text{If } vec{a} = hat{i} + hat{j} + hat{k}, | vedclass

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(B) Given: $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}, \vec{c} = \hat{i} - \hat{j}$.First,calculate the cross pro

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$(\frac{4}{5}, \frac{11}{5}, \frac{19}{5})$

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(B) Given: $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}, \vec{c} = \hat{i} - \hat{j}$.First,calculate the cross products:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & -1 & 3 \end{vmatrix} = \hat{i}(3+1) - \hat{j}(3-2) + \hat{k}(-1-2) = 4\hat{i} - \hat{j} - 3\hat{k}$.$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 3 \ 1 & -1 & 0 \end{vmatrix} = \hat{i}(0+3) - \hat{j}(0-3) + \hat{k}(-2+1) = 3\hat{i} + 3\hat{j} - \hat{k}$.$\vec{c} 	imes \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 0 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-1-0) - \hat{j}(1-0) + \hat{k}(1+1) = -\hat{i} - \hat{j} + 2\hat{k}$.Now,equate $6\hat{i} + 2\hat{j} + 3\hat{k} = \lambda_1(4\hat{i} - \hat{j} - 3\hat{k}) + \lambda_2(3\hat{i} + 3\hat{j} - \hat{k}) + \lambda_3(-\hat{i} - \hat{j} + 2\hat{k})$.Comparing coefficients of $\hat{i}, \hat{j}, \hat{k}$:$4\lambda_1 + 3\lambda_2 - \lambda_3 = 6$ $(i)$$-\lambda_1 + 3\lambda_2 - \lambda_3 = 2$ (ii)$-3\lambda_1 - \lambda_2 + 2\lambda_3 = 3$ (iii)Subtracting (ii) from $(i)$: $5\lambda_1 = 4 \Rightarrow \lambda_1 = \frac{4}{5}$.Substitute $\lambda_1$ into (ii) and (iii):$3\lambda_2 - \lambda_3 = 2 + \frac{4}{5} = \frac{14}{5}$ (iv)$-\lambda_2 + 2\lambda_3 = 3 + 3(\frac{4}{5}) = \frac{27}{5}$ $(v)$Multiply (iv) by $2$ and add to $(v)$: $6\lambda_2 - 2\lambda_3 + (-\lambda_2 + 2\lambda_3) = \frac{28}{5} + \frac{27}{5} \Rightarrow 5\lambda_2 = \frac{55}{5} = 11 \Rightarrow \lambda_2 = \frac{11}{5}$.From (iv): $\lambda_3 = 3(\frac{11}{5}) - \frac{14}{5} = \frac{33-14}{5} = \frac{19}{5}$.Thus,$(\lambda_1, \lambda_2, \lambda_3) = (\frac{4}{5}, \frac{11}{5}, \frac{19}{5})$.

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