Question

Find the distance from the origin to the plane ABC, where a=(1,1,0) b=(0,1,3) c= (-1,3,3)

I got my answer to be 9/7, im not sure if im right or not

Answer 1

consider the equation of the plane is\[ax+by+cz+d=0----------(1)\]
now for point a(1,1,0)
equation of the plane is
\[a+b+d=0--------(2)\\for~~~b~(0,1,3)\\b+3c+d=0--------(3)\\for~~c~(-1,3,3)\\-a+3b+3c+d=0-------(4)\\ now\\(2)-(3)~gives\\a=3c\\(3)-(4)~gives\\a=2b\\2b=3c\\b=3c/2\]
\[(2)-(4)~gives\\2a-2b-3c=0\\d=\frac{ -9c }{ 2 }\\so ~the~equation~of~the~plane~is \\3cx+(3c/2)y+cz-(9c/2)=0\\6x+3y+2z=9\]
\[distence~from~origin,d=\frac{ 9 }{ \sqrt{6^2+3^3+2^2} }=\frac{ 9 }{7 }\]

Answer 2

yes y're completely correct :)

Answer 3

Thank you very much, just wanted to check i was doing it right before i do them all using the wrong method :)