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Find the equations of the planes determined by the given conditions.

The plane contains the origin and is normal to the vector4,−1,5

Short Answer

Expert verified

The equation of the plane that is determined by the given conditions is 4x-1y+5z=0

Step by step solution

01

Given information

The plane that is normal to the vector4,−1,5 and contains the origin.

02

Calculation

Consider the plane that is normal to the vector⟨4,-1,5⟩ and contains the origin.

The goal is to determine the plane equation that is determined by the given conditions.

The equation of the plane with r0 a point that lies in the plane and n a vector normal to the plane is given by:

n·r-r0=0

The normal vector is:

n=⟨4,-1,5⟩

The point r0is:

r0=(0,0,0)

03

Calculation

The plane that contains the origin and is normal to the vector ⟨4,-1,5⟩ has the equation:

⟨4,-1,5⟩·(⟨x,y,z⟩-⟨0,0,0⟩)=0(Substitution)⟨4,-1,5⟩·⟨x,y,z⟩=0(Simplify)4x-1y+5z=0

Thus, the equation of the plane that is determined by the given conditions is 4x-1y+5z=0

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