Find the equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to $x-y+z=0$.

Question

Accepted Answer

>

> [!TLDR]
> Use the pencil of planes through the intersection line, then force the plane to be perpendicular to $x-y+z=0$.
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> Any plane through the line of intersection of $x+y+z=1$ and $2x+3y+4z=5$ can be written as
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> $$(x+y+z-1) + \lambda(2x+3y+4z-5) = 0,$$
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> i.e.
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> $$(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z - (1+5\lambda) = 0.$$
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> Its normal vector is $\vec n = (1+2\lambda,\ 1+3\lambda,\ 1+4\lambda)$.
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> This plane is perpendicular to $x - y + z = 0$ (normal $(1,-1,1)$) when the two normals are perpendicular:
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> $$\vec n\cdot(1,-1,1) = 0 \;\Rightarrow\; (1+2\lambda)(1) + (1+3\lambda)(-1) + (1+4\lambda)(1) = 0.$$
> $$1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0 \;\Rightarrow\; 1 + 3\lambda = 0 \;\Rightarrow\; \lambda = -\frac13.$$
>
> Substitute $\lambda = -	frac13$:
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> - coefficient of $x$: $1 + 2(-	frac13) = 	frac13$, …

Q.Find the equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to $x-y+z=0$ .

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

Q.Find the equation of the plane through the line of intersection of the planes x+y+z=1x+y+z=1x+y+z=1 and 2x+3y+4z=52x+3y+4z=52x+3y+4z=5 which is perpendicular to x−y+z=0x-y+z=0x−y+z=0.

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Q.Find the equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to $x-y+z=0$ .

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