Let points be A(0,0,0) and B(1,1,1) and Q point on plane $x-y+z=3$ We have to find the value of Q for Maximum value for $|QA-QB|$.
Ithought the Q will be that point where the line AB intersects the plane .
From that i got Q is (3,3,3)
but it is wrong
Can anybody explain me why
I got the same answer as you. So maybe I am unable to help.
As $0-0+0-3<0$ and $1-1+1-3<0$, $A$ and $B$ lies on the same side of the plane.
By triangle inequality, $QA+AB\ge QB$ and $QB+AB\ge QA$. So
$$AB\ge QA-QB\ge -AB$$
$$|QA-QB|\le AB$$
So the maximum value is $\sqrt{(1-0)^2+(1-0)^2+(1-0)^2}=\sqrt{3}$.
It happens when $A$, $B$ and $Q$ are collinear. So $Q=(t,t,t)$ for some $t\in\mathbb{R}$.
$$t-t+t=3$$
$Q=(3,3,3)$.