Let \(\overline{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) and \(\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\) be two vectors. If \(\overline{\mathrm{c}}\) is a vector such that \(\overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}}\) and \(\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=0\), then \(\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}\) is

\(\text {Given, } \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}} \)
\( \Rightarrow \overline{\mathrm{b}} \times(\overline{\mathrm{c}}-\overline{\mathrm{a}})=\overline{0}\)
\(\Rightarrow \overline{\mathrm{b}}\) is parallel to \((\overline{\mathrm{c}}-\overline{\mathrm{a}})\).
\(\Rightarrow \overline{\mathrm{c}}-\overline{\mathrm{a}}=\lambda \overline{\mathrm{b}} \text { for some scalar } \lambda \)
\( \Rightarrow \overline{\mathrm{c}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}... (i)) \)
\( \Rightarrow \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{a}}+\lambda(\overline{\mathrm{b}} \cdot \overline{\mathrm{a}})\)
\(\Rightarrow 0=|\overline{\mathrm{a}}|^2+\lambda(\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}), \quad \cdots[\because \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=0\) (given) \(]\)
\(\Rightarrow 0=6 \pm 4 \lambda \)
\( \Rightarrow \lambda=-\frac{3}{2}\)
Substitutifing the value of \(\lambda\) in (i), we get
\(\bar{c} =(\hat{i}+2 \hat{j}-\hat{k})-\frac{3}{2}(\hat{i}+\hat{j}-\hat{k}) \)
\( = -\frac{1}{2}(\hat{i}-\hat{j}-\hat{k}) \)
\( \therefore \bar{c} \cdot \bar{b} =-\frac{1}{2}(\hat{i}-\hat{j}-\hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k}) \)
\( =-\frac{1}{2}(1-1+1)=-\frac{1}{2}\)