If overrightarrow{a}=hat{i}+hat{j}+hat{k},ove | vedclass

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(A) Given vectors are $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,and $\overrightarrow{c}=\hat{i}$.First,calcula

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(A) Given vectors are $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,and $\overrightarrow{c}=\hat{i}$.First,calculate the cross product $\overrightarrow{a} 	imes \overrightarrow{b}$:$\overrightarrow{a} 	imes \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(0-1) + \hat{k}(1-1) = -\hat{i} + \hat{j}$.Now,calculate $(\overrightarrow{a} 	imes \overrightarrow{b}) 	imes \overrightarrow{c}$:$(-\hat{i} + \hat{j}) 	imes \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 0 \ 1 & 0 & 0 \end{vmatrix} = \hat{k}(0-1) = -\hat{k}$.We are given $(\overrightarrow{a} 	imes \overrightarrow{b}) 	imes \overrightarrow{c} = \lambda \overrightarrow{a} + \mu \overrightarrow{b}$.Substituting the vectors:$-\hat{k} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j})$$-\hat{k} = (\lambda + \mu)\hat{i} + (\lambda + \mu)\hat{j} + \lambda\hat{k}$.Comparing the coefficients of $\hat{i}$,$\hat{j}$,and $\hat{k}$:$1$) $\lambda + \mu = 0$$2$) $\lambda + \mu = 0$$3$) $\lambda = -1$From equation $(1)$,since $\lambda = -1$,we have $-1 + \mu = 0$,which implies $\mu = 1$.Therefore,$\lambda + \mu = -1 + 1 = 0$.

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,$\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$,then $\lambda+\mu$ is equal to:

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