# Find λ and μ if

Question:

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

Solution:

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

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

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

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 $\mu=\frac{27}{2}$.