Find lambda and mu if (2 hat{i}+6 hat{j}+27 h | vedclass

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Question

(A) Given the cross product of two vectors is the zero vector: $(2 \hat{i}+6 \hat{j}+27 \hat{k}) 	imes(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrigh

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the cross product of two vectors is the zero vector: $(2 \hat{i}+6 \hat{j}+27 \hat{k}) 	imes(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$.We express the cross product as a determinant:$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 2 & 6 & 27 \ 1 & \lambda & \mu\end{array}ight|=0 \hat{i}+0 \hat{j}+0 \hat{k}$.Expanding the determinant:$\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=0 \hat{i}+0 \hat{j}+0 \hat{k}$.Comparing the components on both sides:$1) \ 6 \mu-27 \lambda=0$$2) \ -(2 \mu-27)=0 \Rightarrow 2 \mu=27 \Rightarrow \mu=\frac{27}{2}$$3) \ 2 \lambda-6=0 \Rightarrow 2 \lambda=6 \Rightarrow \lambda=3$.Substituting $\lambda=3$ into the first equation: $6(\frac{27}{2}) - 27(3) = 81 - 81 = 0$,which is consistent.Thus,the values are $\lambda=3$ and $\mu=\frac{27}{2}$.

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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