Find the equation of the plane passing through the intersection of line $$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$$ and the plane $$x-y+z=5$$

and parallel to a vector with direction ratios $<2,3,-2>$

Now point of intersection of given plane and given line is $(2,-1,2)$ and direction normal of required plane will be perpendicular to $<2,3,-2>$ but how would I get a unique equation of required plane?

  • $\begingroup$ Why would you you expect a "unique" such equation? THere are infinite planes through that intersection and parallel to the given vector. $\endgroup$ – DonAntonio Mar 7 '16 at 16:14
  • $\begingroup$ @Joanpemo So we will get infinite planes under given information? $\endgroup$ – Mathematics Mar 7 '16 at 16:17
  • 1
    $\begingroup$ Yes. After all, if you find such plane then you can spin it around that point in the desired direction and that way you get infinite planes. $\endgroup$ – DonAntonio Mar 7 '16 at 16:30

So you want a plane through $\;(2,-1,2)\;$ and parallel to $\;(2,3,-2)\;$ , so for any $\;(a,b,c)\in\Bbb R^3\;$ the following fulfills the conditions:

$$(2,-1,2)+t(2,3,-2)+s(a,b,c)\;,\;\;t,s\in\Bbb R\;$$


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