# Let A = (1, 1, 1, 0), B = (-1, 0, 1, 1,), C = (3, 2, -1, 1) and let D = {Q in R^{4} | Q perp A, Q perp B, Q perp C}

Let $A=\left(1,1,1,0\right),B=\left(-1,0,1,1,\right),C=\left(3,2,-1,1\right)$
and let $D=\left\{Q\in {R}^{4}|Q\perp A,Q\perp B,Q\perp C\right\}$.
Convince me that D is a subspace of ${R}^{4}$. Write D as span of a basis. Write D as a span of an orthogonal basis

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We will denote the scalar product as $⟨-,-⟩$
Let .
Then $⟨\alpha v+\beta w,A⟩=\alpha ⟨v,A⟩+\beta ⟨w,A⟩=0$
This is because, since v, w in D, we have that .
So, $\alpha v+\beta w\perp A$
Similarly, we get that

Thus, $\alpha v+\beta w\in D.$
This proves that D is a subspace of ${R}^{4}$
. Now let .
Then we have that $⟨\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right),A⟩=0⇒{x}_{1}+{x}_{2}+{x}_{3}=0$
$⟨\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right),B⟩=0⇒-{x}_{1}+{x}_{2}+{x}_{3}=0$
$⟨\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right),C⟩=0⇒-3{x}_{1}-2{x}_{2}-{x}_{3}+{x}_{4}=0$
Thus, we have a system of equations
${x}_{1}+{x}_{2}+{x}_{3}=0$
$-{x}_{1}+{x}_{3}+{x}_{4}=0$
$-3{x}_{1}-2{x}_{2}-{x}_{3}+{x}_{4}=0$
The matrix of this homogeneous system is $\left[\begin{array}{cccc}1& 1& 1& 0\\ -1& 0& 1& 1\\ -3& -2& -1& 1\end{array}\right]$
We will now use the following elementary row operations:
1. Interchange the ith and jth row: ${R}_{i}↔{R}_{j}$
. 2, Multiply the eth row by a constant $c\ne 0:c{R}_{i}$
3. Multiply the jth row by a constant c and add it to the ith row:

${R}_{j}+c{R}_{j}$ $\left[\begin{array}{cccc}1& 1& 1& 0\\ -1& 0& 1& 1\\ -3& -2& -1& 1\end{array}\right]$
$\left({R}_{3}+3{R}_{1}\right)\to \left[\begin{array}{cccc}1& 1& 1& 0\\ 0& 1& 2& 1\\ 0& 1& 2& 1\end{array}\right]$
$\left({R}_{3}+3{R}_{1}\right)\to \left[\begin{array}{cccc}1& 0& -1& -1\\ 0& 1& 2& 1\\ 0& 1& 2& 1\end{array}\right]$
$\left({R}_{3}+3{R}_{1}\right)\to \left[\begin{array}{cccc}1& 0& -1& -1\\ 0& 1& 2& 1\\ 0& 0& 0& 0\end{array}\right]$
This means that we have to have 2 parameters: ${x}_{3}=s,{x}_{4}=t,$ and
${x}_{1}=s+t$
${x}_{2}=-2s-t$
This means that