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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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The unit normal for the given plane is <5,2,-1>.
The equation of the plane parallel to the given plane passing through (5,5,4) is therefore
5(x-5)+2(y-5)-1(z-4)=0
simplify =>
5x+2y-z=25+10-4=31

Answer: the plane through (5,5,4) parallel to 5x+2y-z=-6 is 5x+2y-z=31
User Carlsborg
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