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一条美国的微积分的数学问题

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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
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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)