 ,a 21 ) T and a 2 a 12 22 ) T , then by part (a) h 2 ± a 2 ± 2 a 2 11 + a 2 21 )( a 2 12 + a 2 22 ) - ( a 11 a 12 + a 21 a 22 ) 2 a 2 11 a 2 22 - 2 a 11 a 22 a 12 a 21 + a 2 21 a 2 12 ) a 11 a 22 - a 21 a 12 ) 2 Therefore Area of P = h ± a 2 ± = | a 11 a 22 - a 21 a 12 | = | det( A ) |
Section 2 77 15. (a) It θ is the angle between x and y , then cos θ = x T y ± x ±± y ± = 20 8 · 5 = 1 2 = π 3 (b) The distance between the vectors is given by ± x - y ± = ± 0 2 +2 2 +( - 6) 2 +3 2 =7 16. (a) Let α = x T y y T y and β = ( x T y ) 2 y T y In terms of these scalars we have p = α y and p T x = β . Furthermore p T p = α

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This note was uploaded on 04/18/2010 for the course FINANCE 1231854365 taught by Professor Wuyiling during the Spring '10 term at Nashville State Community College.

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HW6_sol - Chapter 5 SECTION 1 14 1(c cos = 10.65 221 46(d...

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