1
$\begingroup$

Can someone please help me understand the difference between the following:

  1. Find the equation of the plane that passes through the point $(2,-1,3)$ and is parallel to the plane $5x-2y+z-5=0$.

In this case, is the plane's normal vector equal to the other plane's $(5x-2y+z-5=0)$ normal vector? i.e. $n=(5,-2,1)$ Since they are parallel?
∴ $(x-2,y+1,z-3).(5,-2,1)=0$, would then be the equation in vector form?

  1. Find the equation of the plane that passes through the point $(1,-1,2)$ and is orthogonal/perpendicular to the plane $x+2y-3z-4=0$

In this case, where the two planes are orthogonal, is the plane's normal vector also equal to the other plane's $(x+2y-3z-4=0)$ normal vector? i.e. $n=(1,2,-3)$
∴ $(x-1,y+1,z-2).(1,2,-3)=0$, would then be the equation in vector form?

  1. Find the equation of the plane that passes through the point $(1,-3,-4)$ and is orthogonal/perpendicular to the line: $x=4+2t, y=-2+3t, z=-5t$

In this case, where the plane is now orthogonal to a line, is the plane's normal vector equal to the line's $(x=4+2t, y=-2+3t, z=-5t)$ vector? i.e. $n=v=(2,3,-5)$
∴ $(x-1,y+3,z+4).(2,3,-5)=0$, would then be the plane's equation in vector form?

$\endgroup$
  • $\begingroup$ There may be an error at #2 in the questions I have received in a study guide. I think another orthogonal plane should be given? Not just one. So that the cross product can be found. $\endgroup$ – bb411 Jun 4 '17 at 15:54
  • $\begingroup$ While there is certainly an infinite number of planes that satisfy the description in #2, $(x-1,y+1,z-2)\cdot(1,2,-3)=0$ isn’t one of them. $\endgroup$ – amd Jun 5 '17 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.