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
Help please!
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