# Vector Geometry

• Oct 19th 2012, 07:07 PM
bastiante
Vector Geometry
Determine the equation of the plane which distance to the origin is equal to 3 and passing through the intersection of the planes x+y+z-11 = 0 and
x-4y+5z-10=0

• Oct 19th 2012, 08:29 PM
chiro
Re: Vector Geometry
Hey bastiante.

Can you show us what you have tried? (Hint: If it goes through both planes then the area of intersection will be where P1 = P2 or Plane1 = Plane2).
• Oct 19th 2012, 09:33 PM
MaxJasper
Re: Vector Geometry
Equation of system of planes containing the line of intersection of the two planes:

$\displaystyle x+y+z-11+k(x-4y+5z-10)=0$

Expanding and collecting terms planes are:

$\displaystyle -11-10 k+(1+k) x-(-1+4 k) y+(1+5 k) z=0$

Distance of point (0,0,0) from these planes:

$\displaystyle d=\left|\frac{-10 k+0 (k+1)+0 (5 k+1)+0 (1-4 k)-11}{\sqrt{(k+1)^2+(5 k+1)^2+(1-4 k)^2}}\right| = 3$

Solve to find k={1, -47/139}. Hence equation of two planes with distance 3 and containing line of intersection of the two given planes are:

$\displaystyle -21+2 x-3 y+6 z=0$

and:

$\displaystyle -1059+92 x+327 y-96 z = 0$

NB.: Check the 2 planes if both are solutions(Shake)
• Oct 20th 2012, 10:30 AM
bastiante
Re: Vector Geometry
Thank you very much! I did not know that equation for intersection of two planes