Find the value of λ and μ if (2 i + 6 j + 27 k ) × ( i +λ j +μ k ) = 0

We have, (2\(\hat{i}\)+ 6\(\hat{j}\)+ 27 \(\hat{k}\)) × (\(\hat{i}\) +λ\(\hat{j}\) +μ\(\hat{k}\)) =\(\vec{0}\)

⇒ (2\(\hat{i}\)+ 6\(\hat{j}\)+ 27 \(\hat{k}\)) × (\(\hat{i}\) +λ\(\hat{j}\) +μ\(\hat{k}\)) =\(\vec{0}\) = (0\(\hat{i}\)+ 0\(\hat{j}\)+ 0 \(\hat{k}\))

⇒ (6μ-27λ)\(\hat{i}\) + (2μ - 27)\(\hat{j}\)+(2λ-6)\(\hat{k}\) = (0\(\hat{i}\)+ 0\(\hat{j}\)+ 0 \(\hat{k}\)) 

on comparing the corresponding components , we have:

(6μ-27λ) = 0,   (2μ - 27) = 0 ,      (2λ-6) =0

solving the above equation, we have    

λ = 3 and  μ= 27/2