• # question_answer If $\mathbf{a}\times \mathbf{r}=\mathbf{b}+\lambda \mathbf{a}$ and $\mathbf{a}\,\,.\,\,\mathbf{r}=3,$ where $\mathbf{a}=2\mathbf{i}+\mathbf{j}-\mathbf{k}$ and $\mathbf{b}=-\mathbf{i}-2\mathbf{j}+\mathbf{k},$ then r and l are equal to  A) $\mathbf{r}=\frac{7}{6}\mathbf{i}+\frac{2}{3}\mathbf{j},\,\,\lambda =\frac{6}{5}$ B) $\mathbf{r}=\frac{7}{6}\mathbf{i}+\frac{2}{3}\mathbf{j},\,\,\lambda =\frac{5}{6}$ C) $\mathbf{r}=\frac{6}{7}\mathbf{i}+\frac{2}{3}\mathbf{j},\,\,\lambda =\frac{6}{5}$ D) None of these

• Given, $\mathbf{a}\times \mathbf{r}=\mathbf{b}+\lambda \,\mathbf{a}\Rightarrow (\mathbf{a}\times \mathbf{r})\,.\,\mathbf{a}=\mathbf{b}\,.\,\mathbf{a}+\lambda \,\mathbf{a}\,.\,\mathbf{a}$
• $\Rightarrow 0=\mathbf{b}\,.\,\mathbf{a}+\lambda \,|\mathbf{a}{{|}^{2}}\Rightarrow \lambda =-\frac{\mathbf{b}\,.\,\,\mathbf{a}}{|\mathbf{a}{{|}^{2}}}=\frac{5}{6}$
• Also, $(\mathbf{a}\times \mathbf{r})\times \mathbf{a}=\mathbf{b}\times \mathbf{a}+\lambda \,\mathbf{a}\times \mathbf{a}\Rightarrow \mathbf{r}=\frac{7}{6}\mathbf{i}+\frac{2}{3}\mathbf{j}$.