Question

一条美国的微积分的数学问题

Find an equation of the plane.
The plane that passes through the point
(−2, 3, 3)
and contains the line of intersection of the planes
x + y − z = 3 and 4x − y + 5z = 5

只要答案，如果写过程，我就奖赏分吧！

Answer 1

Solution:

The nomal vector of plane L₁x + y - z = 3 is (1, 1, -1)
The nomal vector of plane L₂4x + y - z = 3 is (4, 1, -1)

Therefore, the directional vector N of the intersection of L₁and L₂
= (1, 1, -1)×(4, -1, 5) = (4, -9, -5)

We can find a point on the line of intersection of L₁and L₂:
Let x = 0, we have y - z = 3, and y - 5z = -5, solution is z = 2, and y = 5
Hence, point P(0, 5, 2) is a point on the line of intersection.

A is a given point (-2, 3, 3), vector AP = (2, 2, -1)

The vector of the normal direction of the plane required
= N cross times AP
= (4, -9, -5)×(2, 2, -1) = (19, -6, 26)

Sub (19, -6, 26) and the given point (-2, 3, 3) into ax + by + cz = d, we obtain
- 38 - 18 + 78 = d, d = 22

∴The equation of the plane required is 19x - 6y + 26z = 22 (Ans)

Answer 2

题意：求一个过直线（x+y-z=3与4x-y+5z=5交线）和不在此直线上的点（-2,3,3）的平面方程。
解：在直线（x+y-z=3与4x-y+5z=5交线）上取不同的两点（0,5,2）和（4,-4,-3）
│(x+2) (y-3) (z-3)│
根据空间解析几何三点确定平面公式，得行列式│(0+2) (5-3) (2-3)│=0
│(4+2) (-4-3) (-3-3)│
解此行列式方程，得19x-6y+26z=22
故所求平面方程是19x-6y+26z=22。

Answer 3

这是一道空间几何的题：
已知一个平面A过点M(−2, 3, 3)，且包含平面B:x+y−z=3与平面C:4x−y+5z=5的交线l，求平面A的方程?
解：设平面A的方程是：ax+by+cz=d
任取平面B与平面C交线l上的两点：
令z=0,解B和C的方程得点N(8/5, 7/5, 0)
令y=0,解B和C的方程得点O(20/9, 0, -7/9)
由平面A过点M, N, O，可建立方程组：
-2a+3b+3c=d
(8/5)a+(7/5)b=d
(20/9)a-(7/9)c=d
可解得：
a=(19/22)d
b=(-3/11)d
c=(13/11)d
令d=22，可得平面A的方程：
19x-6y+26z=22

Answer 4

这哪里是微积分了。