{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 12 Review - Stewart Calculus ET 5e 0534393217;12...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1, ! 1,2 ( ) 3 ( x ! 1) 2 +( y +1) 2 +( z ! 2) 2 =9 ( x +2) 2 +( y +3) 2 +( z ! 5) 2 = ! 2+4+9+25=36 ! 2, ! 3,5 ( ) 6. u " v = u v cos 45 # =(2)(3) 2 2 =3 2 u $ v = u v sin 45 # =(2)(3) 2 2 =3 2 ! u $ v 2 a +3 b =2 i +2 j ! 4 k +9 i ! 6 j +3 k =11 i ! 4 j ! k b = 9+4+1 = 14 a " b =(1)(3)+(1)( ! 2)+( ! 2)(1)= ! 1 a $ b = i 1 3 j 1 ! 2 k ! 2 1 = 1 ! 4 ( ) i ! 1+6 ( ) j + ! 2 ! 3 ( ) k = ! 3 i ! 7 j ! 5 k b $ c = i 3 0 j ! 2 1 k 1 ! 5 =9 i +15 j +3 k b $ c =3 9+25+1 =3 35 1. (a) By the formula for an equation of a sphere (see Section 13.1 [ ET 12.1]), an equation of the sphere with center and radius is . (b) Completing squares gives . Thus, the sphere is centered at and has radius 2. (a) (b) (c) (d) 3. . . By the right hand rule, is directed out of the page. 4. (a) (b) (c) (d) (e) , (f) 1 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; Review: Exercises
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
a " b $ c ( ) = 1 3 0 1 ! 2 1 ! 2 1 ! 5 = ! 2 1 1 ! 5 ! 3 0 1 ! 5 ! 2 3 0 ! 2 1 =9+15 ! 6=18 c $ c = 0 c a $ b $ c ( ) = a $ 9 i +15 j +3 k ( ) = i 1 9 j 1 15 k ! 2 3 = 3+30 ( ) i ! 3+18 ( ) j + 15 ! 9 ( ) k =33 i ! 21 j +6 k comp a b = b cos ! = a " b / a = ! 1 6 proj a b = ! 1 6 a / a ( ) = ! 1 6 i + j ! 2 k ( ) cos ! = a " b a b = ! 1 6 14 = ! 1 2 21 ! =cos ! 1 ! 1 2 21 % 96 # . 3,2, x " 2 x ,4, x =0 (3)(2 x )+(2)(4)+( x )( x )=0 x 2 +6 x +8=0 ( x +2)( x +4)=0 x = ! 2 x = ! 4. j +2 k ( ) $ i ! 2 j +3 k ( ) =[3 ! ( ! 4)] i ! (0 ! 2) j +(0 ! 1) k =7 i +2 j ! k 7 i +2 j ! k 7 2 +2 2 +( ! 1) 2 = 1 3 6 7 i +2 j ! k ( ) 7 3 6 i + 2 3 6 j ! 1 3 6 k ! 7 3 6 i ! 2 3 6 j + 1 3 6 k . u $ v ( ) " w = u " v $ w ( ) =2 u " w $ v ( ) = u " ! v $ w ( ) = ! u " v $ w ( ) = ! 2 v " u $ w ( ) = v $ u ( ) " w = ! u $ v ( ) " w = ! 2 u $ v ( ) " v = u " v $ v ( ) = u " 0 =0 (g) for any . (h) From part (e), . (i) The scalar projection is . (j) The vector projection is . (k) and 5. For the two vectors to be orthogonal, we need or 6. We know that the cross product of two vectors is orthogonal to both. So we calculate . Then two deg vectors orthogonal to both given vectors are , that is, and 7. (a) (b) (c) (d) 2 Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; Review: Exercises
Background image of page 2
a $ b ( ) " b $ c ( ) $ c $ a ( ) = a $ b ( ) " b $ c ( ) " a c ! b $ c ( ) " c a ( ) = a $ b ( ) " b $ c ( ) " a c = a " b $ c ( ) a $ b ( ) " c = a " b $ c ( ) a " b $ c ( ) = a " b $ c ( ) 2 (0, 0, 0) (1, 1, 1) (1, 0, 0) (0, 1, 1) <1, 1, 1> < ! 1, 1, 1>. ! <1, 1, 1>
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 10

Chapter 12 Review - Stewart Calculus ET 5e 0534393217;12...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online