If the three planes x=5, 2x-5ay+3z-2=0 and 3bx+y-3z=0 contain a common line, then (a,b) is…

Question

If the three planes $x=5$, $2x-5ay+3z-2=0$ and $3bx+y-3z=0$ contain a common line, then $(a,b)$ is equal to

Accepted Answer

Common line direction $(l,m,n)\perp(1,0,0)\Rightarrow l=0$. From the other two normals: $-5am+3n=0$ and $m-3n=0$, giving $n=	frac{5am}3=	frac m3\Rightarrow m(5a-1)=0$. Since $m
eq0$, $a=	frac15$.
On $x=5$: from plane 2, $8-y+3z=0\Rightarrow y=8+3z$; from plane 3, $15b+y-3z=0\Rightarrow y=3z-15b$. Equating: $8=-15b\Rightarrow b=-	frac8{15}$.

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