356
CHAPTER 40
40.80
40.81
40.82
40.83
40.84
40.85
40.86
40.87
where the last step follows from Problem 40.60. Finally, d = \OP\.
Find an equation of the plane containing the point ^,(3, —2,5) and perpendicular to the vector N = (4, 2, -7).
For any point P(x, y, z), P is in the plane if and only if P,,P 1 N, that is, if and only if P.P-N =
4(;e-3) + 2(>> + 2)-7(z - 5) = 0, which can also be written as 4x + 2y - Iz = -27.
Find an equation of the plane containing the point /
> ,(4,3, -2) and perpendicular to the vector (5, -4,6).
We know (from Problem 40.80) that the plane has an equation of the form 5x — 4y + 6z = d. Since P, lies
in the plane, 5(4) — 4(3) + 6(-2) = d. So, d = — 4 and the plane has the equation 5x - 4y + 6z = -4.
Find an equation for the plane through the points P(l, 3,5), Q(-l, 2, 4), and R(4,4,0).
is normal to the plane. N = (-2, -1, -1) x (3,1, -5) = (6, -13,1). So, the equation has
the form 6* - 13y + z = d. Since R is in the plane, 6(4) - 13(4) + 0 = d, d = —28. Thus, the equation is
6x - I3y + z = -28.
Find an equation for the plane through the points (3,2, -1), (1, —1,3), and (3, -2,4).
We shall use a method different from the one used in Problem 40.82. The equation of the plane has the form
ax + by + cz = d. Substitute the values corresponding to the three given points: (1) 3cz + 2b - c = d;
(2) a-b + 3c = d; (3) 3a-2b + 4c = d. Eliminating a from (1) and (2), we get (4) 5b-lOc=-2d.
Eliminating a from (2) and (3), we get (5) b — 5c = —2d. Eliminating d from (4) and (5), we get
5b-Wc = b-5c, 4b = 5c, b=\c. From (5), -2rf=|c-5c, d=$c. From (2), a = d + b-3c =
i fc+ jc-3c = c/8. So, the equation of the plane is (c/8)x + %cy + cz = ^c. Multiplying by 8/c yields
x + Wy + Sz = 15.
Find the cosine of the angle 6 between the planes 4x + 4_y-2z=9 and 2x + y + z =-3.
0 is the angle between the normal vectors (4,4, -2) and (2,1,1). So,
Of course, there is another angle between the planes, the supplement of the angle whose cosine was just found.
The cosine of the other angle is -cos 6 = —5V6/18.
Find parametric equations for the line 2£ that is the intersection of the planes in Problem 40.84.
The line «SPis perpendicular to the normal vectors of the planes, and is, therefore, parallel to their cross product
(4,4, —2) x (2,1,1) = (6, —8, —4). We also need a point on the intersection. To find one, set x = 0 and
solve the two resulting equations, 4y — 2z = 9, y + z = -3. Multiplying the second equation by 2 and
adding, we get 6y = 3, y=\. So, z = -\. Thus, the point is (0, 3,-1) and we obtain the line x=6t,
y= i -8t, z--\-*t.
Find an equation of the plane containing the point P( 1,3,1) and the line !£: x = t, y - t, z = t + 2.
The point Q(0,0,2) is on the given line .5? (set r = 0). The vector = (1,3,-1) lies in the
sought plane. Since the vector (1,1,1) is parallel to &, the cross product (1,3,—l)x(l,l,l) =
(4, -2, —2) is normal to the plane. So, an equation of the plane is 4* — 2y — 2z = d. Since the point
Q(0,0, 2) is in the plane, -4 = d. Thus, the plane has the equation 4x - 2y - 2z = —4, or 2x — y — z =
-2.
(a) Express vectorially the distance from the origin to the intersection of two planes, (b) Check the result of (a)
geometrically.
(a) Figure 40-8(«) indicates two planes, &. and 0>,, with respective normals N. and N 2 , and the common
(cf. Problem 40.85). If P is the
point A. The line of intersection, £, has the vector equation OX = OA + IN
point of j? closest to O, then OP IN, or 0= OP-N = (OA + < P N)-N = O4-N + f P N-N.
and
Hence,
N = PQ x PR
CHAPTER 40
40.80
40.81
40.82
40.83
40.84
40.85
40.86
40.87
where the last step follows from Problem 40.60. Finally, d = \OP\.
Find an equation of the plane containing the point ^,(3, —2,5) and perpendicular to the vector N = (4, 2, -7).
For any point P(x, y, z), P is in the plane if and only if P,,P 1 N, that is, if and only if P.P-N =
4(;e-3) + 2(>> + 2)-7(z - 5) = 0, which can also be written as 4x + 2y - Iz = -27.
Find an equation of the plane containing the point /
> ,(4,3, -2) and perpendicular to the vector (5, -4,6).
We know (from Problem 40.80) that the plane has an equation of the form 5x — 4y + 6z = d. Since P, lies
in the plane, 5(4) — 4(3) + 6(-2) = d. So, d = — 4 and the plane has the equation 5x - 4y + 6z = -4.
Find an equation for the plane through the points P(l, 3,5), Q(-l, 2, 4), and R(4,4,0).
is normal to the plane. N = (-2, -1, -1) x (3,1, -5) = (6, -13,1). So, the equation has
the form 6* - 13y + z = d. Since R is in the plane, 6(4) - 13(4) + 0 = d, d = —28. Thus, the equation is
6x - I3y + z = -28.
Find an equation for the plane through the points (3,2, -1), (1, —1,3), and (3, -2,4).
We shall use a method different from the one used in Problem 40.82. The equation of the plane has the form
ax + by + cz = d. Substitute the values corresponding to the three given points: (1) 3cz + 2b - c = d;
(2) a-b + 3c = d; (3) 3a-2b + 4c = d. Eliminating a from (1) and (2), we get (4) 5b-lOc=-2d.
Eliminating a from (2) and (3), we get (5) b — 5c = —2d. Eliminating d from (4) and (5), we get
5b-Wc = b-5c, 4b = 5c, b=\c. From (5), -2rf=|c-5c, d=$c. From (2), a = d + b-3c =
i fc+ jc-3c = c/8. So, the equation of the plane is (c/8)x + %cy + cz = ^c. Multiplying by 8/c yields
x + Wy + Sz = 15.
Find the cosine of the angle 6 between the planes 4x + 4_y-2z=9 and 2x + y + z =-3.
0 is the angle between the normal vectors (4,4, -2) and (2,1,1). So,
Of course, there is another angle between the planes, the supplement of the angle whose cosine was just found.
The cosine of the other angle is -cos 6 = —5V6/18.
Find parametric equations for the line 2£ that is the intersection of the planes in Problem 40.84.
The line «SPis perpendicular to the normal vectors of the planes, and is, therefore, parallel to their cross product
(4,4, —2) x (2,1,1) = (6, —8, —4). We also need a point on the intersection. To find one, set x = 0 and
solve the two resulting equations, 4y — 2z = 9, y + z = -3. Multiplying the second equation by 2 and
adding, we get 6y = 3, y=\. So, z = -\. Thus, the point is (0, 3,-1) and we obtain the line x=6t,
y= i -8t, z--\-*t.
Find an equation of the plane containing the point P( 1,3,1) and the line !£: x = t, y - t, z = t + 2.
The point Q(0,0,2) is on the given line .5? (set r = 0). The vector = (1,3,-1) lies in the
sought plane. Since the vector (1,1,1) is parallel to &, the cross product (1,3,—l)x(l,l,l) =
(4, -2, —2) is normal to the plane. So, an equation of the plane is 4* — 2y — 2z = d. Since the point
Q(0,0, 2) is in the plane, -4 = d. Thus, the plane has the equation 4x - 2y - 2z = —4, or 2x — y — z =
-2.
(a) Express vectorially the distance from the origin to the intersection of two planes, (b) Check the result of (a)
geometrically.
(a) Figure 40-8(«) indicates two planes, &. and 0>,, with respective normals N. and N 2 , and the common
(cf. Problem 40.85). If P is the
point A. The line of intersection, £, has the vector equation OX = OA + IN
point of j? closest to O, then OP IN, or 0= OP-N = (OA + < P N)-N = O4-N + f P N-N.
and
Hence,
N = PQ x PR