Calculus Stewart ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
1. (a) True; each of the first two lines has a direction vector parallel to the direction vector of the third line, so these vectors are each scalar multiples of the third direction vector. Then the first two direction vectors are also scalar multiples of each other, so these vectors, and hence the two lines, are parallel. (b) False; for example, the x and y axes are both perpendicular to the z axis, yet the x and y axes are not parallel. (c) True; each of the first two planes has a normal vector parallel to the normal vector of the third plane, so these two normal vectors are parallel to each other and the planes are parallel. (d) False; for example, the xy and yz planes are not parallel, yet they are both perpendicular to the xz plane. (e) False; the x and y axes are not parallel, yet they are both parallel to the plane z=1 . (f) True; if each line is perpendicular to a plane, then the lines' direction vectors are both parallel to a normal vector for the plane. Thus, the direction vectors are parallel to each other and the lines are parallel. (g) False; the planes y=1 and z=1 are not parallel, yet they are both parallel to the x axis. (h) True; if each plane is perpendicular to a line, then any normal vector for each plane is parallel to a direction vector for the line. Thus, the normal vectors are parallel to each other and the planes are parallel. (i) True; see Figure 9 and the accompanying discussion. (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 , 0 <90 , and the line will intersect the plane at an angle 90 . 2. For this line, we have r =i 3 k and v=2 i 4 j+5 k , so a vector equation is
0
r=r +t v=(i 3 k)+t(2 i 4 j+5 k)=(1+2t) i 4t j+( 3+5t) k and parametric equations are x=1+2t , y= 4t ,
0
z= 3+5t . 3. For this line, we have r = 2 i+4 j+10 k and v=3 i+ j 8 k , so a vector equation is
0
r=r +t v= ( 2 i+4 j+10 k ) +t(3 i+ j 8 k)=( 2+3t) i+(4+t) j+(10 8t) k and parametric equations are
0
x= 2+3t , y=4+t , z=10 8t . 4. This line has the same direction as the given line, v=2 i j+3 k . Here r =0 i+0 j+0 k , so a vector
0
equation is r=(0 i+0 j+0 k)+t(2 i j+3 k)=2t i t j+3t k and parametric equations are x=2t , y= t , z=3t . 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 =i+6 k , and we can take v=i+3 j+k . Then a vector equation is
0
r=(i+6 k)+t(i+3 j+k)=(1+t) i+3t j+(6+t) k , and parametric equations are x=1+t , y=3t , z=6+t . 6. The vector v= 1 0,2 0,3 0 = 1,2,3 is parallel to the line. Letting P = ( 0,0,0 ) , parametric
0
1
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
equations are x=0+1 t=t , y=0+2 t=2t , z=0+3 t=3t , while symmetric equations are x=
y z = . 2 3
7. The vector v= 4 1,3 3,0 2 = 5,0, 2 is parallel to the line. Letting P = ( 1,3,2 ) , parametric
0
equations are x=1 5t , y=3+0t=3 , z=2 2t , while symmetric equations are here that the direction number b=0 , so rather than writing write the equation y=3 separately.
x 1 z 2 = ,y=3 . Notice 5 2
y 3 in the symmetric equation we must 0
8. v= 2 6,4 1,5 ( 3) = 4,3,8 , and letting P = ( 6,1, 3) , parametric equations are x=6 4t , y=1+3t 0 , z= 3+8t , while symmetric equations are 1 , 3 1 2 1 , 4 2 x 6 y 1 z+3 = = . 4 3 8
9. v= y=1+
2 0,1
=
2,
, and letting P = ( 2,1, 3) , parametric equations are x=2+2t ,
0
1 x 2 y 1 z+3 x 2 z+3 t , z= 3 4t , while symmetric equations are = = or =2y 2= . 2 2 1/2 4 2 4
10. v=(i+ j) ( j+k)=
0
i j k 1 1 0 =i j+k is the direction of the line perpendicular to both i+ j 0 1 1 and j+k . With P = ( 2,1,0 ) , parametric equations are x=2+t , y=1 t , z=t and symmetric equations are y 1 x 2= =z or x 2=1 y=z . 1 11. The line has direction v= 1,2,1 . Letting P = ( 1, 1,1 ) , parametric equations are x=1+t , y= 1+2t
0
, z=1+t and symmetric equations are x 1=
y+1 =z 1 . 2
12. Setting x=0 , we see that ( 0,1,0 ) satisfies the equations of both planes, so they do in fact have a line of intersection. v=n n = 1,1,1 1,0,1 = 1,0, 1 is the direction of this line. Taking the point
1 2
( 0,1,0 ) as P0 , parametric equations are x=t , y=1 , z= t , and symmetric equations are x= z , y=1 .
13. Direction vectors of the lines are v = 2 ( 4),0 ( 6), 3 1 = 2,6, 4 and
1
v = 5 10,3 18,14 4 = 5, 15,10 , and since v =
2 2
5 v , the direction vectors and thus the lines are 2 1
2
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
parallel. 14. Direction vectors of the lines are v = 2,4,4 and v = 8, 1,4 . Since v v = 16 4+16 0 , the
1 2 1 2
vectors and thus the lines are not perpendicular. 15. (a) A direction vector of the line with parametric equations x=1+2t , y=3t , z=5 7t is v= 2,3, 7 and the desired parallel line must also have v as a direction vector. Here P = ( 0,2, 1 ) , so symmetric
0
equations for the line are
x y 2 z+1 = = . 2 3 7 x y 2 1 2 11 = = or x= , y= . Thus 2 3 7 7 7 . Similarly for the yz plane, we need
(b) The line intersects the xy plane when z=0 , so we need the point of intersection with the xy plane is x=0 0= y 2 z+1 = 3 7 2 11 , ,0 7 7
y=2 , z= 1 . Thus the line intersects the yz plane at ( 0,2, 1 ) . For the xz x 2 z+1 = = 2 3 7 x= 4 11 , z= . So the line intersects the xz plane at 3 3
plane, we need y=0 4 11 ,0, 3 3 .
16. (a) A vector normal to the plane 2x y+z=1 is n= 2, 1,1 , and since the line is to be perpendicular to the plane, n is also a direction vector for the line. Thus parametric equations of the line are x=5+2t , y=1 t , z=t . (b) On the xy plane, z=0 . So z=t=0 in the parametric equations of the line, and therefore x=5 and 5 7 y=1 , giving the point of intersection ( 5,1,0 ) . For the yz plane, x=0 which implies t= , so y= 2 2 5 7 5 and z= and the point is 0, , . For the xz plane, y=0 which implies t=1 , so x=7 and z=1 2 2 2 and the point of intersection is ( 7,0,1 ) . 17. From Equation 4, the line segment from r =2 i j+4 k to r =4 i+6 j+k is
0 1
r(t)=(1 t) r +t r =(1 t)(2 i j+4 k)+t(4 i+6 j+k)=(2 i j+4 k)+t(2 i+7 j 3 k) , 0 t
0 1
1.
18. From Equation 4, the line segment from r =10 i+3 j+k to r =5 i+6 j 3 k is
0 1
r(t) =(1 t) r0+t r1=(1 t)(10 i+3 j+k)+t(5 i+6 j 3 k) =(10 i+3 j+k)+t( 5 i+3 j 4 k) , 0 t 1
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
The corresponding parametric equations are x=10 5t , y=3+3t , z=1 4t , 0 t
1 2 1
1.
2
19. Since the direction vectors are v = 6,9, 3 and v = 2, 3,1 , we have v = 3v so the lines are parallel. 20. The lines aren't parallel since the direction vectors 2,3, 1 and 1,1,3 aren't parallel. For the lines to intersect we must be able to find one value of t and one value of s that produce the same point from the respective parametric equations. Thus we need to satisfy the following three equations: 1+2t= 1+s , 3t=4+s , 2 t=1+3s . Solving the first two equations we get t=6 , s=14 and checking, we see that these values don't satisfy the third equation. Thus L and L aren't parallel and don't
1 2
intersect, so they must be skew lines. 21. Since the direction vectors 1,2,3 and 4, 3,2 are not scalar multiples of each other, the lines are not parallel, so we check to see if the lines intersect. The parametric equations of the lines are L :
1
x=t , y=1+2t , z=2+3t and L : x=3 4s , y=2 3s , z=1+2s . For the lines to intersect, we must be able
2
to find one value of t and one value of s that produce the same point from the respective parametric equations. Thus we need to satisfy the following three equations: t=3 4s , 1+2t=2 3s , 2+3t=1+2s . Solving the first two equations we get t= 1 , s=1 and checking, we see that these values don't satisfy the third equation. Thus the lines aren't parallel and don't intersect, so they must be skew lines. 22. Since the direction vectors 2,2, 1 and 1, 1,3 aren't parallel, the lines aren't parallel. Here the parametric equations are L : x=1+2t , y=3+2t , z=2 t and L : x=2+s , y=6 s , z= 2+3s . Thus, for the
1 2
lines to intersect, the three equations 1+2t=2+s , 3+2t=6 s , and 2 t= 2+3s must be satisfied simultaneously. Solving the first two equations gives t=1 , s=1 and, checking, we see that these values do satisfy the third equation, so the lines intersect when t=1 and s=1 , that is, at the point ( 3,5,1 ) . 23. Since the plane is perpendicular to the vector 2,1,5 , we can take 2,1,5 as a normal vector to the plane. ( 6,3,2 ) is a point on the plane, so setting a= 2 , b=1 , c=5 and x =6 , y =3 , z =2 in
0 0 0
Equation 7 gives 2(x 6)+1(y 3)+5(z 2)=0 or 2x+y+5z=1 to be an equation of the plane. 24. j+2 k= 0,1,2 is a normal vector to the plane and ( 4,0, 3) is a point on the plane, so setting a=0 , b=1 , c=2,x =4 , y =0 , z = 3 in Equation 7 gives 0(x 4)+1(y 0)+2[z ( 3)]=0 or y+2z= 6 to be an
0 0 0
equation of the plane. 25. i+ j k= 1,1, 1 is a normal vector to the plane and ( 1, 1,1 ) is a point on the plane, so setting a=1 , b=1 , c= 1 , x =1 , y = 1 , z =1 in Equation 7 gives 1( x 1 ) +1[y ( 1)] 1(z 1)=0 or x+y z= 1 to be
0 0 0
an equation of the plane.
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
26. Since the line is perpendicular to the plane, its direction vector 1,2, 3 is a normal vector to the plane. An equation of the plane, then, is 1[x ( 2)]+2(y 8) 3(z 10)=0 or x+2y 3z= 16 . 27. Since the two planes are parallel, they will have the same normal vectors. So we can take n= 2, 1,3 , and an equation of the plane is 2(x 0) 1(y 0)+3(z 0)=0 or 2x y+3z=0 . 28. Since the two planes are parallel, they will have the same normal vectors. So we can take n= 1,1,1 , and an equation of the plane is 1[x ( 1)]+1(y 6)+1[z ( 5)]=0 or x+y+z=0 . 29. Since the two planes are parallel, they will have the same normal vectors. So we can take n= 3,0, 7 , and an equation of the plane is 3(x 4)+0[y ( 2)] 7(z 3)=0 or 3x 7z= 9 . 30. First, a normal vector for the plane 2x+4y+8z=17 is n= 2,4,8 . A direction vector for the line is v= 2,1, 1 , and since n v=0 we know the line is perpendicular to n and hence parallel to the plane. Thus, there is a parallel plane which contains the line. By putting t=0 , we know the point ( 3,0,8 ) is on the line and hence the new plane. We can use the same normal vector n= 2,4,8 , so an equation of the plane is 2(x 3)+4(y 0)+8(z 8)=0 or x+2y+4z=35 . 31. Here the vectors a= 1 0,0 1,1 1 = 1, 1,0 and b= 1 0,1 1,0 1 = 1,0, 1 lie in the plane, so a b is a normal vector to the plane. Thus, we can take n=a b= 1 0,0+1,0+1 = 1,1,1 . If P is the
0
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 . 32. Here the vectors a= 2, 4,6 and b= 5,1,3 lie in the plane, so n=a b= 12 6,30 6,2+20 = 18,24,22 is a normal vector to the plane and an equation of the plane is 18(x 0)+24(y 0)+22(z 0)=0 or 18x+24y+22z=0 . 33. Here the vectors a= 8 3,2 ( 1),4 2 = 5,3,2 and b= 1 3, 2 ( 1), 3 2 = 4, 1, 5 lie in the plane, so a normal vector to the plane is n=a b= 15+2, 8+25, 5+12 = 13,17,7 and an equation of the plane is 13(x 3)+17[y ( 1)]+7(z 2)=0 or 13x+17y+7z= 42 . 34. 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= 3,1, 1 is one vector in the plane. We can verify that the given point ( 1,2,3) 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 ( 0,1,2 ) is on the line, so b= 1 0,2 1,3 2 = 1,1,1 and n=a b= 1+1, 1 3,3 1 = 2, 4,2 . Thus, an equation of the plane is 2(x 1) 4(y 2)+2(z 3)=0 or 2x 4y+2z=0 . (Equivalently, we can write x 2y+z=0 .) 35. If we first find two nonparallel vectors in the plane, their cross product will be a normal vector to
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
the plane. Since the given line lies in the plane, its direction vector a= 2,5,4 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= 6 4,0 3, 2 7 = 2, 3, 9 and n=a b= 45+12,8 18,6 10 = 33, 10, 4 . Thus, an equation of the plane is 33(x 6) 10(y 0) 4[z ( 2)]=0 or 33x+10y+4z=190 . y z 1 1 = , lies in the plane, its direction vector a= 1, , is 1/2 1/3 2 3 parallel to the plane. The point ( 0,0,0 ) is on the line (put t=0 ), and we can verify that the given point ( 1, 1,1 ) in the plane is not on the line. The vector connecting these two points, b= 1, 1,1 , is therefore parallel to the plane, but not parallel to 1,2,3 . Then 36. Since the line x=2y=3z , or x= 1 1 1 1 5 2 3 + , 1, 1 = , , is a normal vector to the plane, and an equation of 2 3 3 2 6 3 2 5 2 3 the plane is (x 0) (y 0) (z 0)=0 or 5x 4y 9z=0 . 6 3 2 a b= 37. A direction vector for the line of intersection is a=n n = 1,1, 1
1 2
2, 1,3 = 2, 5, 3 , and a 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 7 3 planes reduce to y z=2 and y+3z=1 with simultaneous solution y= and z= . So a point on the 2 2 3 1 7 3 line is 0, , 1, , and another vector parallel to the plane is . Then a normal vector 2 2 2 2 to the plane is n= 2, 5, 3 3 1 , = 2,4, 8 and an equation of the plane is 2 2 2(x+1)+4(y 2) 8(z 1)=0 or x 2y+4z= 1 . 1,
38. n = 1,0, 1 and n = 0,1,2 . Setting z=0 , it is easy to see that ( 1,3,0 ) is a point on the line of 1 2 intersection of x z=1 and y+2z=3 . The direction of this line is v =n n = 1, 2,1 . A second vector
1 1 2
parallel to the desired plane is v = 1,1, 2 , since it is perpendicular to x+y 2z=1 . Therefore, a
2
normal of the plane in question is n=v v = 4 1,1+2,1+2 = 3,3,3 , or we can use 1,1,1 . Taking
1 2
( x ,y ,z ) = ( 1,3,0) , the equation we are looking for is ( x 1 ) + ( y 3) +z=0
0 0 0
x+y+z=4 .
39. Substitute the parametric equations of the line into the equation of the plane: (3 t) (2+t)+2(5t)=9
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Stewart Calculus ET 5e 0534393217;12. Vectors and the of Geometry Space; 12.5 Equations of Lines and Planes
8t=8 t=1 . Therefore, the point of intersection of the line and the plane is given by x=3 1=2 , y=2+1=3 , and z=5(1)=5, that is, the point ( 2,3,5) . 40. Substitute the parametric equations of the line into the equation of the plane: (1+2t)+2(4t) (2 3t)+1=0 13t=0 t=0 . Therefore, the point of intersection of the line and the plane is given by x=1+2(0)=1 , y=4(0)=0 , and z=2 3(0)=2, that is, the point ( 1,0,2 ) . 41. Parametric equations for the line are x=t , y=1+t , z= plane gives 4(t) (1+t)+3 intersection is ( 2,3,1 ) . 42. A direction vector for the line through ( 1,0,1 ) and ( 4, 2,2 ) is v= 3, 2,1 and, taking P = ( 1,0,1 ) , 0 parametric equations for the line are x=1+3t , y= 2t , z=1+t . Substitution of the parametric equations into the equation of the plane gives 1+3t 2t+1+t=6 t=2 . Then x=1+3(2)=7 , y= 2(2)= 4 , and z=1+2=3 so the point of intersection is ( 7, 4,3) . 43. Setting x=0 , we see that ( 0,1,0 ) satisfies the equations of both planes, so that they do in fact have a line of intersection. v=n n = 1,1,1 1,0,1 = 1,0, 1 is the direction of this line. Therefore,
1 2
1 t =8 2
9 t=9 2
1 t and substituting into the equation of the 2 1 t=2 . Thus x=2 , y=1+2=3 , z= (2)=1 and the point of 2
direction numbers of the intersecting line are 1 , 0 , 1 . 44. The angle between the two planes is the same as the angle between their normal vectors. The normal vectors of the two planes are 1,1,1 and 1,2,3 . The cosine of the angle between these 1,1,1 1,2,3 1+2+3 6 6 = = = two planes is cos = . 7 1,1,1 1,2,3 1+1+1 1+4+9 42 45. Normal vectors for the planes are n = 1,4, 3 and n = 3,6,7 , so the normals (and thus the
1 2
planes) aren't parallel. But n n = 3+24 21=0 , so the normals (and thus the planes) are
1 2
perpendicular. 46. Normal vectors for the planes are n = 1,4, 2 and n = 3, 12,6 . Since n = 3n , the normals
1 2 2 1
(and thus the planes) are parallel. 47. Normal vectors for the planes are n = 1,1,1 and n = 1, 1,1 . The normals are not parallel, so
1 2
neither are the planes. Furthermore, n n =1 1+1=1 0 , so the planes aren't perpendicular. The
1 2
angle between them is given by
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
n n cos =
1
2 2
n
1
n
=
1 1 = 3 3 3
=cos
1
1 3
70.5 .
48. The normals are n = 2, 3,4 and n = 1,6,4 so the planes aren't parallel. Since
1 2
n n =2 18+16=0 , the normals (and thus the planes) are perpendicular.
1 2
49. The normals are n = 1, 4,2 and n = 2, 8,4 . Since n =2n , the normals (and thus the planes)
1 2 2 1
are parallel. 50. The normal vectors are n = 1,2,2 and n = 2, 1,2 . The normals are not parallel, so neither are
1 2
the planes. Furthermore, n n =2 2+4=4 0 , so the planes aren't perpendicular. The angle between
1 2
n n them is given by cos =
1
2 2
n
1
n
=
4 4 = 9 9 9
=cos
1
4 9
63.6 .
51. (a) To find a point on the line of intersection, set one of the variables equal to a constant, say z=0 . (This will only work if the line of intersection crosses the xy plane; otherwise, try setting x or y equal to 0 .) Then the equations of the planes reduce to x+y=2 and 3x 4y=6 . Solving these two equations gives x=2 , y=0 . So a point on the line of intersection is ( 2,0,0 ) . The direction of the line y z is v=n n = 5 4, 3 5, 4 3 = 1, 8, 7 , and symmetric equations for the line are x 2= = . 1 2 8 7 n n 6 3 4 5 1 2 . Therefore (b) The angle between the planes satisfies cos = = = 5 3 50 n n
1 2
=cos
1
6 5
119 (or 61 ). n = 1, 2,1 and 2x+y+z=1
1 1
52. (a) x 2y+z=1
n = 2,1,1 . The vector that gives the direction of
2 2
the line of intersection of these two planes is v=n n = 2 1,2 1,1+4 = 3,1,5 . Setting x=y=0 , we see that both planes contain ( 0,0,1 ) so that this point must lie on their line of intersection. Then x z 1 =y= . symmetric equations for this line are 3 5 n n 2 2+1 1 1 1 1 2 (b) cos = = = =cos 80 . 6 1+4+1 4+1+1 6 n n
1 2
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
53. Setting x=0 , the equations of the two planes become z=y and 5y+z= 1 , which intersect at y= and z=
1 6
1 1 1 . Thus we can choose x ,y ,z = 0, , . The vector giving the direction of this 0 0 0 6 6 6 intersecting line, v , is perpendicular to the normal vectors of both planes. So v=n n = 2, 5, 1 1,1, 1 = 6,1,7 . Therefore, by Equations 2, parametric equations for this line
(
)
1
2
are x=6t , y=
1 1 +t , z= +7t . 6 6
54. Setting y=0 , the equations of the two planes become 2x+5z= 3 and x+z= 2 , which intersect 7 1 7 1 and z= . Thus we can choose x ,y ,z = ,0, . The vector giving the direction of at x= 0 0 0 3 3 3 3 this intersecting line, v , is perpendicular to the normal vectors of both planes. So v=n n = 2,0,5 1, 3,1 = 15,5 2, 6 =3 5,1, 2 . Therefore, by Equations 2, parametric equations
(
)
1
2
of the line of intersection of the two planes are x=
7 1 +5t , y=t , z= 2t . 3 3
55. The plane contains all perpendicular bisectors of the line segment joining ( 1,1,0 ) and ( 0,1,1 ) . All 1 1+1 1 1 1 of these bisectors pass through the midpoint of this segment , , = ,1, . The 2 2 2 2 2 direction of this line segment 1 0,1 1,0 1 = 1,0, 1 is perpendicular to the plane so that we can 1 1 choose this to be n . Therefore the equation of the plane is 1 x +0(y 1) 1 z =0 x=z . 2 2 56. The plane will contain all perpendicular bisectors of the line segment joining the two points. Thus, a point in the plane is P = ( 1, 1,2 ) , the midpoint of the line segment joining the two given points,
0
and a normal to the plane is n= 6, 6,2 , the vector connecting the two points. So an equation of the plane is 6(x+1) 6(y+1)+2(z 2)=0 or 3x 3y+z=2 . 57. The plane contains the points ( a,0,0 ) , ( 0,b,0 ) and ( 0,0,c ) . Thus the vectors a= a,b,0 and b= a,0,c lie in the plane, and n=a b= bc 0,0+ac,0+ab = bc,ac,ab is a normal vector to the plane. The equation of the plane is therefore bcx+acy+abz=abc+0+0 or bcx+acy+abz=abc . Notice x y z that if a 0 , b 0 and c 0 then we can rewrite the equation as + + =1 . This is a good a b c equation to remember! 58. (a) For the lines to intersect, we must be able to find one value of t and one value of s satisfying the three equations 1+t=2 s , 1 t=s and 2t=2 . From the third we get t=1 , and putting this in the second gives s=0 . These values of s and t do satisfy the first equation, so the lines intersect at the
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
point P = ( 1+1,1 1,2(1) ) = ( 2,0,2 ) . 0 (b) The direction vectors of the lines are 1, 1,2 and 1,1,0 , so a normal vector for the plane is 1,1,0 1, 1,2 = 2,2,0 and it contains the point ( 2,0,2 ) . Then the equation of the plane is 2(x 2)+2(y 0)+0(z 2)=0 x+y=2 . 59. Two vectors which are perpendicular to the required line are the normal of the given plane, 1,1,1 , and a direction vector for the given line, 1, 1,2 . So a direction vector for the required line is 1,1,1 1, 1,2 = 3, 1, 2 . Thus L is given by x,y,z = 0,1,2 +t 3, 1, 2 , or in parametric form, x=3t , y=1 t , z=2 2t . 60. Let L be the given line. Then ( 1,1,0 ) is the point on L corresponding to t=0 . L is in the direction of a= 1, 1,2 and b= 1,0,2 is the vector joining ( 1,1,0 ) and ( 0,1,2 ) . Then 1, 1,2 1,0,2 1 3 1 b proj b= 1,0,2 , ,1 is a direction 1, 1,2 = 1,0,2 1, 1,2 = a 2 2 2 2 2 2 1 +( 1) +2 3 1 , ,1 = 3,1,2 is also a direction vector, and the line has vector for the required line. Thus 2 2 2 parametric equations x= 3t , y=1+t , z=2+2t . (Notice that this is the same line as in Exercise 59.) 61. Let P have normal vector n . Then n = 4, 2,6 , n = 4, 2, 2 , n = 6,3, 9 , n = 2, 1, 1 .
i i 1 2 3 4
2 n , so n and n are parallel, and hence P and P are parallel; similarly P and P are 1 1 3 1 3 2 4 3 3 1 parallel because n =2n . However, n and n are not parallel. 0,0, lies on P , but not on P , 2 4 1 2 1 3 2 so they are not the same plane, but both P and P contain the point ( 0,0, 3) , so these two planes are Now n =
2 4
identical. 62. Let L have direction vector v . Then v = 1,1, 5 , v = 1,1, 1 , v = 1,1, 1 , v = 2,2, 10 . v
i i 1 2 3 4 2
and v are equal so they're parallel. v =2v , so L and L are parallel. L contains the point ( 1,4,1 ) , 3 4 1 4 1 3 but this point does not lie on L , so they're not equal. ( 2,1, 3) lies on L , and on L , with t=1 . So 2 4 1 L and L are identical.
1 4
63. Let Q= ( 2,2,0 ) and R= ( 3, 1,5) , points on the line corresponding to t=0 and t=1 . Let P= ( 1,2,3) . Then a=QR= 1, 3,5 , b=QP= 1,0,3 . The distance is
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
a b d= = a
1, 3,5
1,0,3
1, 3,5
=
9, 8, 3 1, 3,5
=
9 +8 +3 1 +3 +5
2 2
2
2
2
=
154 35
=
2
22 . 5
64. Let Q= ( 5,0,1 ) and R= ( 4,3,3) , points on the line corresponding to t=0 and t=1 . Let P= ( 1,0, 1 ) . Then a=QR= 1,3,2 and b=QP= 4,0, 2 . The distance is a b d= = a 1,3,2 4,0, 2 1,3,2 = 6, 10,12 1,3,2 = 2 3 +5 +6 1 +3 +2
2 2 2 2 2
=
2 70 14
=2 5 .
2
65. By Equation 9, the distance is D=
1 25 (1)(2)+( 2)(8)+( 2)(5) 1 = . 3 1+4+4 1 26 4(3)+( 6)( 2)+1(7) 5 = . 16+36+1 53
66. By Equation 9, the distance is D=
67. Put y=z=0 in the equation of the first plane to get the point ( 1,0,0 ) on the plane. Because the planes are parallel, the distance D between them is the distance from ( 1,0,0 ) to the second plane. By 7 6 7 3( 1)+6(0) 3(0) 4 Equation 9, D= = or . 18 3 6 2 2 2 3 +6 +( 3) 68. Put y=z=0 in the equation of the first plane to get the point 4 ,0,0 3 on the plane. Because the 4 ,0,0 3 to the second plane.
planes are parallel the distance D between them is the distance from 1 By Equation 9, D= 4 3
2
+2(0) 3(0) 1 = 1 +2 +( 3)
2 2
1 . 3 14
69. The distance between two parallel planes is the same as the distance between a point on one of the planes and the other plane. Let P = x ,y ,z be a point on the plane given by ax+by+cz+d =0 . Then
0
(
0 0 0
)
1
ax +by +cz +d =0 and the distance between P and the plane given by ax+by+cz+d =0 is, from
0 0 0 1 0 2
ax +by +cz +d Equation 9, D=
0 0 0
2
d +d =
2 1 2
2 2
d d =
1 2
2 2
.
a +b +c
2
2
2
a +b +c
a +b +c
2
70. The planes must have parallel normal vectors, so if ax+by+cz+d=0 is such a plane, then for some
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
t 0 , a,b,c =t 1,2, 2 = t,2t, 2t . So this plane is given by the equation x+2y 2z+e=0 , where 1 e e=d/t . By Exercise 69, the distance between the planes is 2= 6= 1 e e=7 or 5 . 2 2 2 1 +2 +( 2) So the desired planes have equations x+2y 2z=7 and x+2y 2z= 5 . 71. L : x=y=z
1
x=y (1). L : x+1=y/2=z/3
2
x+1=y/2 (2). The solution of (1) and (2) is x=y= 2 .
However, when x= 2 , x=z z= 2 , but x+1=z/3 z= 3 , a contradiction. Hence the lines do not intersect. For L , v = 1,1,1 , and for L , v = 1,2,3 , so the lines are not parallel. Thus the lines are
1 1 2 2
skew lines. If two lines are skew, they can be viewed as lying in two parallel planes and so the distance between the skew lines would be the same as the distance between these parallel planes. The common normal vector to the planes must be perpendicular to both 1,1,1 and 1,2,3 , the direction vectors of the two lines. So set n= 1,1,1 1,2,3 = 3 2, 3+1,2 1 = 1, 2,1 . From above, we know that ( 2, 2, 2 ) and ( 2, 2, 3) are points of L and L respectively. So in the notation of Equation 8,
1 2
1( 2) 2( 2)+1( 2)+d =0
1
d =0 and 1( 2) 2( 2)+1( 3)+d =0
1 2
d =1 .
2
0 1 1 = . 1+4+1 6 Alternate solution (without reference to planes): A vector which is perpendicular to both of the lines is n= 1,1,1 1,2,3 = 1, 2,1 . Pick any point on each of the lines, say ( 2, 2, 2 ) and ( 2, 2, 3) , and form the vector b= 0,0,1 connecting the two points. The distance between the two skew lines is n b 1 0 2 0+1 1 1 the absolute value of the scalar projection of b along n , that is, D= = = . n 1+4+1 6 By Exercise 69, the distance between these two skew lines is D= 72. First notice that if two lines are skew, they can be viewed as lying in two parallel planes and so the distance between the skew lines would be the same as the distance between these parallel planes. The common normal vector to the planes must be perpendicular to both v = 1,6,2 and v = 2,15,6 ,
1 2
the direction vectors of the two lines respectively. Thus set n=v v = 36 30,4 6,15 12 = 6, 2,3 .
1 2
Setting t=0 and s=0 gives the points ( 1,1,0 ) and ( 1,5, 2 ) . So in the notation of Equation 8, 6 2+0+d =0 d = 4 and 6 10 6+d =0 d =10 . Then by Exercise 69 , the distance between the
1 1 2 2
4 10 14 = =2 . 7 36+4+9 Alternate solution (without reference to planes): We already know that the direction vectors of the two lines are v = 1,6,2 and v = 2,15,6 . Then n=v v = 6, 2,3 is perpendicular to both lines. two skew lines is given by D=
1 2 1 2
Pick any point on each of the lines, say ( 1,1,0 ) and ( 1,5, 2 ) , and form the vector b= 0,4, 2 connecting the two points. Then the distance between the two skew lines is the absolute value of the scalar projection of b along n , that is,
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Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes
D=
n b 1 14 = 0 8 6 = =2 . n 7 36+4+9
73. If a 0 , then ax+by+cz+d=0 a ( x+d/a ) +b(y 0)+c(z 0)=0 which by (7) is the scalar equation of the plane through the point ( d/a,0,0 ) with normal vector a,b,c . Similarly, if b 0 (or if c 0 ) the equation of the plane can be rewritten as a(x 0)+b ( y+d/b ) +c(z 0)=0 which by (7) is the scalar equation of a plane through the point ( 0, d/b,0 ) with normal vector a,b,c . 74. (a) The planes x+y+z=c have normal vector 1,1,1 , so they are all parallel. Their x , y , and z intercepts are all c . When c>0 their intersection with the first octant is an equilateral triangle and when c<0 their intersection with the octant diagonally opposite the first is an equilateral triangle.
(b) The planes x+y+cz=1 have x intercept 1 , y intercept 1 , and z intercept 1/c . The plane with c=0 is parallel to the z axis. As c gets larger, the planes get closer to the xy plane. , which are perpendicular to the (c) The planes ycos +zcos =1 have normal vectors 0,cos ,sin x axis, and so the planes are parallel to the x axis. We look at their intersection with the yz plane. These are lines that are perpendicular to cos ,sin and pass through ( cos ,sin ) , since cos +sin =1 . So these are the tangent lines to the unit circle. Thus the family consists of all planes tangent to the circular cylinder with radius 1 and axis the x axis.
2 2
13