$Q)$ Let $W = \langle w_1, w_2, w_3 \rangle (\subset \mathbb{R^4})$
(Here the $w_1 = (1,0,1,-3), w_2=(-1,2,0,0)$, $w_3=(1,-8,2,9)$ )
Find the $v =(a,b,c,2) \in W^{\bot}$
The answer is at the bottom of this post.
My trial) Since $W^{\bot} \oplus W = \mathbb{R}^4$, $dim(W^{\bot}) = 1$
Hence $\langle v,w_i \rangle = 0$ for $i= \{1,2,3\}$
Therefore only solving the below equation, we can get the value of the $(a,b,c)$
$$ \left\{ \begin{array}{}a+c-6=0 \\ -a+2b=0 \\ a-8b+2c+18=0 \end{array} \right. $$
But the answer was $(a,b,c) = (6,3,0)$
So, Which the point do I have mistake in my trial?
Thanks.
You were on the right track.
Your system is equivalent to $c=6-a$, $a=2b$, and $a-8b+2c+18=0$,
which means $a-4a+12-2a+18=0$, which means $-5a+30=0$ or $a=6$.
Can you now solve for $b$ and $c$ ?