Solução_Calculo_Stewart_6e

F.
TX.10 SECTION 13.5 EQUATIONS OF LINES AND PLANES ET SECTION 12.5 ¤ 121
29. Since the two planes are parallel, they will have the same normal vectors. So we can take n = h3, 0, −7i, and an equation of
the plane is 3(x − 4) + 0[y − (−2)] − 7(z − 3) = 0 or 3x − 7z = −9.
31. Here the vectors a = h1 − 0, 0 − 1, 1 − 1i = h1, −1, 0i and b = h1 − 0, 1 − 1, 0 − 1i = h1, 0, −1i lie in the plane, so
a × b is a normal vector to the plane. Thus, we can take n = a × b = h1 − 0, 0+1, 0+1i = h1, 1, 1i. IfP 0 is the point
(0, 1, 1), an equation of the plane is 1(x − 0) + 1(y − 1) + 1(z − 1) = 0 or x + y + z =2.
33. Here the vectors a = h8 − 3, 2 − (−1), 4 − 2i = h5, 3, 2i and b = h−1 − 3, −2 − (−1), −3 − 2i = h−4, −1, −5i lie in
the plane, so a normal vector to the plane is n = a × b = h−15 + 2, −8+25, −5+12i = h−13, 17, 7i andanequationof
the plane is −13(x − 3) + 17[y − (−1)] + 7(z − 2) = 0 or −13x +17y +7z = −42.
35. If we first find two nonparallel vectors in the plane, their cross product will be a normal vector to the plane. Since the given
line lies in the plane, its direction vector a = h−2, 5, 4i is one vector in the plane. We can verify that the given point (6, 0, −2)
does not lie on this line, so to find another nonparallel vector b which lies in the plane, we can pick any point on the line and
find a vector connecting the points. If we put t =0, we see that (4, 3, 7) is on the line, so
b = h6 − 4, 0 − 3, −2 − 7i = h2, −3, −9i and n = a × b = h−45 + 12, 8 − 18, 6 − 10i = h−33, −10, −4i. Thus,an
equation of the plane is −33(x − 6) − 10(y − 0) − 4[z − (−2)] = 0 or 33x +10y +4z = 190.
37. A direction vector for the line of intersection is a = n 1 × n 2 = h1, 1, −1i×h2, −1, 3i = h2, −5, −3i,anda is parallel to the
desired plane. Another vector parallel to the plane is the vector connecting any point on the line of intersection to the given
point (−1, 2, 1) in the plane. Setting x =0, the equations of the planes reduce to y − z =2and −y +3z =1with
simultaneous solution y = 7 and z = 3 .Soapointonthelineis 2 2
0, 7 , 3
2 2 and another vector parallel to the plane is
−1, −
3
, − 1
2 2 . Then a normal vector to the plane is n = h2, −5, −3i× −1, −
3
, − 1
2 2 = h−2, 4, −8i and an equation of
the plane is −2(x +1)+4(y − 2) − 8(z − 1) = 0 or x − 2y +4z = −1.
39. To find the x-intercept we set y = z =0in the equation 2x +5y + z =10
and obtain 2x =10 ⇒ x =5so the x-intercept is (5, 0, 0). When
x = z =0we get 5y =10 ⇒ y =2,sothey-intercept is (0, 2, 0).
Setting x = y =0gives z =10,sothez-intercept is (0, 0, 10) and we
graph the portion of the plane that lies in the first octant.
41. Setting y = z =0in the equation 6x − 3y +4z =6gives 6x =6 ⇒
x =1,whenx = z =0we have −3y =6 ⇒ y = −2,andx = y =0
implies 4z =6 ⇒ z = 3 2
, so the intercepts are (1, 0, 0), (0, −2, 0),and
(0, 0, 3 ).Thefigure shows the portion of the plane cut off by the coordinate
2
planes.