Match the following

(i) Solved the two equations, say, i.e. \(x+y=|a|\) and \(a x-y=1\), we get \(x=\frac{|a|+1}{a+1}>0\) and \(y=\frac{|a|-1}{a+1}>0\)
when \(a+1>0\); we get \(a>1\)
\(\therefore \quad a_0=1\)
(ii) We have,
\(\mathbf{a}=\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+\gamma \hat{\mathbf{k}}\)
\(\Rightarrow \quad \mathbf{a} \cdot \hat{\mathbf{k}}=\gamma\)
Now, \(\hat{\mathbf{k}} \times(\hat{\mathbf{k}} \times \mathbf{a})=(\hat{\mathbf{k}} \cdot \mathbf{a})-(\hat{\mathbf{k}} \cdot \hat{\mathbf{k}}) \mathbf{a}=\gamma \hat{\mathbf{k}}-(\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+\gamma \hat{\mathbf{k}})\) \(\Rightarrow \quad \alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}=0 \Rightarrow \alpha=\beta=0\)
Also, \(\quad \alpha+\beta+\gamma=2 \Rightarrow \gamma=2\)


(iv) \(\sin A \sin B \sin C+\cos A \cos B \leq \sin A \sin B+\cos A \cos B=\cos (A-B)\)
\[
\begin{array}{lc}
\Rightarrow & \cos (A-B) \geq 1 \Rightarrow \cos (A-B)=1 \\
\Rightarrow & \sin C=1
\end{array}
\]