Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 12.3 THE DOT PRODUCT 0 1-19
49. a+ (b + c) = (at, a2} + ((bt , b2} + (c1, c2}) = (at , a2} + (b1 + c1, ~ + c2)
= (a1 + b1 + c1, a2 + ~ + c2} = ((a1 + bt) + c1, (a2 + ~) + c2)
= (a1 + b1 ,a2 + b2} + (ct, c2} = ((a1,a2} + (b1 , b2}) + (c1,c2}
.= (a+ b} + c
_____. -+ -+
51 . Consider triangle ABC, where D and E are the midpoints of AB and BC. We know that AB. + BC = AC (1) and
---+ ---+ _. I .-;. ---+ ---+ -+ . . ----+ ---t
DB +BE= DE (2). However, DB = ~AB, and BE = ~BC. Substituting these expressions for DB and BE into
---+ ---+ -+ ---t ----+ . .-. --+
· (2) gives ~AB + ~BC = DE. Comparing this with (1) gives DE = ~AC. Therefore·AC and DE are parallel and
12.3 The Dot Product
1. (a) a · b is a scalar, and the dot product is defined only for vectors, so (a · b) · c has no meaning.
(b) (a . b) c is a scalar multiple of a vector, so it does have meaning.
(c) Both [a[ and b · care scalars, so [al (b ·c) is an ordinary product of real numbers, and has meaning.
(d) Both a and b +care vectors, so th~ dot product a · (b +c) has meaning.
(e) a · b is a scalar, but cis a vector, and so the two quantities cannot be added and a · b -1:' c has no meaning.
(f) Ia ! is a scalar, and the dot product is defined only for vectors, so lal · (b + c ) has no meaning.
3. a· b = ( - 2, ~) · (- 5,1 2} = (-2)(-5} + (~)(12) = 10 + 4 = 14
5. a · b = (4, 1, i ) · (6, -3, - 8) = (4)(6) + (1)(- 3) + (i) (- 8) = 19
7. a· b = (2 i + j ) · (i - j + k ) = (2)(1) + (1)( - 1) + (0)(1) = 1
'-1
' '• \
9. By Theorem 3, a · b = lal lb1 cos 0 = (6)(5) cos 2 ; = 30 ( - ~) = -15.
11. u , v , aJ)d ware all unit vectors, so the triangle is an equilateral triangle. Thus the angle between u and v is 60° and
u · v = lullvl cos60° = (1 )( 1) (~) = ~. lfw is moved so it has the same initial point as u , we can see that the angle
between them is 120° and we have u · w = lul lw l cos 120° = (1)(1) ( -~) = - ~.
13. (a) i · j = (1, 0, 0) · (0, 1, 0) = (1)(0) + (0)(1) + (0)(0) = 0. Similarly, j · k = (0)(0) + (1)(0) + (0)(1) ,;, 0 and
k . i = {0){1) + {0){0) + (1)(0) = 0.
Another method: Because i, j , and k are mutually perpendicular, the cosine factor in each dot product (see Theorem 3)
is cos~= 0.
(b) By Pr~perty I of the dot product, i · i = lil 2 = 1 2 = 1 since i is a unit vector. Similarly, j · j = lj l 2 = 1 and
k . k = lkl 2 = 1.
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