Question

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

Answer 1

$(2\hat{i}+6\hat{j}+27\hat{k})\times (\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$

$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 6& 27\\ 1& \lambda & \mu \end{array}\right|=0\hat{i}+0\hat{j}+0\hat{k}$

$\Rightarrow \hat{i}(6\mu -27\lambda )-\hat{j}(2\mu -27)+\hat{k}(2\lambda -6)=0\hat{i}+0\hat{j}+0\hat{k}$

On comparing the corresponding components, we have:

$6\mu -27\lambda =0$

$2\mu -27=0$

$2\lambda -6=0$

Now,
$2\lambda -6=0\Rightarrow \lambda =3$

$2\mu -27=0\Rightarrow \mu =\frac{27}{2}$

Hence, $\lambda =3$ and $\mu =\frac{27}{2}$.