For vec a=\1,1,1\ , vec b=\1,1,0\ , vec c=\1,-1,0\ find vec…
For $\vec a=\{1,1,1\}$, $\vec b=\{1,1,0\}$, $\vec c=\{1,-1,0\}$ find $\vec x$ from $\vec x\cdot\vec a=3$ and $\vec x\times\vec b=\vec c$.
Explanation of Method
The vector x from one scalar-product and one vector-product condition.
Step 1: Existence
\[ \vec c\cdot\vec b=1-1+0=0\ \checkmark \]
Step 2: Solutions of $\vec x\times\vec b=\vec c$ (Problem 105)
\[ \vec b\times\vec c=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&0\\1&-1&0\end{vmatrix}=\{0,0,-2\},\qquad |\vec b|^2=2, \]
\[ \vec x=\frac{\vec b\times\vec c}{|\vec b|^2}+\mu\vec b=\{0,0,-1\}+\mu\{1,1,0\}=\{\mu,\ \mu,\ -1\}. \]
Step 3: The scalar condition
\[ \vec x\cdot\vec a=\mu+\mu-1=3\ \Longrightarrow\ \mu=2. \]
\[\vec x=\{2,\ 2,\ -1\}\]
Step 4: Check
\[ \vec x\cdot\vec a=2+2-1=3\ \checkmark,\qquad \vec x\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\2&2&-1\\1&1&0\end{vmatrix}=\vec i(0+1)-\vec j(0+1)+\vec k(2-2)=\{1,-1,0\}=\vec c\ \checkmark \]
Result
\[\boxed{\,\vec x=\{2,\ 2,\ -1\}\,}\]
Remark
Correction. The source uses the sign-reversed formula from Problem 105 and gets $\{\mu,\mu,1\}$ and $\vec x=\{1,1,1\}$. Its own check then gives $\vec x\times\vec b=\{-1,1,0\}=-\vec c$, after which it concludes that the problem is inconsistent and needs a sign change. The problem is consistent as stated. With the correct formula the unique answer is $\vec x=\{2,2,-1\}$, which satisfies both conditions.