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Find an equation of the plane. The plane that passes through the line of intersection of the planes $x − z = 3$ and $y + 4z = 2$ and is perpendicular to the plane $x + y − 3z = 5$

I did the cross product of the first two vectors and then cross product of that vector with the third vector but not sure if I am doing it correctly.

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  • $\begingroup$ you have actually a vector parallel to your plane, hence you can find a vector perpadicular to the plane that you are looking for and take a point on the intersection of the two given lines. Then you have eveything that you need. $\endgroup$
    – user321268
    Sep 19, 2016 at 19:38
  • $\begingroup$ Could you please write out the steps because I am very confused $\endgroup$
    – Cantwait428
    Sep 19, 2016 at 19:40

1 Answer 1

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A plane passing through the intersection of planes $H_1=0$ and $H_2=0$ belongs to the pencil generated by these planes, i.e. it has equation $\lambda H_1+\mu H_2=0$ for some $\lambda, \mu$.

In the present case, the equation will be $$\lambda(x-z)+\mu(y+4z)=\lambda x +\mu y+ (4\mu-\lambda)z=3\lambda+2\mu, $$ and it will be perpendicular to the plane $x-y+3z=5$ if they have orthogonal normal vectors.

As a vector normal to the plane is $(\lambda, \mu, 4\mu-\lambda)$, the orthogonality condition is $$\lambda-\mu+3(4\mu-\lambda)=-2\lambda+11\mu=0.$$ A solution (defined within a non-zero factor) is $\lambda=11$, $\mu=2$, whence the equation $$ 11x+2y-3z=37. $$

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