Problem source

Prove that: 
\begin{questionparts}
\item if $a+2b+3c=7x$, then 
\[
a^{2}+b^{2}+c^{2}=\left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2};
\]
\item if $2a+3b+3c=11x$, then 
\[
a^{2}+b^{2}+c^{2}=\left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2}.
\]
\end{questionparts}
Give a general result of which $\textbf{(i) }$and $\textbf{(ii) }$are
special cases.

Solution source

\begin{questionparts}
\item \begin{align*}
\left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2} &= x^2-2ax+a^2 + 4x^2 -4bx+b^2 + 9x^2-6cx + c^2 \\
&= (1^2 + 2^2 + 3^2)x^2 - 2x(a+2b+3c) +a^2+b^2 + c^2 \\
&= 14x^2 - 2x(7x) + a^2 + b^2 + c^2 \\
&=  a^2 + b^2 + c^2
\end{align*}

\item 
\begin{align*}
\left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2} &= (2^2+3^2+3^2)x^2 - 2x(2a+3b+3c) + (a^2 + b^2+c^2) \\
&= 22x^2 - 2x(11x) + a^2+b^2+c^2 \\
&= a^2+b^2+c^2
\end{align*}
\end{questionparts}

The general result is:

If $\frac{A^2+B^2+C^2}{2}x =Aa+Bb+Cc$ then $(Ax-a)^2 + (Bx-b)^2 + (Cx-c)^2 = a^2+b^2+c^2$

Alternatively, if $\lambda = \frac{2\mathbf{x} \cdot \mathbf{y}}{\Vert x \Vert^2}$ then $\Vert \lambda \mathbf{x} - \mathbf{y}\Vert^2 = \Vert \mathbf{y} \Vert^2$ which is easy to see is true.