If (2 hat{i} + 6 hat{j} + 27 hat{k}) times (h | vedclass

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(C) Given the cross product of two vectors is the zero vector,$\vec{a} 	imes \vec{b} = \vec{0}$,which implies the vectors are collinear.We can write

Vedclass · Accepted Answer

$3, \frac{27}{2}$

Vedclass · Answer

(C) Given the cross product of two vectors is the zero vector,$\vec{a} 	imes \vec{b} = \vec{0}$,which implies the vectors are collinear.We can write the cross product as a determinant:$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 6 & 27 \ 1 & \lambda & \mu \end{vmatrix} = \vec{0}$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 coefficients of $\hat{i}, \hat{j}, 	ext{ and } \hat{k}$:$6\mu - 27\lambda = 0 \quad \dots(1)$$2\mu - 27 = 0 \quad \dots(2)$$2\lambda - 6 = 0 \quad \dots(3)$From equation $(3)$,$2\lambda = 6 \implies \lambda = 3$.From equation $(2)$,$2\mu = 27 \implies \mu = \frac{27}{2}$.Substituting these values into equation $(1)$: $6(\frac{27}{2}) - 27(3) = 3(27) - 81 = 81 - 81 = 0$. The values satisfy the equation.Thus,$\lambda = 3$ and $\mu = \frac{27}{2}$.

If $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = \vec{0}$,then $\lambda$ and $\mu$ are respectively:

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