Find the equation of the plane that passes through the point ((3,,-2,,1)) and is parallel to the plane (2x-y+3z=4).
Answer
(2x-y+3z=11)
The correct answer is (2x-y+3z=11) — Therefore the equation of the plane is (2x-y+3z=11).
Step-by-step solution
- The normal vector of a plane is unchanged if the plane is translated; hence the required plane has the same normal vector ((2,-1,3)).
- Use the point ((3,-2,1)) to find the constant term: (2(3)-(-2)+3(1)=6+2+3=11).
- Therefore the equation of the plane is (2x-y+3z=11).
Common mistake: Substituting the point into the original plane equation instead of the new one
Concept tested: Equation of a plane parallel to a given plane
Published · How this solution was prepared and checked
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