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Find an orthonormal basis in the subspace $Bbb R^4$ spanned by all solutions of $x+2y+3z-6j=0$. Then express vector $b = (1,1,1,1)$ to this basis.

I'm very confused on how to start this question. I know I would have to use
The Gram-Schmidt process but I'm not sure how.

Thank you

First solve the equation parametrically, say: $x=-2y-3z+6j$. This shows the subspace of solutions is isomorphic to $mathbf R^3$.

Then find a basis of the hyperplane, for instance:
$$u=begin{bmatrix}2\-1\0\0end{bmatrix},;v=begin{bmatrix}3\0\-1\0end{bmatrix},;w=begin{bmatrix}6\0\0\1end{bmatrix}. $$

Apply Gram-Schmidt to find an orthogonal basis:
$$e_1=u,quad e_2=v-frac{langle u,vrangle}{langle u,urangle},u, quad e_3=w-frac{langle u,wrangle}{langle u,urangle},u-frac{langle vu,wrangle}{langle v,vrangle},v$$
(i.e. subtract from $v$ the orthogonal projection of $v$ onto the line directed by u, and subtract from $w$ the projection of $w$ onto the subspace generated by $u$ and $v$).

Finally, normalise:
$$e'_1=frac{e_1}{langle e_1,e_1rangle}, quad e'_2=frac{e_2}{langle e_2,e_2rangle}, quad e'_3=frac{e_3}{langle e_3,e_3rangle}.$$

From x+ 2y+ 3z- 6j= 0 we have x= 6j- 3z- 2y. So any vector in that subspace can be written as = <6j- 3z- 2y, y, z, j> = j<6, 0, 0, 1>- z<-3, 0, 1, 0>+ y<-2, 1, 0, 0>. {<6, 0, 0, 1>, <-3, 0, 1, 0>, <-2, 1, 0, 0>} is a basis for that three dimensional vector space.

Using the Gran-Schmidt process, <6, 0, 0, 1> has length $sqrt{7}$ and a unit vector in that direction is $left<frac{6}{sqrt{7}}, 0, 0, frac{1}{sqrt{7}}right>$. Now, note that the dot product of that and <-3, 0, 1, 0> is $frac{-18}{sqrt{7}}$ so that $frac{-18}{sqrt{7}}left<frac{6}{sqrt{7}}, 0, 0, frac{1}{sqrt{7}}right>= left<-frac{108}{7}, 0, 0, -frac{18}{7}right>$ is its projection on $left<frac{6}{sqrt{7}}, 0, 0, frac{1}{sqrt{7}}right>$ and then $left<-3, 0, 1, 0right>- left<frac{6}{sqrt{7}}, 0, 0, frac{1}{sqrt{7}}right>= left<-3- frac{6}{sqrt{7}}, 0, 1- frac{1}{sqrt{7}}, 0right>= left<frac{-3sqrt{7}- 6}{sqrt{7}}, 0, frac{sqrt{7}- 1}{sqrt{7}}, 0right>$ is perpendicular to it. Divide that by its length to get a second unit vector perpendicular to the first. Continue!

You could circumvet part 2 if you chose $begin{bmatrix} frac 12\frac 12\frac 12\frac 12end{bmatrix}$ as your first basis vector.

i.e. A non-orthonormal basis is

$begin{bmatrix} 1\1\1\1end{bmatrix},begin{bmatrix} -2\1\0\0end{bmatrix},begin{bmatrix} 0\0\2\1end{bmatrix}$

Then apply Graham-Schmidt. Normalize the first vector

$begin{bmatrix} frac 12\frac 12\frac 12\frac 12end{bmatrix}$

Then take the inner product of the normalized first vector and the second vector

$begin{bmatrix} frac 12&frac 12&frac 12&frac 12end{bmatrix}begin{bmatrix}-2\1\0\0end{bmatrix} = -1$

and

$begin{bmatrix}-2\1\0\0end{bmatrix} - (-1)begin{bmatrix} frac 12\frac 12\frac 12\frac 12end{bmatrix} = begin{bmatrix}-frac32\frac 12\frac 12\frac 12end{bmatrix}$

and normalize it:

$begin{bmatrix}-frac 3{sqrt {12}}\frac 1{sqrt {12}}\frac 1{sqrt {12}}\frac 1{sqrt {12}}end{bmatrix}$

and finally
$begin{bmatrix} frac 12&frac 12&frac 12&frac 12end{bmatrix}begin{bmatrix} 0\0\2\1end{bmatrix} = frac 32,begin{bmatrix}-frac 3{sqrt {12}}&frac 1{sqrt {12}}&frac 1{sqrt {12}}&frac 1{sqrt {12}}end{bmatrix}begin{bmatrix} 0\0\2\1end{bmatrix} = frac {3}{sqrt 12}$

$begin{bmatrix} 0\0\2\1end{bmatrix} + frac 32begin{bmatrix} frac 12\frac 12\frac 12\frac 12end{bmatrix} - frac {3}{sqrt12} begin{bmatrix}-frac 3{sqrt {12}}\frac 1{sqrt {12}}\frac 1{sqrt {12}}\frac 1{sqrt {12}}end{bmatrix} = begin{bmatrix} 0\-1\1\0end{bmatrix}$

and normalize: $begin{bmatrix} 0\-frac 1{sqrt 2}\frac 1{sqrt 2}\0end{bmatrix}$

And since we cheated part 2 is easy.

If that is too cheating... Suppose we start with

$begin{bmatrix} -2\1\0\0end{bmatrix},begin{bmatrix} 0\0\2\1end{bmatrix}, begin{bmatrix} 1\1\1\1end{bmatrix}$

Now the first two vectors are already orthogonal, they just need to be normalized, and the last vector

$begin{bmatrix} 1\1\1\1end{bmatrix} + frac 15begin{bmatrix} -2\1\0\0end{bmatrix} - frac {3}{5}begin{bmatrix} 0\0\2\1end{bmatrix} = begin{bmatrix} 0.6\1.2\-0.2\0.4end{bmatrix}$

I took a short cut. Rather than normalizing the first two basis vectors, I divided the scalar by the square of the norm.

We still need to normalize though.

$begin{bmatrix} -frac {2}{sqrt5}\frac 1{sqrt 5}\0\0end{bmatrix},begin{bmatrix} 0\0\frac2{sqrt 5}\frac1{sqrt 5}end{bmatrix}, begin{bmatrix} frac {3}{5sqrt2}\frac {6}{5sqrt2}\-frac {1}{5sqrt2}\frac {2}{5sqrt2}end{bmatrix}$

And now to find our $b$ in terms of this basis.

Take the inner product of $b$ with each vector and that is the weight in the new basis.

$begin{bmatrix}1&1&1&1end{bmatrix}begin{bmatrix} -frac {2}{sqrt5}&0&frac {3}{5sqrt2}\frac 1{sqrt 5}&0&frac {6}{5sqrt2}\0&frac{2}{sqrt5}&-frac{1}{5sqrt 2}\0&frac {1}{sqrt 5}&frac {2}{5sqrt2}end{bmatrix}= begin{bmatrix} frac {1}{sqrt5}&frac {3}{sqrt5}&sqrt2end{bmatrix}$

First of all you need three vectors, perpendicular to $v=(1,2,3,-6)^t$, which are also mutually perpendicular.

Start with $(2,-1,0,0)^t$ ans $(0,0,6,3)^t$. Now obviously the vector $(3,6,-1,2)^t$ matches the conditions. Normalizing gives
$${b_1,b_2,b_3}=left{frac{sqrt5}{5}begin{pmatrix}2\-1\0\0end{pmatrix},
frac{sqrt5}{5}begin{pmatrix}0\0\2\1end{pmatrix},
frac{sqrt2}{10}begin{pmatrix}3\6\-1\2end{pmatrix}
right}$$
as an orthonormal basis.
Let $w=(1,1,1,1)^t$, then
$$w=langle b_1,wrangle b_1+langle b_2,wrangle b_2+
langle b_3,wrangle b_3.$$