Find the equation of the plane that passes through the points $(2,-3,1)$ and $(-1,1,-7)$, and perpendicular to the plane $x-2y+5z=-1$.

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Accepted Answer

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> [!TLDR]
> Setting up the plane through $(2,-3,1)$ with unknown normal $(a,b,c)$, then using the second point and the perpendicularity condition, gives normal $(4,7,2)$ and the plane $4x+7y+2z+11=0$.
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> Let the required plane be $a(x-2)+b(y+3)+c(z-1)=0$ (it passes through $(2,-3,1)$ by construction).
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> **Passes through $(-1,1,-7)$:**
> $$a(-1-2)+b(1+3)+c(-7-1)=0\ \Rightarrow\ -3a+4b-8c=0.$$
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> **Perpendicular to the plane $x-2y+5z=-1$** means the normals $(a,b,c)$ and $(1,-2,5)$ are perpendicular:
> $$a-2b+5c=0.$$
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> Solve the two equations. From the second: $a=2b-5c$. Substitute into the first:
> $$-3(2b-5c)+4b-8c=0\ \Rightarrow\ -6b+15c+4b-8c=0\ \Rightarrow\ -2b+7c=0\ \Rightarrow\ b=\dfrac{7c}{2}.$$
> …

Q.Find the equation of the plane that passes through the points $(2,-3,1)$ and $(-1,1,-7)$ , and perpendicular to the plane $x-2y+5z=-1$ .

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

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Q.Find the equation of the plane that passes through the points (2,−3,1)(2,-3,1)(2,−3,1) and (−1,1,−7)(-1,1,-7)(−1,1,−7), and perpendicular to the plane x−2y+5z=−1x-2y+5z=-1x−2y+5z=−1.

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Q.Find the equation of the plane that passes through the points $(2,-3,1)$ and $(-1,1,-7)$ , and perpendicular to the plane $x-2y+5z=-1$ .

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