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Unformatted text preview: n2
=
= 4/21, θ = cos−1 (4/21) ≈ 79◦
cos θ =
− n1 n2
(3)(7)
(Note: −n1 is used instead of n1 to get a value of θ in the range [0, π/2])
21. 4, −2, 7 is normal to the desired plane and (0,0,0) is a point on it; 4x − 2y + 7z = 0 22. v = 3, 2, −1 is parallel to the line and n = 1, −2, 1 is normal to the given plane so
v × n = 0, −4, −8 is normal to the desired plane. Let t = 0 in the line to get (−2, 4, 3) which is
also a point on the desired plane, use this point and (for convenience) the normal 0, 1, 2 to ﬁnd
that y + 2z = 10. 471 Chapter 13 23. Find two points P1 and P2 on the line of intersection of the given planes and then ﬁnd an equation
of the plane that contains P1 , P2 , and the given point P0 (−1, 4, 2). Let (x0 , y0 , z0 ) be on the
line of intersection of the given planes; then 4x0 − y0 + z0 − 2 = 0 and 2x0 + y0 − 2z0 − 3 = 0,
eliminate y0 by addition of the equations to get 6x0 − z0 − 5 = 0; if x0 = 0 then z0 = −5, if x0 = 1
then z0 = 1. Substitution of these values of x0 and z0 into either of the equations of the planes
gives the corresponding values y0 = −7 and y0 = 3 so P1 (0, −7, −5) and P2 (1, 3, 1) are on the
−→ −→ line of intersection of the planes. P0 P1 × P0 P2 = 4, −13, 21 is normal to the desired plane whose
equation is 4x − 13y + 21z = −14.
24. 1, 2, −1 is parallel to the line and hence normal to the plane x + 2y − z = 10 25. n1 = 2, 1, 1 and n2 = 1, 2, 1 are normals to the given planes, n1 × n2 = −1, −1, 3 so 1, 1, −3
is normal to the desired plane whose equation is x + y − 3z = 6.
−→ 26. n = 4, −1, 3 is normal to the given plane, P1 P2 = 3, −1, −1 is parallel to the line,
−→ n × P1 P2 = 4, 13, −1 is normal to the desired plane whose equation is 4x + 13y − z = 1.
27. n1 = 2, −1, 1 and n2 = 1, 1, −2 are normals to the given planes,
n1 × n2 = 1, 5, 3 is normal to the desired plane whose equation is x + 5y + 3z = −6.
28. Let t = 0 and t = 1 to get the points P1 (−1, 0, −4) and P2 (0, 1, −2) that lie on the line. Denote the
−→ −→ given point by P0 , then P0 P1 × P0 P2 = 7, −1, −3 is normal to the desired plane whose equation
is 7x − y − 3z = 5.
29. The plane is the perpendicular bisector of the line segment that joins P1 (2, −1, 1) and P2 (3, 1, 5).
−→ The midpoint of the line segment is (5/2, 0, 3) and P1 P2 = 1, 2, 4 is normal to the plane so an
equation is x + 2y + 4z = 29/2.
30. n1 = 2, −1, 1 and n2 = 0, 1, 1 are normals to the given planes, n1 × n2 = −2, −2, 2 so
n = 1, 1, −1 is parallel to the line of intersection of the planes. v = 3, 1, 2 is parallel to the
given line, v × n = −3, 5, 2 so 3, −5, −2 is normal to the desired plane. Let t = 0 to ﬁnd the
point (0,1,0) that lies on the given line and hence on the desired plane. An equation of the plane
is 3x − 5y − 2z = −5.
31. The line is parallel to the line of intersection of the planes if it is parallel to both planes. Normals
to the given planes are n1 = 1, −4, 2 and n2 = 2, 3, −1 so n1 × n2 = −2, 5, 11 is parallel to
the line of intersection of the planes and hence parallel to the desired line whose equations are
x = 5 − 2t, y = 5t, z = −2 + 11t.
−→ −→ 32. Denote the points by A, B , C , and D, respectively. The points lie in the same plane if AB × AC
−→ −→ −→ −→ −→ −→ and AB × AD are parallel (method 1). AB × AC = 0, −10, 5 , AB × AD = 0, 16, −8 , these
vectors are parallel because 0, −10, 5 = (−10/16) 0, 16, −8 . The points lie in the same plane
−→ −→ if D lies in the plane determined by A, B, C (method 2), and since AB × AC = 0, −10, 5 , an
equation of the plane is −2y + z + 1 = 0, 2y − z = 1 which is satisﬁed by the coordinates of D.
33. v = 0, 1, 1 is parallel to the line.
(a) For any t, 6 · 0 + 4t − 4t = 0, so (0, t, t) is in the plane.
(b) n = 5, −3, 3 is normal to the plane, v · n = 0 so the line is parallel to the plane. (0,0,0) is
on the line, (0, 0, 1/3) is on the plane. The line is below the plane because (0,0,0) is below
(0, 0, 1/3). Exercise Set 13.6 472 (c) n = 6, 2, −2 , v · n = 0 so the line is parallel to the plane. (0,0,0) is on the line, (0, 0, −3/2)
is on the plane. The line is above the plane because (0,0,0) is above (0, 0, −3/2).
−→ −→ 34. The intercepts correspond to the points A(a, 0, 0), B (0, b, 0), and C (0, 0, c). AB × AC = bc, ac, ab
is normal to the plane so bcx + acy + abz = abc or x/a + y/b + z/c = 1.
35. v1 = 1, 2, −1 and v2 = −1, −2, 1 are parallel, respectively, to the given lines and to each
other so the lines are parallel. Let t = 0 to ﬁnd the points P1 (−2, 3, 4) and P2 (3, 4, 0) that lie,
−→ respectively, on the given lines. v1 × P1 P2 = −7, −1, −9 so 7, 1, 9 is normal to the desired plane
whose equation is 7x + y + 9z = 25.
36. The system 4t1 − 1 = 12t2 − 13, t1 + 3 = 6t2 + 1, 1 = 3t2 + 2 has the solution (Exercise 26,
Section 13.5) t1 = ...
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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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