Question

Let $S=x^2+y^2+z^2-2x+3y-4z-5=0$ and $A(1,0,-1)$ and $B(-1,0,-2)$, then

• A. $A$ lies inside and $B$ lies outside of $S=0$
• B. $A$ lies outside and $B$ lies inside of $S=0$
• C. $A$ and $B$ lies inside $S=0$
• D. $A$ lies on $S=0$ and $B$ outside of $S=0$

Answer 1

The correct option is A $A$ lies inside and $B$ lies outside of $S=0$

Given $S=x^2+y^2+z^2-2x+3y-4z-5=0$
Now $S\left(A\right)=1^2+0^2+(-1{)}^2-2(1)+3(0)-4(-1)-5=-1<0$
and $S\left(B\right)=(-1{)}^2+0^2+(-2{)}^2-2(-1)+3\left(0\right)-4(-2)-5=10>0$
Hence, point $A$ lie inside and point B lie outside the given sphere.