How do I prove that the distance between two points is longer than one point in the normal and the other point in a plane? So, imagine $P$ is a plane with the normal $n$, and $A = (a1, a2, a3)$ is a point above the plane $P$ that does not lie on the plane. $B$ is a point collinear on the line $L$: $r = (a_1, a_2, a_3) + \lambda n$.
$C$ is a point on $P$ other than $B$.
How do I prove $\|A-C \| > \|A-B \|$ for all points $C \ne B$?
I tried drawing a diagram with points $A$ and $B$ in a line $L$ parallel to the normal, since the shortest distance from $A$ to $P$ is $L$ perpendicular. I drew a line from $B$ to $C$. Since the $L$ is parallel to the normal $n$, $L$ is perpendicular to the plane $P$.
Hence, this would be a right-angled triangle $ABC$.
Now, a hypothenuse is longer than the other two sides.
What's the best way to approach this question?
 A: You have $\mathbf{B} = \mathbf{A} + \lambda \mathbf{n} $ for some $\lambda$.  Point $\mathbf{C}$ is in the plane and $\mathbf{C} \ne \mathbf{B}$, therefore,
$ \mathbf{n} \cdot (\mathbf{C} - \mathbf{B}) = 0 $
Now
$ \mathbf{A} - \mathbf{C} = (\mathbf{A} - \mathbf{B}) + (\mathbf{B} - \mathbf{C}) = -\lambda \mathbf{n} + \mathbf{CB} $
It follows from this that
$\begin{equation} \begin{split}
\| \mathbf{A - C} \|^2 &= (\mathbf{A - C}) \cdot (\mathbf{A - C} ) \\&= (-\lambda \mathbf{n} + \mathbf{CB} ) \cdot (-\lambda \mathbf{n} + \mathbf{CB}) \\
&= \lambda^2 \| \mathbf{n} \|^2 + \| \mathbf{CB} \|^2 \end{split} \end{equation} $
Because $\mathbf{n} \cdot \mathbf{CB} = 0 $
Therefore,
$\| \mathbf{A - C} \|^2 \gt \| \mathbf{B} - \mathbf{C} \|^2 $
because $\lambda^2 \| \mathbf{n} \|^2 \gt 0$ for $\lambda \ne 0 $
Taking the square root of both sides of the above inequality, we deduce that
$\| \mathbf{A - C} \| \gt \| \mathbf{B} - \mathbf{C} \| $
A: I will write vectors A,B,C without overset arrows.
...
We want to show that $||(A-C)||>||(A-B)||$ or after squaring, we want to show that $(A-C)\bullet(A-C)> (A-B)\bullet(A-B)$. This simplifes as
$$A\bullet A-2A\bullet C+C\bullet C\overset{?}> A\bullet A-2A\bullet B+B\bullet B,$$
$$-2A\bullet C+C\bullet C\overset{?}> -2A\bullet B+B\bullet B,$$
$$2A\bullet B-2A\bullet C\overset{?}> B\bullet B-C\bullet C,$$
$$2A\bullet(B-C)\overset{?}> B\bullet B-C\bullet C.  \tag{1}$$
On the other hand, $(B-A)\perp (B-C)$, so $(B-A)\bullet(B-C)=0$ giving
$$A\bullet(B-C)=B\bullet(B-C)\tag{2}$$
Substituting $(2)$ in $(1)$:
$$2B\bullet(B-C)\overset{?}> B\bullet B-C\bullet C$$
$$2B\bullet B-2B\bullet C\overset{?}> B\bullet B-C\bullet C$$
$$B\bullet B-2B\bullet C+C\bullet C\overset{?}> 0$$
$$||B-C||^2> 0$$
which is true since $C\neq B$.
