question_answer

Find the values of \[\lambda \] and \[\mu \] for which \[(2\hat{i}+6\hat{j}+27\hat{k})\times (\hat{i}+\lambda \hat{j}+\mu \hat{k})=\vec{0}\]

Answer:

Let\[\overrightarrow{a}=(2\hat{i}+6\hat{j}+27\hat{k})\]and \[\overrightarrow{b}=\hat{i}+\lambda \hat{j}+\mu \hat{k}\] Then, \[=(6\mu -27\lambda )\,\hat{i}-(2\mu -27)\hat{j}+(2\lambda -6)\hat{k}\] Now, \[(\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{0}\Leftrightarrow 2\lambda -6=0\Rightarrow \lambda =3\] and \[2\mu -27=0\Rightarrow \mu =\frac{27}{2}\]