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Find the equation of the plane through the point p=(5,5,4)p=(5,5,4) and parallel to the plane 5x+2y−z=−6.
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Find the equation of the plane through the point p=(5,5,4)p=(5,5,4) and parallel to the plane 5x+2y−z=−6.

User Markuscosinus
by
8.0k points

1 Answer

5 votes
If the required plane is parallel to the plane
Π : 5x+2y-z=-6
then they both have the normal vector <5,2,-1>

Given the required plane passes through the point (5,5,4), the new plane is defined by
5(x-5)+2(y-5)-1(z-4)=0
=>
plane: 5x+2y-z=31
User Emanresu
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8.0k points
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