Find and if (2 i+6 j+27 k) (i+ j+ k)=0 - Vidyadip

Question

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

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Answer

(b) $3, \frac{27}{2}$
Explanation: It is given that:
$(2 \hat{i}+6 \hat{j}+27 \hat{k}) X(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$
$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu\end{array}\right|=\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=\overrightarrow{0}$, equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$ on both sides, we get
$(6 \mu-27 \lambda)=0,(2 \mu-27)=0,(2 \lambda-6)=0$.
solving, we get $\lambda=3, \mu=\frac{27}{2}$

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