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

SECTION 12.5 EQUATIONS OF LINES AND PlANES 0 129
(j) False; they can be skew, as in Example 3.
(k) True. Consider any normal vector for the plane and any direction vector for the line. If the normal vector is perpendicular
to the direction vector, the line and plane are parallel. Otherwise, the vectors meet at an angle B, 0° :S B < goo., and the
line will intersect the plane at an angle goo -B.
3. For this line, we have r 0 = 2 i + 2.4j + 3.5 k and v = 3 i + 2 j - k, so a vector equation is
r = ro + t v = (2 i + 2.4 j + 3.5 k ) + t(3 i + 2j - k) = (2 + 3t) i + {2.4 + 2t) j + (3.5 - t) k and parametric equations are
X = 2 + 3t, y = 2.4 + 2t, Z = 3.5 - t.
5. A Line perpendicular to the given plane has the same direction as a normal vector to the plane, such as
n = (1, 3, 1). So r 0 = i + 6 k, and we can take v = i + 3j + k. Then a vector equation is
. '
r = {i + 6 k) + t(i + 3j + k ) = (1 + t) i + 3tj + (6 + t) k, and parametric equations are x = 1 + t, y = 3t, z = 6 + t.
7. The vector v = (2 - 0, 1 - ~ . - 3- 1) = (2, ~ . -4) is parallel to the line. Letting Po = (2, 1, -3), parametric equations
2 1
1 3 4 h'l · . X - 2 y - 1 Z + 3
2 - = 112
are x = 2 + t, y = + 2 t, z =- - .t, w 1 e symmetnc equat1ons are -
X - 2 = 2 y _ 2 = Z + 3.
2 - 4
= ---=4 or
9. v = (3- ( -8), - 2- 1,4- 4) = (11, - 3,·0), and letting Po= (- 8, 1, 4), parametric equations are x = -8 + llt,
4 0 4 I ·1 . . x + 8 y - 1 4 N . h h h d' .
y = 1 - 3t, z = + t = , w 11 e symmetnc equations are ----u- :;::
_ 3
, z = . ottcc ere t at t e trectton number
c = 0, so rather than writing z ~ 4 in the symmetric equation we must write the equation z = 4 separately.
11. The line has direction v = (1, 2, 1). Letting Po = (1, - 1, 1), parametric equations are x = 1 + t, y = -1 + 2t, z = 1 + t
. . 1 y + 1 1
an d symmetriC equatiOns are x - = - - = z - .
2
13. Direction vectors of the lines are v 1 = (- 2 - ( - 4), 0 - ( - 6), - 3 - 1) = (2, 6, - 4) and
v 2 = (5 - 10,3 - 18, 14 - 4) = ( -5, - 15, 10), and since v 2 = -:-~ v1 ,
the direction vectors and thus the lines are parallel.
15. (a) The li ne passes through the point (1, - 5, 6) and a direction vector for the line is ( - 1, 2, - 3), so symmetric equations for
. x- 1 y+5 z -6
the line are ---=1 = - 2
- = --=3.
x- 1 y+5 0-6 x - 1
(b) The line intersects the xy-plane when z = 0, so we need -=1' = - 2
- = --=3 or -=1' = 2 => x = - 1,
Y ~ 5 = 2 => y = -1. Thus the point of intersection with the xy-plane is ( - 1, - 1, 0). Similarly for the yz-plane,
we need x = 0 => 1 = y ~ 5 = z ~ 3 6 => y = - 3, z = 3. Thus the line intersects the yz-plane at {0, - 3, 3). For
the xz-plane, we need y = 0 => => x = - ~, z = - ~ . So the line intersects the xz-plane
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