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

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(A) Given $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}$, $\overrightarrow{c}=\hat{i}$.First, calculate the cross

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(A) Given $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}$, $\overrightarrow{c}=\hat{i}$.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 on both sides:For $\hat{i}$: $\lambda + \mu = 0$.For $\hat{j}$: $\lambda + \mu = 0$.For $\hat{k}$: $\lambda = -1$.Since $\lambda + \mu = 0$, the value of $\lambda + \mu$ is $0$.

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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