4

Find an equation of a plane that passes through $p(1,5,1)$ and is perpendicular to planes $2x+y-2z = 2$ and $x+3z=4$.

I basically need the 2 other points to make the vector and perform the cross product.

Since $ax+by+cz+d=0$ is the form of a plane. Can I obtain the points as $(a,b,c)$ of the 2 planes given? Also, how can I set them up to obtain the points?

Didier
  • 19,132
John
  • 349
  • Its normal is $(2,1,-2)\times (1,0,3)=(3,-8,-1)$ and so it is $3x-8y-z=-38$. – Bob Dobbs Nov 04 '22 at 22:15
  • @Ketan please be a little more careful when applying the algebraic-geometry tag - from the tag description, "[t]his tag should not be used for elementary problems which involve both algebra and geometry." – KReiser Nov 04 '22 at 22:25
  • I didn't see -38 in the answer. I so commented.@KReiser – Bob Dobbs Nov 04 '22 at 22:26

1 Answers1

5

you have two vectors(normals of the given planes) $$u=(2,1,-2), v=(1,0,3)$$ then $n=u×v$
The plane equation is then $$[(x,y,z)-P].n=0$$

Community
  • 1
Semsem
  • 7,651
  • I got $n = 5i-8j-k$, but what is $(x,y,z)-P$ suppose to be? and what is the period on $n$? – John Feb 04 '14 at 18:48
  • The equation of the plane is $[r-r_0].n=0$ where r is a general point on the plane and $r_0$ is a fixed point on the plane and so $r-r_0$ would be perpendicular to $n$ and their scalar product is zero – Semsem Feb 04 '14 at 18:51
  • oh i see. from the $n$ i obtained above. I get: $5(x-1)-8(y-5)-(z-1)=0$ – John Feb 04 '14 at 18:53
  • @John I think $n$ should be $3i-8j-k$ – Pratyush Nov 28 '15 at 06:30