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Unformatted text preview: 4, t2 = −1/3 so (−17, −1, 1) is the point of intersection. v1 = 4, 1, 0 and
v2 = 12, 6, 3 are (respectively) parallel to the lines, v1 × v2 = 3, −12, 12 so 1, −4, 4 is normal
to the desired plane whose equation is x − 4y + 4z = −9.
37. n1 = −2, 3, 7 and n2 = 1, 2, −3 are normals to the planes, n1 × n2 = −23, 1, −7 is parallel
to the line of intersection. Let z = 0 in both equations and solve for x and y to get x = −11/7,
y = −12/7 so (−11/7, −12/7, 0) is on the line, a parametrization of which is
x = −11/7 − 23t, y = −12/7 + t, z = −7t.
38. Similar to Exercise 37 with n1 = 3, −5, 2 , n2 = 0, 0, 1 , n1 × n2 = −5, −3, 0 . z = 0 so
3x − 5y = 0, let x = 0 then y = 0 and (0,0,0) is on the line, a parametrization of which is
x = −5t, y = −3t, z = 0.
√
39. D = 2(1) − 2(−2) + (3) − 4/ 4 + 4 + 1 = 5/3
√
40. D = 3(0) + 6(1) − 2(5) − 5/ 9 + 36 + 4 = 9/7
√
√
41. (0,0,0) is on the ﬁrst plane so D = 6(0) − 3(0) − 3(0) − 5/ 36 + 9 + 9 = 5/ 54.
√
√
42. (0,0,1) is on the ﬁrst plane so D = (0) + (0) + (1) + 1/ 1 + 1 + 1 = 2/ 3.
43. (1,3,5) and (4,6,7) are on L1 and L2 , respectively. v1 = 7, 1, −3 and v2 = −1, 0, 2 are,
respectively, parallel to L1 and L2 , v1 × v2 = 2, −11, 1 so the plane 2x − 11y + z + 51 = 0
√
√
contains L2 and is parallel to L1 , D = 2(1) − 11(3) + (5) + 51/ 4 + 121 + 1 = 25/ 126.
44. (3,4,1) and (0,3,0) are on L1 and L2 , respectively. v1 = −1, 4, 2 and v2 = 1, 0, 2 are parallel to
L1 and L2 , v1 × v2 = 8, 4, −4 = 4 2, 1, −1 so 2x + y − z − 3 = 0 contains L2 and is parallel to
√
√
L1 , D = 2(3) + (4) − (1) − 3/ 4 + 1 + 1 = 6.
√
√
45. The distance between (2, 1, −3) and the plane is 2 − 3(1) + 2(−3) − 4/ 1 + 9 + 4 = 11/ 14 which
is the radius of the sphere; an equation is (x − 2)2 + (y − 1)2 + (z + 3)2 = 121/14.
46. The vector 2i + j − k is normal to the plane and hence parallel to the line so parametric equations
of the line are x = 3 + 2t, y = 1 + t, z = −t. Substitution into the equation of the plane yields
2(3 + 2t) + (1 + t) − (−t) = 0, t = −7/6; the point of intersection is (2/3, −1/6, 7/6).
47. v = 1, 2, −1 is parallel to the line, n = 2, −2, −2 is normal to the plane, v · n = 0 so v is
parallel to the plane because v and n are perpendicular. (−1, 3, 0) is on the line so
√
√
D = 2(−1) − 2(3) − 2(0) + 3/ 4 + 4 + 4 = 5/ 12 473 Chapter 13 48. (a) (b) n r – r0 P(x0, y0) n · (r − r0 ) = a(x − x0 ) + b(y − y0 ) = 0 P(x, y)
r r0 O (c) See the proof of Theorem 13.6.1. Since a and b are not both zero, there is at least one point
(x0 , y0 ) that satisﬁes ax + by + d = 0, so ax0 + bx0 + d = 0. If (x, y ) also satisﬁes ax + by + d = 0
then, subtracting, a(x − x0 ) + b(y − y0 ) = 0, which is the equation of a line with n = a, b
as normal.
√
(d) Let Q(x1 , y1 ) be a point on the line, and position the normal n = a, b , with length a2 + b2 ,
so that its initial point is at Q. The distance is the orthogonal projection of
−→ QP0 = x0 − x1 , y0 − y1 onto n. Then
−→ D = projn QP 0 −→ ax0 + by0 + d
QP0 · n
√
=
n=
.
2
n
a2 + b2 √
√
49. D = 2(−3) + (5) − 1/ 4 + 1 = 2/ 5
50. (a) If x0 , y0 , z0 lies on the second plane, so that ax0 + by0 + cz0 + d2 = 0, then by Theorem
 − d2 + d1 
ax0 + by0 + cz0 + d1 
√
=√
13.6.2, the distance between the planes is D =
2 + b2 + c2
a
a2 + b2 + c2
5
(b) The distance between the planes −2x + y + z = 0 and −2x + y + z + = 0 is
3
5
0 − 5/3
= √.
D= √
4+1+1
36 EXERCISE SET 13.7
1. (a) elliptic paraboloid, a = 2, b = 3
(b) hyperbolic paraboloid, a = 1, b = 5
(c) hyperboloid of one sheet, a = b = c = 4
(d) circular cone, a = b = 1
(e) elliptic paraboloid, a = 2, b = 1
(f ) hyperboloid of two sheets, a = b = c = 1
√
√
2. (a) ellipsoid, a = 2/2, b = 1/2, e = 3/3
(b) hyperbolic paraboloid, a = b = 1
(c) hyperboloid of one sheet, a = 1, b = 3, c = 1
(d) hyperboloid of two sheets, a = 1, b = 2, c = 1
√
√
(e) elliptic paraboloid, a = 2, b = 2/2
√
(f ) elliptic cone, a = 2, b = 3 Exercise Set 13.7 474 3. (a) −z = x2 + y 2 , circular paraboloid
opening down the negative z axis z x y (b) z = x2 + y 2 , circular paraboloid, no change
(c) z = x2 + y 2 , circular paraboloid, no change
(d) z = x2 + y 2 , circular paraboloid, no change z x (e) x = y 2 + z 2 , circular paraboloid
opening along the positive xaxis y (f ) y = x2 + z 2 , circular paraboloid
opening along the positive y axis z z y x x
y z 4. (a)
(b)
(c)
(d) x +y −z
x2 + y 2 − z 2
x2 + y 2 − z 2
x2 + y 2 − z 2
2 2 2 = 1,
= 1,
= 1,
= 1, no
no
no
no change
change
change
change
x
y 475 Chapter 13 (e) −x2 + y 2 + z 2 = 1, hyperboloid
of one sheet with xaxis as axis (f ) x2 − y 2 + z 2 = 1, hyperboloid
of one sheet with y axis as axis
z z x
x 5. (a)
(b)
(c)
(d)
(e)
(f ) y y hyperboloid of one sheet, axis is y axis
hyperboloid of two sheets separated by yz plane
elliptic paraboloid opening along the positive xaxis
elliptic cone with xax...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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