高数英文版

be OA a , OB b and OC c , respectively. Then, just as the last example, we can obtain the equation for the plane
x y z 1. a b c
z
C (0,0, c ) P ( x, y, z )
3 x 2 y 6 z 6.
x
B(2,0,0)
8
Intercept Form of the Equation for a Plane
In general, if the intercepts of the plane with the x-axis, y-axis and z-axis
n1 kn2 for some scalar k.
3
Finding an Equation for a plane
Find an equation for the plane through P0 (3,0,7) perpendicular to
n 5i 2j k.
Solution
10
Some Planes with Special Locations
(1) If a given plane passes through the origin O (0,0,0), then
x y z 0 satisfy the general equation for the plane, so that D 0.
2
Equations for Planes in Space
Equation for a plane The plane through P0 ( x0 , y0 , z0 ) normal to n Ai Bj Ck has Vector equation: n P0 P 0 Component equation:
2 : A2 x B2 y C2 z D2 0.
There normal vectors can be chosen as n1 ( A1 , B1 , C1 ) and
| A1 A2 B1 B2 C1C 2 | A B C
2 1 2 1 2 1
A B C
Find an equation for the plane through A(0,0,1), B(2,0,0) and C (0, 3,0). Solution I We have to find a vector normal to the plane and use it with one of the points (it does not matter which) to write an equation for the plane. The cross product
Leabharlann Baidu
Therefore, the equation of the plane through the origin is z Ax By Cz 0.
y
O
x
11
Some Planes with Special Locations
(2) If a given plane is parallel to the z-axis, then the normal vector
A( x x0 ) B( y y0 ) C ( z z0 ) 0 D Ax0 By0 Cz0
Component equation simplified:
Ax By Cz D ,
where
Two planes are parallel if and only if their normals are parallel, or
A B 0. Therefore, the equation of this plane is
Cz D 0
or
D z z0 . C Similarly, the equations of planes which
z
n
z0
y
O
are orthogonal to the x-axis or y-axis are
Since these three vector are coplanar
AC (0,3, 1)
A(0,0,1),
y
P ( x, y, z ) O C (0, 3,0).
x
B(2,0,0)
if and only if the point P lies in the plane, so we have AP , AB , AC 0,
y
B(0, b,0)
O
x
A( a ,0,0)
Intercept form of the equation
9
General Equations for Planes
General Equation for a plane The equation can be rewritten in the form
2 2 2 2
2 2
.
14
Some Problems about Planes
(2) Position Relationships Between Two Planes
If two planes are parallel or orthogonal, their normal vector also parallel or orthogonal.
z
n
A(0,0,1)
y
C (0, 3,0)
the point-normal form of the equation and obtain
3( x 0) 2( y 0) 6( z 1) 0 3 x 2 y 6 z 6.
x
B(2,0,0)
A( x x0 ) B( y y0 ) C ( z z0 ) 0.
7
Finding an Equation for a Plane Through Three Points
Solution (continued) that is
x 2 0 y 0 3 z 1 1 0. 1
P ( x, y, z ) O
z
A(0,0,1),
y
C (0, 3,0).
Expanded the determinant on the left side of above equation, we have
n ( A, B , C ) is orthogonal to k (0,0,1), and n k C 0.
Therefore, the equation of this plane is
Ax By D 0.
z
Similarly, the equations of planes which
a plane with normal vector n ( A, B , C ), if A, B, C are not all 0. In fact, if C 0, the equation can be written as
D A( x 0) B( y 0) C z 0. C
( Ai Bj Ck) ( x x0 )i ( y y0 )j ( z z0 )k 0
or
Point normal form scalar equation of the plane
A( x x0 ) B( y y0 ) C ( z z0 ) 0.
are parallel to the x-axis or y-axis are
By Cz D 0, Ax Cz D 0,
n
O
y
x
respectively.
12
Some Planes with Special Locations
(3) If a given plane is orthogonal to the z-axis, then n // k and so
Lecture 5
Planes and Lines in Space
Equations for Planes in Space
A plane in space is determined by knowing a point on the plane and its “tilt” or orientation. Suppose that plane M passes through a point P0 ( x0 , y0 , z0 ) and is normal (perpendicular) to the nonzero vector n Ai Bj Ck. Then M is the sets of all points P ( x , y , z ) for which P0 P is orthogonal to n. Thus, the dot product n P0 P 0. This equation is equivalent to
Ax D 0, By D 0,
x
respectively.
13
Some Problems about Planes
(1) Angle Between Planes
The angle between two intersecting
n2
planes is defined to be the (acute)
The component equation is
5( x ( 3)) 2( y 0) ( 1)( z 7) 0.
Simplifying, we obtain
5 x 15 2 y z 7 0 5 x 2 y z 22.
4
Finding an Equation for a Plane through Three Points
Ax By Cz D ,
where
D Ax0 By0 Cz0
Therefore, the equation of any plane is a linear equation in three
variables. Conversely, any linear equation in three variables represents
angle determined by the normal vectors as shown in the figure. Let

n1
2
1

1 : A1 x B1 y C1 z D1 0,
n2 ( A2 , B2 , C2 ), respectively. Then
|| n1 n 2 || cos || n1 || || n 2 ||
6
Finding an Equation for a Plane Through Three Points
Solution II Suppose that P ( x , y , z )
z
is any point in the plane, then
AB (2,0, 1) AP ( x, y, z 1)
AB AC 2 0 1 3i 2j 6k n. 0 3 1 i j k
O
z
n
A(0,0,1)
y
C (0, 3,0)
x
B(2,0,0)
5
Finding an Equation for a Plane Through Three Points
Solution (continued) It is easy to see that n is normal to the plane. We substitute the components of this vector and the coordinate of A(0,0,1) into