Theorem-1

 Let $\pmb{ V}$ be inner product space over a field $\pmb{ F}$ and $\pmb{S=\{x_{1},x_{2},…,x_{k} \}}$ be a orthogonal subset of non null vectors of $\pmb{ V}$. If $\pmb{ x\in V-L(S)}$ and $\pmb{y=x- \displaystyle\sum_{i=1}^{k} c_{i}x_{i}}$ then
 (a)   $\pmb{ x}$ is orthogonal to each $\pmb{ x_{i}}$

 (b)   $\pmb{ L\{x_{1},x_{2},…,x_{k},x \}=L\{x_{1},x_{2},…,x_{k},y \}}$ where $\pmb{ c_{i}=\frac{\lang x,x_{i}\rang}{\lang x_{i},x_{i}\rang}}$, $\pmb{ i=1,2,…,k}$ .

 Proof:   Given that $\pmb{S=\{x_{1},x_{2},…,x_{k} \}}$ be a orthogonal subset of non null vectors of the inner product space $\pmb{ V}$ over the field $\pmb{ F}$.
 Let $\pmb{ x\in V-L(S)}$ and $$\begin{array}{rrrr}& {\displaystyle y=x-{\displaystyle \sum _{i=1}^k{c}_i{x}_i}}& & \end{array}$$$$\begin{align} \pmb{y=x- \displaystyle\sum_{i=1}^{k} c_{i}x_{i}} \end{align}$$   where $\pmb{ c_{i}=\frac{\lang x,x_{i}\rang}{\lang x_{i},x_{i}\rang}}$, $\pmb{ i=1,2,…,k}$ .
 To prove $\pmb{ \lang y,x_{j} \rang=0,~j=1,2,…k}$.  Now for all $\pmb{j}$, $$\begin{align*} \pmb{ \lang y,x_{j} \rang}& = \pmb{ \left\lang x- \displaystyle\sum_{i=1}^{k} c_{i}x_{i}~,~x_{j} \right\rang}\\ & = \pmb{ \left\lang x,~x_{j} \right\rang} – \pmb{ \left\lang \displaystyle\sum_{i=1}^{k} c_{i}x_{i}~,~x_{j} \right\rang}\\ & = \pmb{ \left\lang x,~x_{j} \right\rang} – \pmb{ \displaystyle\sum_{i=1}^{k} c_{i} \left\lang x_{i}~,~x_{j} \right\rang}\\ & = \pmb{ \left\lang x,~x_{j} \right\rang} – \pmb{c_{j} \left\lang x_{j}~,~x_{j} \right\rang~since~\lang x_{i},x_{j}\rang=0~when~i\ne j}\\ & = \pmb{ \left\lang x,~x_{j} \right\rang} – \pmb{\frac{\lang x,x_{j}\rang}{\cancel{\lang x_{j},x_{j}\rang}} \cancel{\left\lang x_{j}~,~x_{j} \right\rang}~since~\lang x_{j},x_{j}\rang\ne 0, x_{j}~ is ~non~null }\\ & = \pmb{ \cancel{\lang x,~x_{j} \rang} – \cancel{\lang x,~x_{j} \rang} }\\ & = \pmb{0} \end{align*}$$  Therefore $\pmb{ \lang y,x_{j} \rang=0,~j=1,2,…k}$ .

 To prove $\pmb{ L\{x_{1},x_{2},…,x_{k},x \}=L\{x_{1},x_{2},…,x_{k},y \}}$

 First we prove that $\pmb{ L\{x_{1},x_{2},…,x_{k},x \}\subset L\{x_{1},x_{2},…,x_{k},y \}}$
 Let $$\begin{align*} & \pmb{z\in L\{x_{1},x_{2},…,x_{k},x \}}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}p_{i}x_{i}+px~for~some~p_{i},p\in F}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}p_{i}x_{i}+p \left(y+\displaystyle\sum_{i=1}^{k} c_{i}x_{i} \right)~using~(1) }\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}p_{i}x_{i}+py+ \left(\displaystyle\sum_{i=1}^{k} pc_{i}~x_{i} \right)}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}\left(p_{i}+pc_{i}\right)x_{i}+py}\\ \implies& \pmb{z\in L\{x_{1},x_{2},…,x_{k},y \}}\\ \end{align*}$$  Therefore $\pmb{ L\{x_{1},x_{2},…,x_{k},x \}=L\{x_{1},x_{2},…,x_{k},y \}}$.

 Lastly we prove that $\pmb{ L\{x_{1},x_{2},…,x_{k},y \}\subset L\{x_{1},x_{2},…,x_{k},x \}}$
 Let $$\begin{align*} & \pmb{z\in L\{x_{1},x_{2},…,x_{k},y \}}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}q_{i}x_{i}+qy~for~some~q_{i},q\in F}\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}q_{i}x_{i}+q \left(x-\displaystyle\sum_{i=1}^{k} c_{i}x_{i} \right)~using~(1) }\\ \implies & \pmb{z=\displaystyle\sum_{i=1}^{k}q_{i}x_{i}+qx- \left(\displaystyle\sum_{i=1}^{k} qc_{i}~x_{i} \right)}\\ \implies &\pmb{z=\displaystyle\sum_{i=1}^{k}\left(q_{i}-qc_{i}\right)x_{i}+qx}\\ \implies& \pmb{z\in L\{x_{1},x_{2},…,x_{k},x \}}\\ \end{align*}$$  Therefore $\pmb{ L\{x_{1},x_{2},…,x_{k},y \}\subset L\{x_{1},x_{2},…,x_{k},x \}}$ .
 Hence $\pmb{ L\{x_{1},x_{2},…,x_{k},y \}= L\{x_{1},x_{2},…,x_{k},x \}}$ .