Question

Find $\lambda$ and $\mu$, if $(2\hat{i}+6\hat{j}+27\hat{k})\times (\hat{i}+\lambda \hat{j}+\mu \hat{k})=0$

Answer 1

Given, $(2\hat{i}+6\hat{j}+27\hat{k})\times (\hat{i}+\lambda \hat{j}+\mu \hat{k})=0\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 6& 27\\ 1& \lambda & \mu \end{array}\right|=0\Rightarrow \hat{i}(6\mu -27\lambda )-\hat{j}(2\mu -27)+\hat{k}(2\lambda -6)=0\hat{i}+0\hat{j}+0\hat{k}$
On comparing the corresponding components, we have
$6\mu -27\lambda =0,-2\mu +27=0,2\lambda -6=0\Rightarrow 2\mu =-9\lambda ,\mu =\frac{27}{2}\text{and}\lambda =3$
Hence, $\lambda =3$ and $\mu =\frac{27}{2}$