If \(\bar{a}=\hat{j}-\hat{k}\) and \(\bar{c}=\hat{i}-\hat{j}-\hat{k}\), then the vector \(\bar{b}\) satisfying \(\bar{a} \times \bar{b}+\bar{c}=\overline{0}\) and \(\bar{a} \cdot \bar{b}=3\) is

Given:
- \(\mathbf{a}=\mathbf{j}-\mathbf{k}\),
- \(\mathbf{c}=\mathbf{i}-\mathbf{j}-\mathbf{k}\),
- Find \(\mathbf{b}\) such that \((\mathbf{a} \times \mathbf{b})+\mathbf{c}=0\) and \(\mathbf{a} \cdot \mathbf{b}=\mathbf{3}\).
Step 1: Write b in component form:
\(\mathbf{b}=p \mathbf{i}+q \mathbf{j}+r \mathbf{k}\)
Step 2: Compute \(\mathbf{a} \times \mathbf{b}\) :
\(a =(0,1,-1), \quad b =(p, q, r)\)
\(a \times b =\left|\begin{array}{ccc} i & j & k \\ 0 & 1 & -1 \\ p & q & r\end{array}\right|= i (1 r-(-1) q)- j (0 r-(-1) p\)\()+ k (0 q-1 p)\)
\(a \times b =(r+q) i -(-p) j -p k\)
\(a \times b =(r+q) i +p j -p k\)
Step 3: Use \((\mathbf{a} \times \mathbf{b})+\mathbf{c}=0\) :
\((r+q) \mathbf{i}+p \mathbf{j}-p \mathbf{k}+(\mathbf{i}-\mathbf{j}-\mathbf{k})=0\)
Equating components:
- \(i\)-component: \(r+q+1=0 \Rightarrow r+q=-1\),
- \(j\)-component: \(p-1=0 \quad \Rightarrow \quad p=1\),
- \(k\)-component: \(-p-1=0 \Rightarrow p=-1\).
Step 4: Solve \(\mathbf{a} \cdot \mathbf{b}=3\) :
\(\mathbf{a} \cdot \mathbf{b}=(0) p+(1) q+(-1) r=q-r=3\)
From \(r+q=-1\) and \(q-r=3\), solve for \(q\) and \(r\) :
- Add equations: \(2 q=2 \Rightarrow q=1\),
- Substitute: \(r+1=-1 \Rightarrow r=-2\).
Final b:
\(\mathbf{b}=-\mathbf{i}+\mathbf{j}-\mathbf{k}\)
Answer: \(\mathbf{- i}+\mathbf{j}-\mathbf{k}\).