HM6-soln

# HM6-soln - Solution for HW6 November 5 2011 7.52 ~ A ~ B ~...

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Unformatted text preview: Solution for HW6 November 5, 2011 7.52 ~ A + ~ B + ~ C = (2 + 1 + 3) ~ i + (- 1 + 1- 2) ~ j + (1 + 2 + 4) ~ j = 6 ~ i- 2 ~ j + 7 ~ k. And the unit vector is ~ A + ~ B + ~ C ~ A + ~ B + ~ C = 6 ~ i- 2 ~ j + 7 ~ k 6 ~ i- 2 ~ j + 7 ~ k = 6 ~ i- 2 ~ j + 7 ~ k √ 89 . 7.56 proj ~ B ( ~ A + ~ C ) = ( ~ A + ~ C ) · ~ B ~ B = (2 + 3 , 1- 4 , 1 + 2) · (1 ,- 2 , 2) p 1 2 + (- 2) 2 + 2 2 = 5 + 6 + 6 √ 9 = 17 3 . 1 7.58 The rhombus has the property that ~ AB = ~ DC, ~ AB = ~ AD . The two diagonals can be expressed as ~ AB + ~ AD and ~ AB- ~ AD . Therefore ( ~ AB + ~ AD ) · ( ~ AB- ~ AD ) = ~ AB · ~ AB- ~ AD · ~ AD = ~ AB 2- ~ AD 2 = 0 shows that the two diagonals are perpendicular to each other. 7.62 Not necessarily true. When ~ A and ~ B- ~ C are parallel we also have ~ A × ~ B = ~ A × ~ C . 7.68 Denote the given four vetices as A,B,C,D. V olumn = 1 6 ~ AB · ( ~ AC × ~ AD ) 2 = 1 6 1- 2- 1- 1 2- 1- 2 1- 1- 1- 1 1- 2- 2- 1 1- 1 = 1 6 · 8 = 4 3 . 7.69 First we have ( ~ A × ~ B ) · ( ~ C × ~ D ) = ~ D · [( ~ A × ~ B ) × ~ C ] = ~ D · [( ~ A · ~ C ) ~ B- ( ~ B · ~ C ) ~ A ] = ( ~ A · ~ C )( ~ D · ~ B )- ( ~ B · ~ C )( ~ D · ~ A ) ....
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HM6-soln - Solution for HW6 November 5 2011 7.52 ~ A ~ B ~...

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