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Find the equation of the plane in xyz-space through the point P=(3,2,3) and perpendicular to the vector n=(3,5,−2).

What I tried:

Equation of the line:

3x + 5y - 2z + D = 0

replace with the point:

3(3) + 5(2) - 2(3) + D = 0
9 + 10 - 6 + D = 0
13 + D = 0
D = -13

Complete equation:

3x + 5y - 2z - 13 = 0

Solve for z:

z = (3x + 5y - 13)/2

The webwork system I'm using says my answer is not good. Please help

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2 Answers

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A very convenient fact about planes and normal vectors to those planes are that if a plane has the form $ax+by+cz=d$, $(a,b,c)$ is normal to the plane.

Hence, we know that $3x+5y-2z=k$ where $k$ is a constant are all perpendicular planes to the vector. Let us plug in $(3,2,3)$ into this equation to find $k=13$. So, our desired plane equation is $$3x+5y-2z=13$$I am not sure why your website is flagging this. Perhaps it's because you wrote the equation in an equivalent but not as recognizable format for the software.

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Note that cartesian equation $3x + 5y - 2z + D = 0$ with the free parameter $D$ represents all the planes with normal $n$ and by the condition that the point $P$ belongs to the plane we find the equation of the plane we are looking for $3x + 5y - 2z -13 = 0$ and we are done.

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