Calculus with Analytic Geometry by edwards & Penney soln ch12

Calculus with Analytic Geometry

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
y x R S P (2,3) R S x y P (–10, –20) (1,–2) (4,2) u v x y w = u + v Section 12.1 C12S01.001: v = * RS = h 3 1 , 5 2 i = h 2 , 3 i . The position vector of the point P (2 , 3) and * RS are shown next. C12S01.002: v = * RS = h 1 ( 2) , 4 ( 3) i = h 3 , 7 i . C12S01.003: v = * RS = h− 5 5 , 10 10 i = 10 , 20 i . The position vector of the point P and * RS are shown next. C12S01.004: v = * RS = h 15 ( 10) , 25 20 i = h 25 , 45 i . C12S01.005: w = u + v = h 1 , 2 i + h 3 , 4 i = h 1+3 , 2+4 i = h 4 , 2 i . The next Fgure illustrates this computation in the form of the triangle law for vector addition (see ±ig. 12.1.6 of the text). C12S01.006: u + v = h 4 , 2 i + 2 , 5 i = h 4 2 , 2+5 i = h 2 , 7 i . C12S01.007: Given: u =3 i +5 j , v =2 i 7 j : 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
(5,–2) (3,5) u v u + v u + v =3 i +5 j +2 i 7 j =(3+2) i +(5 7) j =5 i 2 j . The next fgure illustrates the triangle law For vector addition using u and v . C12S01.008: u + v = h 7 10 , 5+0 i = h− 3 , 5 i = 3 i j . C12S01.009: Given: a = h 1 , 2 i and b = 3 , 2 i . Then: | a | = p (1) 2 +( 2) 2 = 5 , |− 2 b | = |h− 6 , 4 i| = 36 + 16 = 2 13 , | a b | = |h 1 ( 3) , 2 2 = 16 + 16 = 4 2 , a + b = h 1 3 , 2+2 i = 2 , 0 i , 3 a 2 b = h 3 , 6 i−h− 6 , 4 i = h 3 ( 6) , 6 4 i = h 9 , 10 i . C12S01.010: Given: a = h 3 , 4 i and b = 4 , 3 i . Then: | a | = 9+16 = 25 = 5 , 2 b | = 8 , 6 = 64+36 =10 , | a b | = 3 ( 4) , 4 3 = 49 + 1 = 5 2 , a + b = h 3 4 , 4+3 i = 1 , 7 i , 3 a 2 b = h 9 , 12 8 , 6 i = h 9 ( 8) , 12 6 i = h 17 , 6 i . C12S01.011: Given: a = 2 , 2 i and b = 3 , 4 i . Then: | a | = 4+4 = 8=2 2 , 2 b | = 6 , 8 = 36+64 =10 , | a b | = 2 ( 3) , 2 ( 4) = 1 , 2 = 1+4 = 5 , a + b = 2 3 , 2 4 i = 5 , 6 i , 3 a 2 b = 6 , 6 6 , 8 i = 6 ( 6) , 6 ( 8) i = h 0 , 2 i . 2
Background image of page 2
C12S01.012: Given: a = 2 h 4 , 7 i = h− 8 , 14 i and b = 3 4 , 2 i = h 12 , 6 i . Then: | a | = 64 + 196 = 260 = 2 65 , |− 2 b | = |h− 24 , 12 i| = 576 + 144 = 720 = 12 5 , | a b | = 8 12 , 14 6 = 20 , 20 = 400 + 400 = 20 2 , a + b = 8+12 , 14+6 i = h 4 , 8 i , 3 a 2 b = 24 , 42 i−h 24 , 12 i = 24 24 , 42 12 i = 48 , 54 i . C12S01.013: Given: a = i +3 j and b =2 i 5 j . Then: | a | = | i j | = 1+9 = 10 , 2 b | = | 4 i 10 j | = 16 + 100 = 2 29 , | a b | = i +8 j | = 1+64 = 65 , a + b = i j +2 i 5 j =3 i 2 j , 3 a 2 b i +9 j 4 i +10 j = i +19 j . C12S01.014: Given: a i 5 j and b = i 6 j . Then: | a | = | 2 i 5 j | = 4 + 25 = 29 , 2 b | = 2 i +12 j | = 4 + 144 = 2 37 , | a b | = | i + j | = 2 , a + b i 5 j + i 6 j i 11 j , 3 a 2 b =6 i 15 j 2 i j =4 i 3 j . C12S01.015: Given: a i and b = 7 j . Then: | a | = | 4 i | = 16 = 4 , 2 b | = | 14 j | = p (14) 2 =14 , | a b | = | 4 i +7 j | = 16 + 49 = 65 , a + b i 7 j , 3 a 2 b =12 i +14 j . C12S01.016: Given: a = i j and b i j . Then: | a | = 1+1 = 2 , 2 b | = 4 i 4 j | = 32 = 4 2 , 3
Background image of page 3

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

View Full DocumentRight Arrow Icon
| a b | = |− 3 i 3 j | = 18 = 3 2 , a + b = i j +2 i j = i + j , 3 a 2 b = 3 i 3 j 4 i 4 j = 7 i 7 j . C12S01.017: Because | a | = 9+16 =5, u = 1 5 a = 3 5 i 4 5 j and v = 1 5 a = 3 5 i + 4 5 j . C12S01.018: Because | a | = 25 + 144 = 13, u = 1 13 a = 5 13 i 12 13 j and v = 1 13 a = 5 13 i + 12 13 j .
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 147

Calculus with Analytic Geometry by edwards & Penney soln ch12

This preview shows document pages 1 - 5. Sign up to view the full document.

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