A plane $\pi_2$ intersects $\pi_1$: $4x-2y+7z-3=0$ at right angles. Two points lie on $\pi_2$: $A(3,2,0)$ and $B(2,-2,1)$. Write a cartesian equation for $\pi_2$.

I know that the normals of these two planes must be perpendicular since the planes are perpendicular. So if $n_2=(a,b,c)$ is the normal for plane $2$ and $n_1=(4,-2,7)$ is the normal for plane $1$ then the dot product between $n_1$ and $n_2$ is $0$.

Not sure where else to go with this question.

  • $\begingroup$ Hint: the equation for $\pi_2$ is $ax+by+cz+d =0$. $\endgroup$
    – xbh
    Commented Aug 13, 2018 at 18:44

1 Answer 1


The plane $\pi_2$ has a cartesian equation of the type $ax+by+cz=d$. And you know that $(3,2,0),(2,-2,1)\in\pi_2(\iff3a+2b=2a-2b+c=d)$. Furthermore, $\pi_1$ and $\pi_2$ intersect at right angles, which is equivalent to the assertion$$(a,b,c).(4,-2,7)=0(\iff4a-2b+7c=0).$$Can you take it from here?

  • $\begingroup$ ok so I solved the two equations: 4a-2b+7c=0 and 3a+2b=2a-2b+c. I let c=t and I got a=t-(44/18)t, b=11t/18, c=t. How do I then change this to Cartesian form? $\endgroup$
    – VL12345
    Commented Aug 13, 2018 at 19:07
  • $\begingroup$ @VL12345 It sees that you missed the “$=d$” part of $3a+2b=2a-2b+c=d$. Anyway, the answer is $a=\frac{13}{28}d$, $b=-\frac{11}{56}d$, and $c=-\frac9{28}d$. I would take $d=56$, which means that a Cartesian equation of the plane is $26x-11y-18z=56$. $\endgroup$ Commented Aug 13, 2018 at 19:13

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